Guesstimation: Solving the World's Problems on the Back of a Cocktail Napkin
Author: Lawrence Weinstein; John A. Adam
Date: 2009-02-09T05:00:00+00:00
Keywords: Science
Description: Guesstimation is a book that unlocks the power of approximation--it's popular mathematics rounded to the nearest power of ten! The ability to estimate is an important skill in daily life. More and more leading businesses today use estimation questions in interviews to test applicants' abilities to think on their feet. Guesstimation enables anyone with basic math and science skills to estimate virtually anything--quickly--using plausible assumptions and elementary arithmetic. Lawrence Weinstein and John Adam present an eclectic array of estimation problems that range from devilishly simple to q
Publisher: Princeton University Press

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Lawrence Weinstein and John A. Adam

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guess. .sstimation

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Solving the World’s Problems
on the Back of a Cocktail Napkin Princeton University Press Princeton and Oxford iii

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Copyright c 2008 by Lawrence Weinstein and John A. Adam Requests for permission to reproduce material from this work should be sent to Permissions, Princeton University Press Published by Princeton University Press, 41 William Street, Princeton, New Jersey 08540

In the United Kingdom: Princeton University Press, 3 Market Place, Wood-stock, Oxfordshire OX20 1SY

All Rights Reserved Library of Congress Cataloging-in-Publication Data Weinstein, Lawrence, 1960-Guesstimation: solving the world’s problems on the back of a cocktail napkin / Lawrence Weinstein and John A. Adam.

p. cm.

Includes bibliographical references and index.

ISBN 978-0-691-12949-5 (clothbound : alk. paper) 1. Estimation theory– Miscellanea. 2. Problem solving–Miscellanea. I. Adam, John A. II. Title.

QA276.8.W45 2008

519.5 44–dc22

2007033928

British Library Cataloging-in-Publication Data is available This book has been composed in Minion with Scala Sans display Printed on acid-free paper. ∞
press.princeton.edu Printed in the United States of America 10

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To my wife, Carol, and my children, Lee and Rachel, who understand when I get that vacant look in my eyes and start mumbling exponents. [LW]

To four wonderful and cherished nonmathematicians: my wife, Susan, and my children, Rachel, Matthew, and Lindsay, and last, but not least (though he is still very small), to my grandson, John Mark—may you come to love numbers at least as much as I do! [JAA]

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Contents

Acknowledgments
xi

Preface
xiii

1

How to Solve Problems 1

2

Dealing with Large Numbers 11

2.1 Scientific Notation 11

2.2 Accuracy

14

2.3 A Note on Units 16

2.4 Unit Conversion 17

3

General Questions 19

3.1 One big family 21

3.2 Fore!

25

3.3 This is a fine pickle you’ve got us into, Patty 29

3.4 Throwing in the towel 31

3.5 Hey buddy, can you fill a dome?

35

3.6 A mole of cats 39

3.7 Massive MongaMillions 41

3.8 Tons of trash

43

3.9 Mt. Trashmore

47

3.10 Juggling people 51

3.11 Shelving the problem 53

4

Animals and People 55

4.1 More numerous than the stars in the sky 57

4.2 Laboring in vein 61

4.3 Unzipping your skin 65

4.4 Hair today, gone tomorrow 69

4.5 Hot dawg!

73

4.6 Playing the field 75

4.7 Ewww . . . gross!

77
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4.8 Going potty

79

4.9 Let’s get one thing straight!

83

5

Transportation 87

5.1 Driving past Saturn 89

5.2 Drowning in gasoline 91

5.3 Slowly on the highway 95

5.4 Rickshaws and automobiles 99

5.5 Horse exhaust

103

5.6 Tire tracks

107

5.7 Working for the car 109

6

Energy and Work 113

6.1 Energy of height 114

6.1.1 Mountain climbing 115

6.1.2 Flattening the Alps 119

6.1.3 Raising a building 123

6.2 Energy of motion 126

6.2.1 At your service 127

6.2.2 Kinetic trucking 129

6.2.3 Racing continents 131

6.2.4 “To boldly go . . . ”

135

6.3 Work

138

6.3.1 Crash!

139

6.3.2 Spider-Man and the subway car 143

7

Hydrocarbons and Carbohydrates 145

7.1 Chemical energy 145

7.1.1 Energy in gasoline 147

7.1.2 Battery energy 151

7.1.3 Battery energy density 155

7.1.4 Batteries vs. gas tanks 159

7.2 Food is energy 162

7.2.1 Eat here, get gas 163

7.2.2 Farmland for ethanol 167
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7.3 Power!

170

7.3.1 Hot humans

171

7.3.2 Fill ’er up with gasoline 173

7.3.3 Fill ’er up with electricity 175

8

The Earth, the Moon, and Lots of Gerbils 179

8.1 “And yet it moves” (e pur si muove) 181

8.2 Duck!

185

8.3 Super-sized Sun 189

8.4 Sun power

193

8.5 Gerbils 1, Sun 0

197

8.6 Chemical Sun

201

8.7 Nearby supernova 205

8.8 Melting ice caps 209

9

Energy and the Environment 213

9.1 Power to the people 215

9.2 Continental power 219

9.3 Solar energy

223

9.4 Land for solar energy 225

9.5 Tilting at windmills 229

9.6 The power of coal 233

9.7 The power of nuclei 237

9.8 Hard surfaces

239

10 The Atmosphere 243

10.1 Into thin air 245

10.2 Ancient air

247

10.3 Suck it up

251

10.4 CO2 from coal 255

10.5 A healthy glow 259

10.6 CO2 from cars 261

10.7 Turning gas into trees 265

10.8 Turning trees into gas 269
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11 Risk 273

11.1 Gambling on the road 275

11.2 The plane truth 277

11.3 Life’s a beach 279

11.4 Up in smoke

281

12 Unanswered Questions 285

A Needed Numbers and Formulas 289

A.1 Useful Numbers 289

A.2 Handy Formulas 289

A.3 Metric Prefixes 290

B

Pegs to Hang Things On 291

Bibliography

295

Index

299
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Acknowledgments

We would like to thank everyone who helped us, or, at least, who did not slow us down too much.

I (John) would like to thank my Department Chair, J. Mark Dorrepaal, for his enthusiastic support for this venture (and many others). Our editor, Vickie Kearn, once told me that writing a book was like giving birth to a pine cone; her sound advice and sense of humor has enabled this “baby” to be brought to full term in a relatively pain-free fashion (though with occasional cravings for ice cream). My coauthor, Larry, has been a pleasure to work with, and has kept me on my intellectual toes (without treading on them). My delight at “discovering” Patty Edwards and her artistic talents, as she walked her dog and I just walked around our neighborhood, has increased daily. Her superb contributions to this book will no doubt engender a whole new genre of stories: “A physicist, an artist, and a mathematician walked into a bar. . . ” Finally, my wife, Susan, has been a wise and effective sounding board for my book-related mutterings, and I thank her for all her love and advice.

In particular, I (Larry) thank the Old Dominion University Physics Department, who encouraged me to develop the course “Physics on the Back of an Envelope,” on which this book is based. Now that there is a textbook for this course, I will be able to hand it off to someone else to teach. I thank all of my colleagues and students (too numerous to mention by name, for fear of omitting someone) who suggested interesting questions or innovative solutions. I thank our editor at Princeton, Vickie Kearn, who encouraged us to write the book and helped us throughout the writing process. In any successful collaboration, both parties have to be willing to do 90% of the work. I thank xi

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my coauthor, John, for definitely doing more than his share and for making the project fun. I am also grateful to my wife and children, Carol, Lee, and Rachel, for their specific suggestions and corrections, for their general support of this project, and for making life fun and worthwhile.

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Preface

How big IS it?

Numbers are thrown at us all the time. They are frequently used to scare us: “Shark attacks doubled this year!” or “Dozens of lives could be saved by using infant car seats on airplanes!” They are often used to tempt us: “This week’s lottery prize is $100 million!”

They are certainly needed to understand the world around us: “The average American produces 100 cubic feet of garbage every year!” or “Nuclear power plants produce tons of high-level radioactive waste!”

You can make sense of these often confusing and sometimes contradictory numbers with just two tools: (1) an understanding of the meaning of large numbers and (2) an ability to make rough, common-sense, estimates starting from just a few basic facts. We’ll teach you these straightforward skills so you can better understand the world around you and better recognize numerical, political, and scientific nonsense.

You can also use these tools to further your career.

Many top companies use estimation questions in job interviews to judge the intelligence and flexibility of their applicants [1]. Leading software firms, management consultants, and investment banks (for example, Microsoft, Goldman Sachs, and Smith Barney) ask questions such as What’s the size of the market for disposable diapers in China? How many golf balls does it take to fill a 747? and How many piano tuners are there in the world? [2, 3] Companies use these questions as an excellent test of the applicants’ abilities to think on their feet and to apply their mathematical skills to real-world problems.

These problems are frequently called “Fermi problems,” after the legendary physicist Enrico Fermi, who xiii

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delighted in creating and solving them. During one of the first atomic bomb tests, Fermi supposedly dropped a few scraps of paper as the shock wave passed and estimated the strength of the blast from the motion of the scraps as they fell.

In this book, we will help you develop the ability to estimate almost anything, from the amount of landfill space needed to the number of people in the world picking their nose at this instant. As there is no single correct way to analyze these questions, we will indicate some of the many paths to the right answer.

We will start with two short chapters on how to estimate and how to handle large numbers, and then we’ll move on to the heart of the book: interesting questions (with lots of hints if you want them) followed by the answers on the other side of the page. The questions are divided into chapters, each focusing on a particular topic, such as energy and the environment, transportation, and risk. Each chapter will start with easier questions and work up to harder ones. The questions in chapters 6 through 9 will cover energy in its various forms. We will start with mountain climbing and go on to compare gasoline, chocolate chip cookies, batteries, the Sun, gerbils, windmills, and uranium.

The questions cover a large range of phenomena, from the simple to the complex and from the silly to the serious. We will figure out the answers to many fascinating questions, including the following:

• If all the people in the world were crammed together, how much space would we occupy?

• How many batteries would it take to replace your car’s gas tank?

• Could Spider-Man really stop a subway car?

• How much waste do nuclear and coal electric power generators make per year?

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• What does it really cost to drive a car?

• Which is more powerful per pound, the Sun or a gerbil?

• How much more cropland would we need to replace gasoline with corn-based ethanol?

All you will need to answer these questions is a willingness to think and to handle big numbers. We will remind you of the few scientific principles and equations you might need. You will be astonished at how much you can figure out starting with the knowledge you already have.

The new knowledge you will attain can be applied to all other estimation problems you may come across in the future. Oh, and good luck on that job interview!

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How to Solve Problems

Chapter 1

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STEP 1: Write down the answer [4]. In other words, come up with a reasonably close solution. This is frequently all the information you need.

For example, if it is 250 miles from New York to Boston, how long will it take to drive? You would immediately estimate that it should take about four or five hours, based on an average speed of 50–60 mph.

This is enough information to decide whether or not you will drive to Boston for the weekend. If you do decide to drive, you will look at maps or the Internet and figure out the exact route and the exact expected driving time.

Similarly, before you go into a store, you usually know how much you are willing to spend. You might think it is reasonable to spend about $100 on an X-Game2. If you see it for $30, you will automatically buy it. If it sells for $300, you will automatically not buy it.

Only if the price is around $100 will you have to think about whether to buy it.

We will apply the same reasoning here. We’ll try to estimate the answer to within a factor of ten. Why a factor of ten? Because that is good enough to make most decisions.

Once you have estimated the answer to a problem, the answer will fall into one of the three “Goldilocks”

categories:

1. too big

2. too small

3. just right

If the answer is too big or too small, then you know what to do (e.g., buy the item, don’t drive to Boston).

Only if the answer is just right will you need to put more work into solving the problem and refining the answer. (But that’s beyond the scope of this book. We just aim to help you estimate the answer to within a factor of ten.) 2

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If all problems were as simple as that, you wouldn’t need this book. Many problems are too complicated for you to come up with an immediate correct answer.

These problems will need to be broken down into smaller and smaller pieces. Eventually, the pieces will be small enough and simple enough that you can estimate an answer for each one. And so we come to STEP 2: If you can’t estimate the answer, break the problem into smaller pieces and estimate the answer for each one. You only need to estimate each answer to within a factor of ten. How hard can that be?

It is often easier to establish lower and upper bounds for a quantity than to estimate it directly. If we are trying to estimate, for example, how many circus clowns can fit into a Volkswagen Beetle, we know the answer must be more than one and less than 100. We could average the upper and lower bounds and use 50
for our estimate. This is not the best choice because it is a factor of 50 greater than our lower bound and only a factor of two lower than our upper bound.

Since we want our estimate to be the same factor away from our upper and lower bounds, we will use the geometric mean. To take the approximate geometric mean of any two numbers, just average their coefficients and average their exponents.* In the clown case, the geometric mean of one (100)† and 100 (102) is 10 (101) because one is the average of the exponents zero and two. Similarly, the geometric mean of 2×1015

* We use coefficients and exponents to describe numbers in scientific notation. The exponent is the power of ten and the coefficient is the number (between 1 and 9.99) that multiplies the power of ten. If you are not familiar with this notation, please quickly check the section on scientific notation (“Dealing with Large Numbers”) and then come right back. We’ll wait for you here.

† Any number raised to the 0th power is 1.

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and 6 × 103 is about 4 × 109 (because 4 = 2+6 and 2

9 = 15+3 ).* If the sum of the exponents is odd, it is a 2
little more complicated. Then you should decrease the exponent sum by one so it is even, and multiply the final answer by three. Therefore, the geometric mean of one and 103 is 3 × 101 = 30.

EXAMPLE 1: MongaMillions Lottery Ticket Stack Here’s a relatively straightforward example: Your chance of winning the MongaMillions lottery is one in 100 million.† If you stacked up all the possible different lottery tickets, how tall would this stack be? Which distance is this closest to: a tall building (100 m or 300 ft), a small mountain (1000 m), Mt Everest (10,000 m), the height of the atmosphere (105 m), the distance from New York to Chicago (106 m), the diameter of the Earth (107 m), or the distance to the moon (4×108 m)?

Imagine trying to pick the single winning ticket from a stack this high.

Solution: To solve this problem, we need two pieces of information: the number of possible tickets and the thickness of each ticket. Because your chance of winning is one in 100 million, this means that there are 100 million (108) possible different tickets.‡ We can’t reliably estimate really thin items like a single lottery ticket (is it 1/16 in. or 1/64 in.? is it 1 mm or 0.1 mm?) so let’s try to get the thickness of a pack of tickets.

* To be more precise (which this book rarely is), the geometric √
mean of two numbers, b and c , is a =
bc . Our approximate rule
is exact for the exponents and close enough for this book for the coefficients.

† Lottery billboards frequently have the odds of winning in very small print at the bottom.

‡ 100 million = 100,000,000 or 1 followed by eight (count them!) zeros. This can be written in scientific notation as 1 × 108.

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Let’s think about packs of paper in general. One ream of copier or printer paper (500 sheets) is about 1.5 to 2 in. (or about 5 cm since 1 in. = 2.5 cm) but paper is thinner than lottery tickets. A pack of 52 playing cards is also about 1 cm. That’s probably closer. This means that the thickness of one ticket is cm
t =

1 cm

= 0.02

× 1 m

52 tickets
ticket

102 cm

=
m

2 × 10−4 ticket

Therefore, the thickness of 108 tickets is m

T = 2 × 10−4

× 108 tickets = 2 × 104 m
ticket

2 × 104 m is 20 kilometers or 20 km (which is about 15 miles since 1 mi = 1.6 km).

If stacked horizontally, it would take you four or five hours to walk that far.

If stacked vertically, it would be twice as high as Mt Everest (30,000 ft or 10 km) and twice as high as jumbo jets fly.

Now perhaps you used the thickness of regular paper so your stack is a few times shorter. Perhaps you used 1 mm per ticket so your stack is a few times taller. Does it really matter whether the stack is 10 km or 50 km? Either way, your chance of pulling the single winning ticket from that stack is pretty darn small.

EXAMPLE 2: Flighty Americans

These problems are great fun because, first, we are not looking for an exact answer, and second, there are many different ways of estimating the answer. Here is a slightly harder question with multiple solutions.

How many airplane flights do Americans take in one year?

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We can estimate this from the top down or from the bottom up. We can start with the number of airports or with the number of Americans.

Solution 1: Start with the number of Americans and estimate how many plane flights each of us take per year. There are 3 × 108 Americans.* Most of us probably travel once a year (i.e., two flights) on vacation or business and a small fraction of us (say 10%) travel much more than that. This means that the number of flights per person per year is between two and four (so we’ll use three). Therefore, the total number of flights per year is N = 3 × 108 people × 3 flights/person-year = 9 × 108 passengers/year

Solution 2: Start with the number of airports and then estimate the flights per airport and the passengers per flight. There are several reasonable size airports in a medium-sized state (e.g., Virginia has Dulles, Reagan-National, Norfolk, Richmond, and Charlottesville; and Massachusetts has Boston and Springfield). If each of the fifty states has three airports then there are 150
airports in the US. Each airport can handle at most one flight every two minutes, which is 30 flights per hour or 500 flights per 16-hour day. Most airports will have many fewer flights than the maximum. Each airplane can hold between 50 and 250 passengers. This means * This is one of those numbers you need to know to do many estimation questions. Go ahead and write it on your palm so you’ll be prepared for the test at the end of the book.

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that we have about

N = 150 airports × 100 flights × 100 passengers airport-day flight

×365 days = 5 × 108 passengers per year year

Wow! Both methods agree within a factor of two.

The actual number of US domestic airline passengers in 2005 was 6.6 × 108, which is close enough to both answers.

EXAMPLE 3: Piano Tuners in Los Angeles Now let’s work out a harder problem.

How many piano tuners are there in Los Angeles (or New York or Virginia Beach or your own city)? This is the classic example originated by Enrico Fermi [5] and used at the beginning of many physics courses because it requires employing the methods and reasoning used to attack these problems but does not need any physics concepts.

Solution: This is a sufficiently complicated problem that we cannot just estimate the answer. To solve this, we need to break down the problem. We need to estimate (1) how many pianos there are in Los Angeles and (2) how many pianos each tuner can care for.

To estimate the number of pianos, we need (1) the population of the city, (2) the proportion of people that own a piano, and (3) the number of schools, churches, etc. that also have pianos. To estimate the number of pianos each tuner can care for, we need to estimate (1) how often each piano is tuned, (2) how much time it takes to tune a piano, and (3) how much time a piano tuner spends tuning pianos.

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This means that we need to estimate the following: 1. population of Los Angeles 2. proportion of pianos per person

3. how often each piano is tuned per year 4. how much time it takes to tune each piano 5. how much time each piano tuner works per year Let’s take it from the top.

1. The population of Los Angeles must be much less than 108 (since the population of the US is 3 × 108). It must be much more than 106 (since that is the size of an ordinary big city). We’ll estimate it at 107.

2. Pianos will be owned by individuals, schools, and houses of worship. About 10% of the population plays a musical instrument (it’s surely more than 1% and less than 100%). At most 10% of musicians play the piano and not all of them own a piano so the proportion that own a piano is probably 2–3% of the musicians.

This would be 2 × 10−3 of the population. There is about one house of worship per thousand people and each of those will have a piano.

There is about one school per 500 students (or about 1 per 1000 population) and each of those will have a piano. This gives us about 4 or 5

×10−3 pianos per person. Thus, the number of pianos will be about 107 × 4 × 10−3 = 4 × 104.

3. Pianos will be tuned less than once per month and more than once per decade. We’ll estimate once per year.

4. It must take much more than 30 minutes and less than one day to tune a piano (assuming that it is not too badly out of tune). We’ll estimate 8

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2 hours. Another way to look at it is that there are 88 keys. At 1 minute per key, it will take 1.5
hours. At 2 minutes per key, it will take 3 hours.

5. A full-time worker works 8 hours per day, 5 days per week, and 50 weeks per year which gives 8 × 5 × 50 = 2000 hours. In 2000 hours she can tune about 1000 pianos (wow!).

This means that the 4 × 104 pianos need 40 piano tuners.

How close are we? Well, the Yellow Pages for our city of 106 inhabitants (ten times fewer than LA) has 16 entries under the heading of “Pianos—Tuning, Repairing & Refinishing.” There are probably only one or two tuners per entry and they probably do not spend full time tuning. This means that our estimate is probably too low by a factor of five. However, that is a LOT closer than we could get by just guessing.

Remember that we are only trying to estimate the answer within a factor of ten.

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Dealing with Large Numbers

Chapter 2

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2.1 ScientificNotation

As you may have noticed, we used 108 instead of 100,000,000 to represent 100 million. We do this for two reasons. The first is that, if we multiply 3 trillion times 20 quadrillion like this: 3000000000000 × 20000000000000000

= 6000000 . . . . . .

we will almost certainly miscount all those pesky zeros.

If we use a calculator, we will first miscount the pesky zeros and then we will enter the incorrect number of zeros in the calculator so the number will be even more wrong. We’ll get an answer with the correct first digit (6) but the incorrect size. That’s like getting $60 when you should have received $6000. The number of zeros is much more important than the leading digit.

There is an easy and compact way of writing very large and very small numbers. Any number can be written as the product of a number between 1 and 10
and a number that is a power of ten. For example, we can write 257 as 2.57×100 and 0.00257 as 2.57×0.001.

Now we have to count the zeros (but only once per number). One hundred (100) has two zeros so we write it as 102, and 0.001 has three zeros (counting the leading zero) and is less than one so we write it as 10−3.

Therefore, we write 257 as 2.57 × 102 and 0.00257 as 2.57 × 10−3. The exponent tells us how many zeros are in the power of ten (2 and −3 in the previous sentence) and the coefficient multiplies the power of ten. This system is called scientific notation.

Here are some examples to make this clearer: 0.01 = 1 × 10−2

2000 = 2 × 103

3,000,000 = 3 × 106

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The second reason for using scientific notation of the form x × 10 y is that the most important part of the number is the exponent y and not the coefficient x. When we write the 300 million population of the United States as 3 ×108, the 8 is much more important than the 3. This is because if you change the 3 to a 4, you change the population by only 1/3 or 30%.

If you change the 8 to a 9, then you change the population by a factor of 10 (or 1000%). That is a huge change, especially if you think that we are already too crowded here. Thus, we use scientific notation so that the exponent is written explicitly.

The rules for multiplying and dividing with scientific notation are straightforward. When we multiply numbers, we multiply coefficients and add exponents.

For example,

3 × 106 × 4 × 108 = (3 × 4) × 106+8

= 12 × 1014 = 1.2 × 1015

When we divide numbers, we divide coefficients and subtract exponents. For example, 3 × 106 = 3 × 106−8 = 0.75 × 10−2 = 7.5 × 10−3

4 × 108

4

Note that in both of these examples, we had to cope with a coefficient either more than 10 or less than 1. In these cases, we will rewrite the out-of-range coefficient in scientific notation itself. Thus, in the first example the coefficient 12 can be written as 1.2 × 101. This means that what we really did was 12 × 1014 = (1.2 × 101) × 1014 = 1.2 × 1015

In the second case, the coefficient 0.75 is rewritten as 7.5 × 10−1 so that 0.75 × 10−2 = (7.5 × 10−1) × 10−2 = 7.5 × 10−3

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When we add or subtract numbers using scientific notation, the exponents of both numbers must be the same. To add 3×107 and 4×108, we need to convert the number with the smaller exponent into a form where it has the same exponent as the other number. In this case, when we increase the exponent by one (from 7
to 8), we must simultaneously divide the coefficient by 10 (this is because increasing the exponent increases the number by a factor of ten so we must decrease the coefficient to compensate and keep the number the same). Thus, we have 3 × 107 + 4 × 108 = 0.3 × 108 + 4 × 108 = 4.3 × 108

Going back to that first equation, we write 3 trillion (3000000000000) as 3 × 1012 and 20 quadrillion (20000000000000000) as 2×1016 so that the operation becomes 3 × 1012 × 2 × 1016 = (2 × 3) × 1012+16 = 6 × 1028

We no longer need to count zeros; all we have to do is to add exponents. We have a much easier time adding 12 and 16 to get 28 than we do counting 12 zeros and 16 zeros and then writing down 28 zeros.

2.2 Accuracy

The most important part of any number is the exponent. After that, the next most important number is the first digit of the coefficient (the number that multiplies the power of ten). The second and subsequent digits of the coefficient are just small corrections to the first digit.

The number of digits in the coefficient (also called the “number of significant figures”) tells us how well we know that number. For example, if your friend gives you driving directions, there is a tremendous 14

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chapter02.tex
difference between the direction to “drive a couple of dozen miles east and then turn left on Obscure Alley” and the direction to “drive 25.2 miles east and then turn left on Obscure Alley.” The first direction is rather vague and imprecise. You expect to find Obscure Alley somewhere between 12 and 36 miles away. If you miss the left turn, you will probably drive quite far before you turn around to look for it again.

The second direction is very precise and you expect to find Obscure Alley between 25.1 and 25.3 miles away.

If you miss the turn, you will probably turn around by the time you have driven 26 miles. The extra digits in the second set of directions imply that your friend has measured the distance carefully.

Similarly, the silliness of having too many digits is illustrated by the following anecdote. Suppose that you ask a museum guard how old a dinosaur skeleton is. He responds that it is 75 million and 3 (75,000,003) years old. When you look puzzled, he explains that when he started the job three years ago, the skeleton was already 75 million years old.

Many of us make the same kind of mistake with our calculators. Suppose that we used 23.0 gallons of gasoline to drive 327 miles. If we divide 327 by 23 on our calculator we get 14.2173913. . .. But this cannot be the answer to our problem. We did not measure either the miles driven or the fuel consumed to 1 part in a billion so our answer cannot possibly be so precise. Our gas mileage should be 327 mil/23.0 gal = 14.2 mpg.

There are lots of rules for how to deal with significant figures in scientific calculations. Fortunately, we will not need most of them.

In this book, we will estimate quantities to within a factor of ten. Therefore, we will almost always keep only one digit in the coefficient. This means that we will round off 7.2 × 103 to just 7 × 103. Our estimates are just not accurate past the first digit. Keeping more 15

January 2, 2008

Time: 12:54pm
chapter02.tex
digits is lying; it claims we know the answer much better than we actually do.

There is another benefit to keeping only one significant figure: you shouldn’t need a calculator to solve these problems. If you can add or subtract one-or two-digit numbers (the exponents) without a calculator and multiply or divide one-digit numbers without a calculator, then you are all set to proceed.

2.3 ANoteonUnits

We have been using the metric system as students and scientists for decades now, but we still think in inches, feet, pounds, and degrees Fahrenheit (US customary units) because those are the units we use every day.

When we estimate things, we’ll typically estimate them in US units and then convert to metric. When we’re done with the problem, we might convert back to US
units. If we used metric units in daily life, we would have to do this for some quantities anyway, since units such km/hr, liters (L), and cubic centimeters (cm3) are not part of the MKS (meter/kilogram/second) system.

Why would anyone bother to do all that? Well, it’s a lot easier to do the calculations in metric because the conversion factors are (1) merely powers of ten ( you try converting miles to inches without a calculator), and (2) converting quantities like volume is also much easier in the metric system. We know that 1 L =1000 cm3
and 1 m3 = 1000 L. What is a gallon in terms of cubic inches or a cubic foot in terms of cups? We sure don’t know.

In addition to simplifying unit conversions, the metric system (SI or International System of Units) is much easier for doing many calculations. All the quantities are standardized on the meter, second, 16

Chapter 2 Dealing with Large Numbers January 2, 2008

Time: 12:54pm
chapter02.tex

Quantity Metric Unit

US Customary equivalent

Length

1 meter (m)

3 ft

Length

103 m (1 km)

0.6 mi

Length

0.01 m (1 cm)

0.4 in.

Volume

1 liter (L)

1 quart

Mass

1 kilogram (kg) Mass of 2.2 pounds (lb) Mass 103 kg

1 ton

Weight

1 newton (N)

0.2 lb

Speed

1 m/s

2.2 mph

Time
π × 107 s

1 year
and kilogram. Thus, the unit of force, the newton (N), is equal to a kilogram–meter/second2. We will use the following abbreviations for units: meter (m), second (s), kilogram (kg), watt (W), joule (J), newton (N), liter (L), hour (hr), etc. For more information on SI units, see the US National Institute of Standards and Technology (NIST) website: http://physics.nist.gov/cuu/Units/index.html. We will use the following abbreviations for US customary units: mi (mile), ft (foot), in. (inch), gal (gallon), and lb (pound).

We will also use the standard prefixes giga or 109

(G), mega or 106 (M), kilo or 103 (k), centi or 10−2

(c), milli or 10−3 (m), micro or 10−6 (µ), and nano or 10−9 (n). These are tabulated in the appendix. If we need to use pico, tera, or yocto, we’ll warn you first.

2.4 UnitConversion We will frequently need to convert a quantity from one unit to another. For example, to calculate the distance light travels in one year or the energy used by a 100-W light bulb in one year, you need to convert 17

January 2, 2008

Time: 12:54pm
chapter02.tex
time from 1 year to some number of seconds. To do this, we will multiply our original number by various conversion factors that individually are equal to one, e.g., 60 s = 1. Thus, 1 min

24 hr

1 year = 1 year × 365 days 1 year

1 day

× 60 min

60 s

= 3.15 × 107 s

1 hr

1 min

Note that 1 year ≈ π × 107 s. π appears in this because the Earth goes around the Sun in an almost perfect circle and the circumference of a circle c = 2π R*.

Another handy conversion is from miles per hour (mph) to meters per second (m/s). We think of speeds in mph, but it is much easier to do calculations in m/s.

Fortunately, the conversion is straightforward. We first convert meters to miles via kilometers and then we convert seconds to hours via minutes: 0.6 mi 1m/s = 1m/s ×

1km

103 m

1 km

×

60 s

60 min

= 2.2 mph

1 min

1 h

Thus, 1 m/s is a little more than 2 mph.

We’ll use both these handy facts later on. Feel free to write them on your palm also.

* You didn’t really believe that, did you? The π is just a coincidence, but it makes a handy mnemonic. Of course, the exponent “7” is much more important than the leading digit “3”.

18

Chapter 2 Dealing with Large Numbers January 29, 2008

Time:

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chapter03.tex

General Questions Let’s start with some straightforward questions about distance and space. We’ll ask about how much space we need for ourselves, how much space we need for our garbage, and how much space we need for our pickles.

Chapter 3

*

January 29, 2008

Time:

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chapter03.tex

20

January 29, 2008

Time:

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chapter03.tex

3.1

One big family If all the humans in the world were crammed together, how much area would we require?

Compare this to the area of a large city, a state or small country, the US, Asia.

How much area
would we need if we gave every
family a house
and a yard

(i.e., a small
plot of land)?

*

.

m

100

=
m

10

×
m

10

=

A is
yard
your

2
of
area
the
then
width,
the
for
m

10
chose
you

If
m)?

100

(
ft

300
m),

(30
ft

100
m),

(10
ft

30
m),

(3
ft

10
yard?

your
is
ide
w

How
square.

be
to
it
assume
yard,
a
of
area
the
estimate

To

HINT: people.

of
number
the
by
that
multiply
then
person,
each
for
want
you
area
the
on
decided
you’ve

Once

HINT: feet.

square

10
about
is
meter
square
a
so
yard,
a
than
longer
slightly
is
meter
a
that

Recall
meter?

square
a
into
fit
can
people
many

How

HINT: it.

ignore

We’ll
writing.

of
time
the
at
billion)

(0.3
million

300
mere
a
is
case
this
in
tad
he

T
billion.

6
over
tad
a
is
world
the
of
population

The

HINT: 21

January 29, 2008

Time:

06:19pm
chapter03.tex

ANSWER: Okay, 6 billion people is 6 × 109 of us. How many can we cram into a square meter (i.e., 3 ft by 3 ft)? We’re not sure, but it is certainly between 3 and 10. We’ll choose 6. (That ignores the space needed to play, eat, sleep, and well, the porta-potty question is in chapter 4, so can you wait until then?) If there are 6 people per square meter, then 6 billion people will need A = 6 × 109people ×

1 m2

= 109 m2

6 people

We have no idea how big a billion square meters is (although it sure sounds like a lot) so let’s convert it to more reasonable units. We’ll convert it to kilometers (km) and to miles. 1 km = 103 m. A square kilometer is a square with sides of 103 m so that 1 km2 = 103 m×

103 m = 106 m2.

A = 109 m2 × 1 km2 = 103 km2

106 m2

(Just to remind you, when you divide numbers in scientific notation, you divide the coefficients and subtract the exponents. In this case, 109/106 = 109−6 =

103.) Thus, we would occupy an area of 1000 square kilometers. That’s a square 30 km on a side. Since a mile is about 1.5 km, all the people on Earth would fit in a square that is 30 km or 20 mi on a side. That’s the area of a large city such as Los Angeles or Virginia Beach.

Wow! That’s not much at all.

Now let’s give every family a house and a yard (a small piece of land). First we need to estimate the size of the average family. In the US and Europe, the average family is about three people, but in the developing world, it’s a bit more than that. We’ll choose three so that we overestimate the amount of land.

Next, we need to estimate the size of the yard. Since we are not good at estimating area, we’ll assume a 22

Chapter 3 General Questions

January 29, 2008

Time:

06:19pm
chapter03.tex
square yard and estimate its width. It will definitely be smaller than a football field (100 m or 300 ft) and bigger than a house (10 m or 30 ft) so we’ll take the geometric mean and estimate 30 m. This means that each family gets a piece of land of area A = 30 m ×

30 m = 103 m2 (or about 1/4 acre). Therefore, all of us together will use a total land area of A = 6×109people× 1family × 103m2 = 2×1012m2

3people
family

(Just to remind you, when you multiply numbers in scientific notation, you multiply the coefficients and add the exponents. In this case, 109 × 103 =

109+3 = 1012.) This is 2 × 106 km2 (two million square kilometers) or about 1 × 106 mi2 (one million square miles). While it sounds like a lot, it is only the area of Alaska or twice the area of Egypt. That is only 1% of the surface area of the Earth.

That could leave a lot of room for other species!

23

January 29, 2008

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chapter03.tex

24

January 29, 2008

Time:

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chapter03.tex

3.2

Fore!

How many golf balls would it take to circle the Earth at the equator?*

* Thanks to Tom Isenhour for this question [6]. Can we have that raise now, Tom?

*
hours.

6
in
istance
d
that
fly
can

You

LA.

to

NY
from
distance
the
times
eight
is
circumference

Earth’s

The

HINT: total.

zones
time

24
re
a

There

NY.

and

LA
between
difference
time

3-hour
a
is

There

HINT: useful.)

less
lot
a
is
formula
the
but

R
π 2
still
is
c radius,
the
remember
don’t
you

If ( . R
π 2

=
c is
circumference
the
then
radius
the
emember
r
you

If

HINT: Earth?

the
of
circumference
the
is

What

HINT: cm.

2.5
is
inch

1
that

Remember
ball?

golf
a
of
diameter
the

What’s

HINT: 25

January 29, 2008

Time:

06:19pm
chapter03.tex

ANSWER: To answer this question we need the diameter of a golf ball and the circumference of the Earth. Let’s start with the easier part. A golf ball has a diameter of a bit less than 2 inches or about 4 cm.

There are several ways to estimate the circumference of the Earth. For example, there is a three-hour time difference between New York and Los Angeles and there are 24 times zones covering the Earth. Therefore, the circumference of the Earth is about eight times the distance from NY to LA. If you don’t remember that the distance between them is 3000 miles, then you can estimate it from the fact that it takes about six hours to fly from NY to LA and a modern jet flies at about 500 mph. Thus, the circumference is about c = 8 × 3000 mi = 2.4 × 104 mi.

Alternatively, we know that passenger jets fly slower than the Earth rotates (since you always arrive after you leave [in local time]) and that some military jets can fly faster than the Earth rotates. Since passenger jets fly at about 500 mph and military jets can fly up to 2000 mph, we can estimate the Earth’s rotation at 1000 mph. Since the Earth rotates completely in 24 hours, its circumference must be c = 24 ×

1000 mph = 2.4 × 104 mi.

Of course, if you remembered that the circumference is 25,000 miles (or 40,000 km) or that the radius of the Earth is 4000 miles (6400 km) and the circumference is c = 2π R, then you didn’t need to estimate it.

Now the arithmetic is straightforward. First we need to convert the circumference of the Earth from kilometers to centimeters. The number of golf balls needed is N = 4 × 104 km × 103 m × 102 cm × 1 golf ball 1 km 1 m

4 cm

= 109golf balls 26

Chapter 3 General Questions

January 29, 2008

Time:

06:19pm
chapter03.tex The Pacific Ocean is a very large water hazard. It would be extremely irritating to lose a billion golf balls in the water! We’d better use the special kind that floats.

This also provides an interesting peg on which to hang the concept of “parts per billion” (ppb). If the air contains so many ppb of some potentially toxic substance, that is about the number of, say, red golf balls in the otherwise white ones surrounding the Earth. You could walk along the equator for months before finding your first red golf ball.

27

January 29, 2008

Time:

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chapter03.tex 28

January 29, 2008

Time:

06:19pm
chapter03.tex This is a fine pickle 3.3
you’ve got us into, Patty If all the pickles sold in the US last year were placed end-to-end, what distance would they cover?

*

? r a e y
a
in
at
e

American
average
the
does
pickles
many

How

HINT: pickle?

typical
a
is
long

How

HINT: 29

January 29, 2008

Time:

06:19pm
chapter03.tex ANSWER: We will need to estimate the number of pickles the average American eats each year and the length of the average pickle. Our average pickle-in-the-street is definitely more than 1 cm and less than 100 cm (a 1-m pickle? yikes!) so we’ll estimate that it is about 10 cm (4 in.) long. The average American consumes more than one pickle per year (that’s including the pickle slices on hamburgers) and less than one per day (400 per year) so we’ll estimate 20 pickles per year.*

Therefore, the total length of all the pickles consumed in one year is L = 3 × 108 people × 20 pickles × 10 cm person pickle

= 6 × 1010 cm × 1 m 102 cm

= 6 × 108 m That is a distance of 6 × 105 kilometers or 4 × 105
miles. That is more than the distance from the Earth to the Moon! Who needs advanced technology? NASA could just take the space pickle.

* According to the USDA [7], Americans consumed about four pounds of pickles per person (try saying that five times fast!) in 2000 so our estimate of 20 pickles per person is not too bad.

30

Chapter 3 General Questions

January 29, 2008

Time:

06:19pm
chapter03.tex 3.4

Throwing in the towel What is the surface area of a typical bath towel (include the fibers!).

Compare this to the area of a room, a house, a football field.

*
fiber.

each
of
thickness and
length
the

Consider

HINT: fiber?

little
ach
e
of
area
surface
the
is

What

HINT: have?

it
does
fibers
total
many

How
towel?

large
a
of
area
the
is

What

HINT: inch?

square
or
centimeter square
a
in
there
are
many
how
towel;
fluffy
really
a
in
fibers
little
the
about

Think

HINT: 31

January 29, 2008

Time:

06:19pm
chapter03.tex ANSWER: That’s obvious, surely! A large rectangular towel 1 m by 2 m has a total surface area of 4 m2

(including both sides), right? (In US units, a big towel may be as large as 3 ft by 6 ft.) Wrong, actually, unless it is a very worn-out towel.

New towels have many little fibers that can absorb a lot of moisture (recall the old puzzle—what gets wetter the more it dries?). Unless you’re a fan of the Hitchhiker’s Guide to the Galaxy, you won’t have brought your own towel, so nip off to the bathroom and examine one; quickly now, we’re dripping all over the floor.

You don’t need to actually go and count the number of fibers per square inch or per square centimeter; in the latter case there must be more than 10 and fewer than 1000, so we take the geometric mean of 101 and 103, which is 102. In a square inch, being about 6 cm2, we should expect about six times as many. This will of course vary, depending on where you buy your towels; we are assuming that we are describing one of those very nice towels found in one of those very nice hotels.

Back already? Right-oh. Now we need to estimate the surface area of each fiber. We can approximate the fiber as a cylinder or a box. Cylinders are complicated so we’ll use boxes. Each fiber is about 0.5 cm (1/4 in.) long and 1 mm (0.1 cm) wide. Each “boxy” fiber then has four flat surfaces, each 0.5 cm by 0.1 cm. Thus, the surface area of one fiber is A fiber = 4 × 0.5 cm × 1 m 102 cm

× 0.1 cm × 1 m 102 cm

= 2 × 10−5 m2

32

Chapter 3 General Questions

January 29, 2008

Time:

06:19pm
chapter03.tex

Now we can calculate the total surface area of our big bathroom towel: A total = towel area × fibers per area × area per fiber = 4 m2 × 102 fibers × 104 cm2 × 2 × 10−5 m2
cm2

1 m2
fiber

= 80 m2

That is about 800 square feet: the size of a large apartment or a small house.

This problem is similar in some ways to calculating the length of the coastline. Just as the area of the towel is much greater than its simple area, the length of coast from, say, New York to Boston is much more than the 200-mile driving distance.

33

January 29, 2008

Time:

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34

January 29, 2008

Time:

06:19pm
chapter03.tex

3.5

Hey buddy, can you fill a dome?

How long would it take a running
water faucet* to fill the (inverted) dome of the US

Capitol building or St. Paul’s Cathedral?

Give your answer in seconds, days, weeks, or whatever units seem reasonable.

* That’s a “tap” for our British readers (and author).

*

.

L/m

10
and

L/gal

4
are

There

HINT: 3

3
showerhead?

modern
a
of
minute
per
gallons
in
rate
flow
the
is
what

Alternatively,
faucet?

kitchen
your
from
water
of
jug
liter)

4 (
gallon
one
a
fill
to
you
take
it
oes
d
long

How

HINT: .

d 4)

/ (1

≈

Volume

HINT: 3
dome.

the
of
diameter the

Estimate HINT: 35

January 29, 2008

Time:

06:19pm
chapter03.tex ANSWER: We need to estimate the volume of the dome and the flow rate of a water faucet. To estimate the volume of the dome, we need to estimate its diameter.

The diameter of the dome of St. Paul’s or the Capitol Building is more than 10 m (or 30 ft) and less than 100 m (the length of a football field, about 300 ft), so we take the geometric mean and estimate it as √10 × 100m = 30m (100 ft) in diameter.

If we remember that the volume of a sphere is V =

4 π R 3 and that a dome is half a sphere, then we have 3

4

V = 1 π R 3 = 2(15m)3 = 6 × 103m3

2 3

If we forgot the equation of a sphere, then we could pretend that the dome is half of a cube (as Picasso might have done) and approximate the volume as V = 1 d 3 = 0.5 × (30 m)3 = 104m3

2

We would be off by only a factor of two. Not a problem!

Now we need to estimate the flow rate of a water faucet. A typical domestic faucet running full-tilt can fill a one-gallon container in under 30 seconds. A US
low-flow shower is limited to 2.5 gallons per minute so that is about the same. A cubic meter of water is 103 liters, or 250 gallons. Thus, the time to fill the Capitol dome is t = volume of dome flow rate = 6×103 m3 × 2.5 × 102 gal/m3

2 gal/min = 7 × 105 min Seven hundred thousand minutes is not a very helpful quantity. Let’s convert it to more useful units and see what we get.

36

Chapter 3 General Questions

January 29, 2008

Time:

06:19pm
chapter03.tex There are 60 minutes in an hour, or about 60×25 =

1500 minutes in a day.* We can convert minutes to days to get t = 7 × 105 min ×

1 day

= 500 days

1.5 × 103 min That is less than two years.

Of course, we’d have to invert the dome first. . .

* We often wish we had an extra hour in the day. In this case, it’s just for easier calculation.

37

January 29, 2008

Time:

06:19pm
chapter03.tex 38

January 29, 2008

Time:

06:19pm
chapter03.tex 3.6

A mole of cats How massive is a mole of cats?* (A mole is the number of atoms that weigh that element’s atomic weight in grams. For example, a mole of hydrogen weighs 1 gram and a mole of carbon weighs 12 grams. It is used in chemistry to make sure that there are equivalent numbers of atoms for a chemical reaction.) Compare this to the mass of a mountain, a continent, the moon (7 × 1022 kg), the Earth

(6 × 1024 kg).

* Thanks again, Tom [8].

*
cat?

omestic
d
adult
typical
a
of
weight
the
is

What

HINT: items.

those
of

)

10

×

(6
number

23

Avogadro’s
contains
anything
of
mole
one
that

Remember

HINT: 39

January 29, 2008

Time:

06:19pm
chapter03.tex ANSWER: We’ll use fat cats weighing about 8 kg (18 lb) each. We’re using a neighbor’s cat, Quentin, as a model here, though without his permission (or the neighbor’s). There are N A = 6 × 1023 items in a mole, whether you are talking about a mole of atoms or a mole of cats (or a mole of moles). This means the whole bunch of them will have a mass of about M = 8 kg × 6 × 1023 = 5 × 1024 kg This is about the mass of the Earth or 70 times the mass of the Moon. Sheer lunacy! And with nine lives for each . . .

And if you think that an Earth made of cats is absurd, continue reading to find out about a Sun made of gerbils (see question 8.5).

40

Chapter 3 General Questions

January 29, 2008

Time: 06:19pm chapter03.tex 3.7

Massive MongaMillions What would be the mass of all 108 MongaMillions lottery tickets? How many 40-ton trucks would be needed to haul them away?

*
ton).

1
or
kg

10
of
mass
a
has
water of
meter 3
cubic (each kg/m

10
is
water of
density he

T
water?

of
that

3

3
with
say,
compared, ticket a
of
density the
be
might What

HINT: density.

×

volume

=

Mass

HINT:

.

m

10
or
cm

10
approximately
is

2

2

−

2

2
in.)

(4
so
m,

10
or
cm

10
approximately
is
inches

Four
meters.

2

1

−
square
to
inches
square
from

Convert
area?

the
is

What

HINT:

m.

10

×

2
as
thickness
its
estimated
we

Remember

4

−
width?

Its
ticket?

a
such
of
length
the
be
might

What

HINT:

41

January 29, 2008

Time:

06:19pm
chapter03.tex

ANSWER: To estimate the mass of all those tickets, we will need to estimate their volume and their density.

The density of an object is expressed in mass per volume. Air has a very low density (1 kg/m3), water has a medium density (103 kg/m3 or 1 kg/L) and lead has a high density (104 kg/m3 or 10 kg/L).

Volume is length times width times thickness. Earlier, we estimated the thickness of each ticket to be 2 × 10−4 m, so we just need the length and width. A ticket is about 4 inches on a side (more that 1 in. and less than 10 in.). We could choose a slightly different number, but 4 in. = 10 cm, which is a nice round number to calculate squares with. Thus, V = 10 cm × 1 m × 10 cm × 1 m × 2 × 10−4 m 102 cm

102 cm

= 2 × 10−6 m3
so the volume of 108 such tickets is V = 108 × 2 ×

10−6 = 200 m3.

What about the mass of this pile of tickets? As we all learned in science class, mass equals volume times density. If we buy a ticket, and don’t win, we might be tempted to discard it. Of course, we wouldn’t, because it is wrong to litter, no matter how annoyed we might be. But hypothetically, suppose we tossed the ticket into a puddle, would it float or sink? The former, I think, at least until it absorbed some water, and perhaps sank, as some paper products will do after a while. That means that the density of the ticket is reasonably close to the density of water. Since the density of water is 1000 kg/m3 or 1 ton/m3, the total mass is 200 tons. This would require five 40-ton trucks to haul it away.

Another way of looking at this lottery is that you would need to buy five tractor-trailer trucks full of tickets to make sure that you won!

42

Chapter 3 General Questions

January 29, 2008

Time:

06:19pm
chapter03.tex

3.8

Tons of trash How much domestic trash is collected each year in the US (in m3 or tons)?

*

US.

the
in
are
there
homes
many
how

Estimate

HINT: compacted.

be
can
but

L)

50

(
gal

13
is
bag
garbage
kitchen

A

HINT: week?

each
out
throw
you
do
trash
much

How

HINT: 43

January 29, 2008

Time:

06:19pm
chapter03.tex

ANSWER: Before the era of recycling (and when the children lived at home), we used to empty our 13-gallon trash can in the kitchen about every other day.

Now, it’s only about once a week, though the recycling bin now gets emptied once a week (with newspapers, boxes and other cardboard containers, bottles, plastic containers, cans, etc.). On second thought, it’s easier to lump them all together for this problem. Since considering recycling changes the answer by less than a factor of two for our household, we will ignore it for this problem.

OK, if we empty the trash three or four times a week, that’s about 50 gallons of trash for four people.

Now a gallon is about 4 liters, and a liter is 10−3 m3, so our 50 gallons per week is 200 liters or 0.2 m3. In one year (50 weeks), four people produce 50×0.2 = 10 m3
of garbage. That is about 300 cubic feet. Yick!

It’s worse than that. There are 3 × 108 of us, or about 108 households, so we produce 109 m3 of un-compacted garbage.

Now let’s try to figure out the mass of that garbage.

There are two things to consider here: first, trash is mostly not liquid (ooh, you threw the soup away!), and therefore there’s a lot of air space in the trash bag, and second, related to it, the density of the trash is much less than that of water. Let’s estimate the density.

That full 13-gal (50-L) trash bag probably only weighs between 10 and 20 lb (5 and 10 kg). Thus, its density is between 0.1 and 0.2 kg/L (or 0.1 to 0.2 tons/m3 or between 10 and 20% that of water).

Let’s take an average density of 0.2 tons/m3. Thus, in one year, my family produced m = 10 m3 ×

0.2 tons/m3 = 2 tons of garbage (in US units, that is, um, 2 tons).* Note that, since the average density is * There are lots of different tons: metric, short, long, . . . Since they differ by only 10%, we will use them interchangeably.

44

Chapter 3 General Questions

January 29, 2008

Time:

06:19pm
chapter03.tex so low, compacting the trash in garbage trucks should reduce the volume by a factor of about three (more than one and less than five).

But enough about us, let’s look at the whole country. With a population of 3 × 108, we produce a total mass and compacted volume of trash of about M = 108 households × 2 tons/year household = 2 × 108 tons of trash/year V = 108 households × 1 × 10 m3/year 3
household

= 3 × 108 m3of trash/year Now let’s compare to reality. According to the US

Environmental Protection Agency [9], in 2005 the US
generated 245 million (2.45 × 108) tons of municipal solid waste (including recycling).

Now we need to figure out what to do with it all.

But that’s the subject of the next question.

45

January 29, 2008

Time:

06:19pm
chapter03.tex 46

January 29, 2008

Time:

06:19pm
chapter03.tex 3.9

Mt. Trashmore If we put all of that trash (see previous question) in a landfill, how much space will this require? What fraction of the US surface area is this?

*

? . . .

uarter
q
half,
istance,
d

E-W
the
as
same
the
istance
d
north–south the

Is

HINT: A.

L
to

NY
from
zones
time
three
is
it

Alternatively, hours.

6
in
coast

West
the
to
coast

East
the
from
fly
can

You

HINT: US?

the
of
area
the
is

What

HINT: it?

pile
you
can
high

How
trash?

that
all
for
need
you
o
d
area
much

How

HINT: problem?

previous
the
in
found
trash
the
of

volume
the
is

What

HINT: 47

January 29, 2008

Time:

06:19pm chapter03.tex ANSWER: We need to figure out how much area we need for all that garbage and how much area we have available. Let’s start with the area we need for the garbage. In the previous question we estimated that Americans generate 3 × 108 m3 of trash per year. If we pile it 1 m high, we will need 3 × 108 m2 and if we pile it 10 m high, we will need only 3 × 107 m2. Here in Virginia Beach we are very proud of our local Everest, Mt. Trashmore, a landscaped former landfill towering 62 ft (20 m) above sea level. This means that we’ll need an area of A trash = 3 × 108 m3/yr = 107 m2/yr 20m Let’s plan ahead and make our landfill large enough for 100 years. In that case, we will need an area of 109 m2.

One billion square meters sure sounds like a lot. Let’s look more closely. A square kilometer (which is about half of a square mile) is a square 103 m on a side, so that 1 km2 = (103 m)2 = 106 m2. This means that one billion square meters is only(!) one thousand square kilometers (109 m2 = 103 km2). That still sounds like a lot, but it is only the area of Los Angeles* or Virginia Beach and, besides, we have a whole country to dump it in.

What is the US land area? Think of the US as being rectangular in shape. We need to find the east–west (New York to Los Angeles) and north–south (Mexico to Canada) distances. You can fly from NY to LA in six hours. The jet flies at about 500 mph so the distance is about 3000 miles (or 5000 km). Alternatively, we know that there are three time zones from NY to LA so that the distance is 3/24 = 1/8 of the Earth’s circumference at the equator (which we figured out in the golf ball problem). Thus, the east–west distance * Some claim that a landfill would improve LA significantly.

48

Chapter 3 General Questions

January 29, 2008

Time:

06:19pm
chapter03.tex
is 4 × 104 km/8 = 5 × 103 km. Good, we got the same answer.

Wait a minute, you may say, the US land area does not include the equator! That’s no problem (except for ambitious politicians), because the width of three time zones in the US is not that different from that of three time zones at the equator.

We can estimate the “vertical” (i.e., north–south) dimension by just eyeballing the relative proportions of our rectangular US from a map. The north–south span is about one-third of the east–west span or about 1000 miles (1600 km). The land area is then approximately A US = 5 × 103 km × 1.6 × 103 km = 8 × 106 km2

Thus, the fraction of land area needed for our trash is f = A trash =

103 km2

= 10−4

A US

8 × 106 km2

This means that after we have dumped all of our trash for 100 years in a single huge land fill, we will still have 99.99% of the US land area for everything else we want to do. This is just as Penn and Teller told us [10].

49

January 29, 2008

Time:

06:19pm
chapter03.tex

50

January 29, 2008

Time:

06:19pm
chapter03.tex

3.10

Juggling people On average, how many people are airborne over the US at any given moment?

*
time.

given
any
at
flying
people
of
fraction
the
to
qual
e
is
flying
spend
people
time
of
fraction

The

HINT: year.

a
in
ays
d
or
hours
of
number
the
compared
year
per,
fly
you
days
or
hours
of
number
the
e.g.,
flying,
spend
you
time
of
fraction
the
about

Think

HINT: day.

the
uring
d
sometime
choose

AM,

3:00
choose

Don’t

HINT: 51

January 29, 2008

Time:

06:19pm
chapter03.tex

ANSWER: There are two basic ideas here. First, the fraction of time the average person spends flying equals the average fraction of people that are airborne at any instant. This means that if you spend 10% of your time flying, then on average 10% of the population is airborne at any given time.* Note that this only works if there are enough people to average things out.† Second, we can use our own experience to estimate the fraction of time an average person spends in the air, or shopping or sleeping or (you name it). In other words, number flying now = time spent flying US population 1yr

Back in chapter 1 we estimated that the average American takes between two and four flights per year.

The typical flight will take between one and six hours (not counting time spent parking, waiting in lines, consuming the delectable airport comestibles, . . .) so we will estimate three flights per year at three hours per flight, or nine hours per year in flight. Now we insert the numbers we know: number flying now =

9 hr

3 × 108 hr people 400 days × 25 hr/day We rearrange this to get number flying now = 3 × 108 hr people × 9 hr 104 hr = 3 × 105 hr people That means there are about three hundred thousand people airborne over the US at this moment. We hope they all land safely.

* This does not mean that if you spend 10% of your time flying, that 10% of the average person (that’s about one leg) is airborne.

† One other person, or even ten others wouldn’t suffice. There have to be enough people so that at every moment some are in the air.

This is not a problem for this question.

52

Chapter 3 General Questions

January 29, 2008

Time: 06:19pm chapter03.tex 3.11

Shelving the problem During the last big California earthquake, two million books fell off the shelves in a university library. How many students would need to be hired to reshelve all of the books in three weeks?

*
week?

a
in
work
student a
would hours many

How

HINT: hour?

an
in
shelve student a
might

books
many

How

HINT: 53

January 29, 2008

Time:

06:19pm
chapter03.tex ANSWER: The books are not just put back on the shelves in random order. They must be placed according to their Library of Congress classification, so care must be taken to identify exactly where the book should reside. We’ll assume that the books fell relatively close to where they belong so no one needs to walk across the library. If a book is at my feet, and I know immediately where to put it, that will take between a few seconds and a minute. Thus, we can reshelve between 60 and 600 books per hour. Let’s work with an average of 200 per hour (that’s one-third of 600 and three times 60).

This means that in three weeks, working eight hours per day and five days per week, one student can reshelve N =

200 books

× 8 hr × 5 days × 3 weeks student-hour day
week

= 2 × 104 books/student Now, we need to reshelve two million books. This means that we need N students =

2 × 106 books = 102 students 2 × 104 books/student Thus, 100 students would be needed to reshelve all of those books in three weeks (assuming that they do not stop to read the books as they reshelve them).

54

Chapter 3 General Questions

January 19, 2008

Time: 11:57am chapter04.tex Animals and People Let’s continue with some more straightforward questions about animals and people. How big are we, how far do we run, how many porta-potties do we need?

Be warned, the last question is harder.

Chapter 4

*

January 19, 2008

Time: 11:57am chapter04.tex 56

January 19, 2008

Time: 11:57am chapter04.tex More numerous 4.1
than the stars in the sky How many cells are there in the human body?

*
see.

can
you
object
smallest
the
about
is
cell
big
eally
r

A

HINT: box.

ectangular r
a
as

volume
your
estimate

Alternatively,

HINT:

kg/m

0 1
or
kg/L

1
is
water
of

3

3
density

The
mass?

your
is

What
mass/density.

=

Volume

HINT:

volume?

your
stimate
e
you
might

How

HINT:

57

January 19, 2008

Time: 11:57am
chapter04.tex

ANSWER: Sorry to ask such a personal question, since we hardly know one another, but what’s your volume?

It’s on your driver’s licence, along with your surface area . . . oops, we were being futuristic. Since you didn’t answer the question, let’s figure it out. Let’s estimate your mass, Mr. Jones, at a nice even 100 kg (or approximately 200 lb; ladies, you can modify this to suit your own figures). Since we can safely assume that you float,* your average density is rather close to that of water or about 1 kg/L or 103 kg/m3. Thus, 100 kg of water occupies 100 kg × (1 m3/103 kg) = 0.1 m3. So your volume, Sir, is about 0.1 m3.

We can do this another way. Let’s approximate our body as a box of length l , width w, and height h, so our volume is V = l × w × h. What choices shall we make for these quantities? Length is easy, l ≈ 6 ft. Now what about w and h? John’s cross section is decidedly not rectangular, but since we’re not going to say what shape it is,† we’ll pick an average width, w = 1 ft (remember to average over head, neck, torso, legs, and feet). As for h, the front–back dimension, it will be about, say, 6 in. (he’s not barrel-chested). Therefore, his volume is 6 × 1 × 1 = 3 ft3. Now 1 m3 is about 3 × 3 × 3 = 27 ft3, 2
so 3 ft3 is about 0.1 m3.

Now let’s figure out the size of a cell using our own eyes. We cannot generally see individual cells with the unaided eye. If you look at a ruler, the lines on the ruler are a fraction of a millimeter (10−3 m) wide. I can quite easily see something one-tenth of a millimeter (10−4 m) in size. I cannot see cells, so they must be smaller than that. The inventor of the microscope used his first crude microscope (with a magnification between 10 and 100) to see cells.

Thus, a typical cell must be 10 to 100 times smaller * That’s easy for us to say—we’re standing at the side of the pool.

† Note: We are dealing with round numbers.

58

Chapter 4 Animals and People

January 19, 2008

Time: 11:57am
chapter04.tex
than 10−4 m or between 10−5 and 10−6 m (that’s 1 to 10 µm [micrometer]) in size.

Here’s another approach. We can see cells with an ordinary light microscope.* This means that the cells must be larger than the wavelength of visible light or we couldn’t see them. The wavelength of visible light ranges from blue light at about 0.4 µm to red light at about 0.7 µm (400 to 700 nm). Thus, cells must be larger than 1 µm. However, while we can see the major features inside a cell, we cannot see that much detail, so that typical cells must be much smaller than about 100 µm.

A typical human cell with a diameter of 10 µm (10 × 10−6 m = 1 × 10−5 m) will have a volume V cell ≈ diameter3 = (10−5 m)3 = 10−15 m3

Now the number of cells in our body is just the ratio of the volumes: V body N cells =

= 10−1 m3 = 1014

V cell 10−15 m3

This means that you, Sir, and I each have about 100
trillion cells in our bodies. Goodness me! That’s about a thousand times as many stars as reside in our galaxy.

Start counting . . . and Mr. Lucas? It’s time for the first “Cell Wars” trilogy . . .

* A light microscope uses light rather than, say, electrons to “see”

with. It may still be rather heavy.

59

January 19, 2008

Time: 11:57am
chapter04.tex

60

January 19, 2008

Time: 11:57am
chapter04.tex

4.2

Laboring in vein What is the total volume of human blood in the world?

*
population?

world
the
is

What

HINT:

be?

this
might

volume
blood
your
of
fraction

What
sitting?

a
at
onate
d
you
do
pints
many

How

HINT:

blood?

is

volume
your
of
fraction

What

HINT: 61

January 19, 2008

Time: 11:57am
chapter04.tex

ANSWER: We can estimate the amount of blood we contain either as a fraction of our total volume or as a multiple of the blood that we can donate. Let’s start with our volume. We’ve already estimated human volume in the previous question; it’s about 0.1 m3 or 100 L. If we were made only of water, our discussion would be more fluid in nature (or we would be all wet).

Since the blood has to carry all of the oxygen and other nutrients we need, our body must be more than 1%
blood. Similarly, it is almost certainly less than 10%
blood. If we take the average, then we each contain about 5% of 100 L, or 5 L of blood.

Alternatively, we can estimate this from the amount of blood we are allowed to donate. A typical Red Cross donation is about one pint of blood. They would surely not let us donate a dangerous amount of blood, so that this one pint is probably about 10% of our total supply. Now we just need to convert from archaic pints to modern liters. There are two pints in a quart, which is about a liter. Thus, 10 pints is about 5 L, so the two methods give about the same answer. [ Confession: JA lived in the United Kingdom during the period regarded as dangerous for the incidence of mad cow disease, so he cannot donate blood. But thus far he has shown no symptoms (outside of this book, that is).]

There are about 6 × 109 people in the world. Thus, the total volume of blood is V =

5 L

× 6 × 109 people = 3 × 1010 L
person

There are 1000 L in a cubic meter, so that V = 3 ×

107 m3. Now let’s see how large a volume this is.

New York City’s Central Park is about 2 km2 or about 2 × 106 m2. Thus, this would cover Central Park to a depth d = volume = 3 × 107 m3 = 15 m area 2 × 106 m2

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Chapter 4 Animals and People

January 19, 2008

Time: 11:57am
chapter04.tex

This is 50 ft or the height of a 5-story building. Given the number of murders occurring in TV series like CSI New York and Law and Order, one might be forgiven for thinking that most of this blood is already there!

If we want to get biblical, we can compare it to the volume of blood shed at the battle of Armageddon as mentioned in the book of Revelation: “They were trampled in the winepress outside the city, and blood flowed out of the press, rising as high as the horses’

bridles for a distance of 1,600 stadia” Rev. 14:20 (NIV).

A horse’s bridle is about 2 m high. At almost 200 m per stadion, 1600 stadia is 300 km. Now we just need the width. That much liquid will spread out a lot, especially when flowing 300 km. Let’s use a width of 3 km. Thus, the volume of blood predicted to flow at Armageddon is V Armageddon = 2 m × 3 × 105 m × 3 × 103 m = 2 × 109 m3

That is about 15 times more blood than humans currently have. We guess we just need more people.

63

January 19, 2008

Time: 11:57am
chapter04.tex

64

January 19, 2008

Time: 11:57am
chapter04.tex

4.3

Unzipping your skin What is the surface area of a typical human?

(Include only the skin, not the surface area of the digestive tract.)

*
interested).

not
you’re
if
even
h
r π
still
is

volume

(its
h
r π
is

volume
its

2

2
interested,
you’re

If
height.

times
adius
r
times
π 2
is
bottom)

and
top

(ignoring
cylinder
a
of
area
surface

The
adius.

r
and
height
your
stimate
e
and
cylinder
a
are
you
imagine

Alternatively,

HINT: back.

your
of
area
the
forget

Don’t
width.

and
length
your
stimate
e
and
sheet
flat
a
as
yourself
of

Think

HINT: 65

January 19, 2008

Time: 11:57am
chapter04.tex

ANSWER: If you could unzip your skin (like the rhinocerous in Kipling’s Just-So story), how large an area would it cover? The area of a sheet on a double bed? Your backyard? This is actually an important question since it determines how much fabric you need to make a suit of clothing, how much sunlight hits your body, how much force is exerted on you when scuba diving, how much of certain medicines you should take, and how much you can perspire.

You could measure most of your area by taking a shirt and a pair of pants, disassembling them, and measuring their total area, but we are not trying to be that precise. There are at least two reasonable approximations, humans as cylinders and humans as flat sheets.

We are certainly not cylindrical critters of height h and radius r , but if we were, then our surface area would just be A = 2π r h. We can account for the fact that we are not simple cylinders by multiplying the final answer by 1.5 or 2. Let’s estimate r and h. Height is easy, ours is 2 m (well, let’s just say that it rounds up to 2 m [6 ft 7 in.]). Radius is less than 1 m (3 ft) and more than 0.1 m (4 in.) so we’ll estimate 0.5 m. This gives a surface area of A = 2π r h = 3 × 0.5 m × 2 m = 6 m2

We can check to see if this is consistent with our previous volume estimate of 0.1 m3. The volume of a cylinder is the area of its base (π r 2) times its height, or V = π r 2 h. This gives V = 3 × (0.5 m)2 × 2 m =

1.5 m3.

Oops, that is way too much. Let’s decrease the radius to 0.2 m (8 in.). That will give us a volume of 0.24 m3, still a bit too big, and an area A = 2.4 m2.

Since we’re not really cylinders, let’s round the answer up to 3 m2.

66

Chapter 4 Animals and People

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Time: 11:57am
chapter04.tex

That’s about the area of a medium-sized towel (not including the fibers!). Of course, we have lots of curves and corners, so this is just a rough estimate, even if we shaved first.

Alternatively, suppose we were flat like a sheet; front and back would each be about 2 m (6 ft) high by 0.5 m (1.5 ft) wide, so that’s a total of 2 m2, a bit smaller than the first estimate, and much simpler to obtain!

Body surface area (BSA) is important in medicine for calculating the appropriate dose of some medications. You can calculate your own BSA at many websites [11]. I calculated mine to be 2.06 m2, so our estimates are pretty darn good (for purposes of this book, but certainly not for medical purposes).

At ten tons of air pressure on every square meter, there are 20 tons of air pushing on you. If you scuba dive down to 120 ft (40 m), there are another 40 tons of water pressure on every square meter. That adds up to another 80 tons of water pressing on you. It’s amazing that the human body can withstand those forces!

67

January 19, 2008

Time: 11:57am
chapter04.tex

68

January 19, 2008

Time: 11:57am
chapter04.tex

4.4

Hair today, gone tomorrow What is the total length of all the hair on an average woman’s
head?

*
cm.

10

=

. in

4
that
remembering
by
metric
to
convert
you
if
arithmetic
the
do
to
easier
is

It

HINT: hair?

of
strand
ach
e
is
long

How

HINT: hairs?

between
separation
average
the
is

What

HINT: scalp?

her
of
area
the
is

What

HINT: 69

January 19, 2008

Time: 11:57am
chapter04.tex

ANSWER: We will answer this question in three stages.

First we will estimate the area of the scalp, then the number of hairs per square centimeter, and then the length of a typical strand of hair.

My hand spans a distance of about 8 inches or 20 cm. My head is about one hand span in diameter.

Assuming my head to be a sphere (and my scalp to be a hemisphere), the area of my scalp is A scalp = 1 4π r 2 = 6(10 cm)2 = 600 cm2

2

(Alternatively, if you treated your scalp as a square 20 cm on a side, you would have gotten 400 cm2. Close enough!) Now we need the number of hairs per area (per square centimeter). We can count the number of hairs along one centimeter and square it. There are 1–2 hairs per millimeter or 10–20 per centimeter. This gives 100–400 hairs per square centimeter. We’ll use 200.

This means that the total number of hairs on a person’s head is about N = 6 × 102 cm2 × 2 × 102 hairs/cm2

= 105 hairs

Blond hair is typically finer and more closely spaced than black hair so your answers may vary. However, this unnecessary precision amounts to bifurcating rabbits.*

All we need now is the length of a typical hair.

Women’s hair ranges from 1 cm (0.5 in.) to 100 cm (1 m or 3 ft). The geometrical mean of 1 and 100 cm is 10 cm (4 in.) or about shoulder length. This means * Splitting hares.

70

Chapter 4 Animals and People

January 19, 2008

Time: 11:57am
chapter04.tex
that the total length of all the hairs on a woman’s head is about L = 105 hairs×10 cm/hair = 106 cm = 104 m = 10 km This is 10 kilometers or about 6 miles. Short hair would be about 1 km and really long hair would be about 100 km.

It’s enough to make your hair stand on end.

71

January 19, 2008

Time: 11:57am
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72

January 19, 2008

Time: 11:57am
chapter04.tex

4.5

Hot dawg!

How long a hot dog (or sausage or wurst or . . .) can be made from a typical cow?

*
cow’s?

the
as
same
the
is

volume
its
that
so
be
to
have
dog
hot
that
would
long

How

HINT:

dog?

hot
a
of
thickness
the
is

What

HINT:

human?

a
of

volume
the
is

What

HINT:

times?

000

1
times?

100
times?

10
person,
a
than
cow
a
is
larger
much

How

HINT:

73

January 19, 2008

Time: 11:57am
chapter04.tex

ANSWER: We first need to figure out the volume of a cow. A cow is about ten times the weight of a person (about one ton or the weight of a small car). It is definitely less than 100 times the weight of a person (10 tons is the weight of a large truck). Since the densities are about the same, this means that it has ten times the volume of a person or V cow = 10 ×

0.1 m3 = 1 m3.

Now we just need to estimate the volume of a sausage or hot dog. We’ll assume square hot dogs so we don’t to bother with pesky factors of π. The thickness, t, of a typical sausage or “dog” is 1 in. (2 cm or 0.02 m).

Thus, for a hot dog of length L , it will have a volume of V hotdog = L × t 2
and since it must have the same volume as the cow it is made from, its length will be L = V cow =

1 m3
t 2

(2 × 10−2 m)2

= 2 × 103 m

This is 2000 m or 2 km (over 1 mile)! Wow!

Note that this assumes that every little bit of the cow is ground up into hot dog. We really, really hope that this is a bad assumption.

A hot dog made from a human being would be ten times shorter or about 200 m. That is twice the length of a football field!

74

Chapter 4 Animals and People

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Time: 11:57am
chapter04.tex

4.6

Playing the field

How far does a soccer or field hockey player travel during the course of a 90-minute game?

*
moving.

spent
time

×
speed
average

=
covered

Distance

HINT:

running?

spend
she
does
game
the
of
much
how
and
run
she
oes
d
fast
how

Alternatively,

HINT:

field?

the
own
d
and
up
un
r
she
does
times
many

How

HINT:

75

January 19, 2008

Time: 11:57am
chapter04.tex

ANSWER: If you’ve ever watched that most wonderful of sporting events, the World Cup,* you’ll have noticed that players (except the goal keepers) spend most of their time running or walking, only occasionally standing still (such as when some reckless and violent individual playing against England commits a terrible foul against an innocent player who would otherwise have scored a magnificent and game-winning goal).

The players rarely stand still; they sometimes walk and sometimes run quite fast.

Let’s estimate walking and running speeds. The world record for running the mile is about 4 minutes.

That would be about 15 miles per hour, which, at 1 m/s ≈ 2 mph, is about 7 m/s. Alternatively, the world record for the 100-m dash is about 10 s, or about 10 m/s. Walking speeds are closer to 2–4 mph or 1–2 m/s. If we assume that half the time is spent running at full speed and the other half is spent walking, then we’ll have an average speed of (7 + 1)/2 = 4 m/s or 8 mph. (The correct answer is surely between 2 and 8 m/s so our answer must be within a factor of two!) During a standard 90-minute (1.5-hour) game, the players travel about 8 mph times 1.5 hr = 12 mi (or almost 20 km). Quite a workout!

* Disclaimer: This is not the opinion of all the authors and certainly not of the editors.

76

Chapter 4 Animals and People

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Time: 11:57am
chapter04.tex

4.7

Ewww . . . gross!

How many people in the world are picking their nose right now?

*
nose?

her
or
his
picking
spend
person
typical
a
does
them
of
many
how
day;
a
in
minutes

10
about
re
a

There

HINT:

3
activity.

indelicate
that
in
spends
them
each
time
of
fraction
the
equals
nose
their
picking
people
of
fraction

The

HINT:

world.

the
in
people

10

×

6
re
a

There

HINT:

9

77

January 19, 2008

Time: 11:57am
chapter04.tex

ANSWER: As with the flighty Americans problem, the fraction of the time you or I spend in some activity (except for scuba diving: none!) is equal to the fraction of people doing it right now.*

We won’t ask how long you spend picking your nose each day, but how long do you estimate that your friends spend in this activity? Ten seconds is probably far too short, and 1000 seconds or 15 minutes is too long, so 100 seconds (about 2 minutes) seems a reasonable compromise. We’ll further assume that our children are correct when they claim that “Mom, everybody does it!”†

Don’t tell our children, but there is some evidence that ingestion of nasal mucous helps train the immune system to recognize harmful bacteria and viruses.

Anyway, we’ll estimate that all six billion of us spend an average of 2 minutes per day picking our nose (neglecting people with more than one nose). There are about 25 × 60 = 1500 minutes in a day. Thus, we have number picking now = 2min

6 × 109 people

1500 min
or

N pick = 6 × 109 people ×

2 min

1500 min

= 107 people

Thus, ten million people are picking their nose at this very moment. Hmmm . . . so this is why politicians use up gallons of hand sanitizer. They shake way too many hands.

* Assuming, of course, that the authors and readers of this book are typical people. Hah!

† Videotape of an audience of doctors at a medical lecture indicated that our children’s claim is quite correct.

78

Chapter 4 Animals and People

January 19, 2008

Time: 11:57am
chapter04.tex

4.8

Going potty

How much space would
a million people need
at a political rally?

How many porta-potties?

*
loo?

the
in
spend
you
do
time
the
of
fraction

What

HINT:

activity.

indelicate
that
in
spends
them
each
time
of
fraction
the
equals
porta-potty
the
in
people
of
fraction

The

HINT:

meter?

square
typical
a
into
crammed
are
people
many

How
ally?

r
political
a
at
have
you
do
room
lbow
e
much

How

HINT:

79

January 19, 2008

Time: 11:57am
chapter04.tex

ANSWER: In a political rally, people are packed reasonably tightly together. There is still space to move around, so we’re not packed as tightly as in the very first question. Let’s estimate that there’s about two feet between people. Since each person is about one foot in size, each person occupies a space that is about 3 ft by 3 ft or 1 m2. That means that 106 people occupy 106 m2 or 1 km2. That is about the size of the National Mall in Washington, DC or Central Park in New York.

Now that you have all those people there, you need to provide facilities for inputs and outputs, that is to say refreshments and porta-potties.

Again, as with the flighty Americans or the nose picking, the fraction of the time I spend in some activity (except for bungee-jumping: none!) is equal to the fraction of people doing it right now. So: how long do you spend in the bathroom during the daytime?

Ignore time spent grooming or powdering your nose.

One minute is probably too short and 100 minutes is far too long (“Are you going to be in there ALL
day?”). So let’s settle on 10 minutes, shall we? Also, let’s just take a 15-hr day because most people don’t go to political rallies while they are sleeping. Let’s convert 15 hours to 15 × 60 = 103 minutes (so it has the same units as our bathroom interval). Thus, the fraction of daytime spent in the bathroom (and hence the fraction of people in the bathroom at any instant) is 10 minutes out of 103 minutes or 1%.

This means that we need one potty for every 100
people, or a total of 104 potties for 106 people. Wow, that’s a huge logistical effort! Note that if there is one potty for every 100 people, given that I spend 1/100
of my time there, then all the potties will be 100%
occupied during the entire rally. Since the demand for potties is generally not uniform, there will be serious queuing problems (that is British for “really long lines”). So a few more might be advantageous . . .

80

Chapter 4 Animals and People

January 19, 2008

Time: 11:57am
chapter04.tex

Now let’s compare with reality. The Virginia Children’s Festival attracts about 3000 people annually to Norfolk’s Town Point Park. They have about forty or fifty porta-potties there. The number of potties required, according to our estimate, is 3000/100 = 30, so they have a few extra to reduce the waiting times.

81

January 19, 2008

Time: 11:57am
chapter04.tex

82

January 19, 2008

Time: 11:57am
chapter04.tex

4.9

Let’s get one thing straight!

How long is all the DNA
in your body?

How long is the DNA
of all humanity?

*
ago.

uestions
q
several
cells
of
number
the
estimated
just

We

HINT:

cell.

the
of
size
the

10

/ 1 t u o b a s i d n a

A

N

D
h t i w
d
e l l fi
s i l l e c
a
f o
s u
e l c u
n
e
h

T

HINT:

m.

10

9

−
length
side
of
cube
a
approximately
is
pair
base

Each

HINT:

size.

in
m

0 1
about
is
atom

One

HINT:

10

−
atoms.

000

1
about
of
composed
each
pairs,”

“base
of
strings
long
of
composed
is

DNA

HINT:

cell?

one
in
is

DNA
much

How

HINT:

83

January 19, 2008

Time: 11:57am
chapter04.tex

ANSWER: This one is a little complicated. We need to calculate the size of the building blocks of DNA and the total volume of DNA in the cell. Then we can use that to figure out how long the total DNA is.

The nucleus of a cell is filled primarily with long strands of DNA called chromosomes.* Each chromosome is composed of long strings of base pairs (the familiar letters AT G C from a long forgotten biology course). Each base pair is a very complicated molecule itself. This means that it must contain a lot of atoms, certainly more than 100 and less than 104, so we’ll estimate 103 atoms per base pair. We will treat each base pair as a cube, 10 atoms on a side. Since all atoms are about 10−10 m, our base pairs are 10−9 m in length.

The volume of a base pair is then the length cubed: V bp = (10−9 m)3 = 10−27 m3

We estimated earlier that a cell is about 10−5 m in size. The nucleus is about 1/10 of that, or 10−6 m.

Thus, the volume of the nucleus is

V n = (10−6 m)3 = 10−18 m3

This means that each nucleus can contain a number of base pairs N bp = V n = 10−18 m3 = 109

V bp

10−27 m3

That is one billion base pairs. That’s a LOT of information.

Now let’s straighten out all that DNA. There are 109
base pairs at 10−9 m each so the total length of DNA in the cell is about 1 m. According to biology textbooks, the length is between 1 and 3 m [12, 13].

* This is a physicist’s view of a cell and is accurate to within a factor of two or three. If you want more precision, call a friendly biologist.

84

Chapter 4 Animals and People

January 19, 2008

Time: 11:57am
chapter04.tex

Now we estimated previously that there are 1014
cells in our bodies. At 1 m of DNA each, that stretches out to 1014 m. That is 1000 times the distance from the Earth to the Sun, ten times the distance from here to Pluto, or about 1% of a light year.

If we stretched out the DNA of everyone on the planet, the combined DNA strand would extend about 108 light years—about 50 times farther than the next galaxy! But who would be around to appreciate the comparison?

85

January 19, 2008

Time: 11:57am
chapter04.tex

86

January 19, 2008

Time: 12:16pm
chapter05.tex

Transportation

Americans have a love–hate relationship with their cars. We love the freedom and privacy they provide but we hate the traffic and pollution they cause and the expensive fuel they need. In this chapter we look at how far we drive, how much it costs, and what some of the alternatives are.

Chapter 5

*

January 19, 2008

Time: 12:16pm
chapter05.tex

88

January 19, 2008

Time: 12:16pm
chapter05.tex

5.1

Driving past Saturn

How many total miles (or kilometers) do all Americans drive in one year? How does this compare to the circumference of the Earth (2.5 × 104 mi), the distance to the Moon (2.4 × 105 mi), the distance to the Sun (9 × 107 mi), or the distance to Saturn (109 mi)?

*

Americans.

10

×

3
re
a

There

HINT:

8
cover?

warranties
car
new
do
year
per
miles
many

How

HINT:

year?

a
in
drive
person
one
does
far

How

HINT:

89

January 19, 2008

Time: 12:16pm
chapter05.tex

ANSWER: To answer this question we need (1) how many miles each American drives and (2) the total number of Americans who drive. There are many ways to estimate our average mileage. You could look it up, but that would violate the spirit of this book, and besides, it would involve getting up and walking over to the computer. You could use your typical yearly mileage (which would probably be within a factor of two of the national average). Or you could look at new car guarantees. Manufacturers advertise 3-year/

36,000-mile or 10-year/120,000-mile warranties. This indicates that each car is driven about 12 thousand miles per year.

Now we need the number of cars. There are 3 × 108 Americans. Almost all of us drive. There is less than one car per person and more than one car per four people so we will estimate one car for every two people.* Thus, the total miles driven by all Americans is about d = 104 mi ×3 × 108 people× 1 car = 2×1012 mi
car

2 people

That is 2 trillion miles or 3 trillion kilometers.

Wow! That distance would take us around the Earth 108 times, to the Moon and back 4 million (4 × 106) times, or to Pluto and back 2000 times. It takes the Earth 3000 years to travel that distance in its orbit around the Sun. At a speed of 3 × 108 m/s, it takes light one-third of a year to travel that far.†

Who knew it was that far to the corner store!

* In 2000, there were an average of 2.6 people and 1.3 cars per American household. This is fortunate because only the 0.6 person can drive the 0.3 car [14].

† This means this means that Americans travel at a total speed of only one-third that of light.

90

Chapter 5 Transportation

January 19, 2008

Time: 12:16pm chapter05.tex 5.2

Drowning in gasoline What volume of gasoline does a typical automobile (car, SUV, or pickup) use during its lifetime? Note that this question asks about the lifetime of the vehicle, not the time that you own it. Compare the weight of the fuel to the weight of the car.

*
meter.

cubic
per
ton
one
is
water
of
density

The

HINT: meter.

cubic
per
liters

10
and
gallon
per
liters
four
are

There

HINT: 3
gallon?

per
get
it
does
miles
many

How

HINT: ?

0 1

?

0 1

?

0 1

?

0 1

?

0 1

?

10

6

5

4

3

2

1
lifetime?

its
in
rive
d
car
typical
a
does
miles
many

How

HINT: 91

January 19, 2008

Time: 12:16pm chapter05.tex ANSWER: First, let’s estimate how many total miles a vehicle travels in its lifetime. Your vehicle’s odometer typically has six digits. This indicates that manufac-turers expect your vehicle to last at least 100,000 (105) miles. However, most vehicles do not reach much past 200,000 (2 × 105) miles.* Let’s start with 105 miles because it is a nice round number. (Cars made in the 1950s and 1960s did not last as long and frequently had only five-digit odometers.) Now let’s estimate the gas mileage. There are several ways to do this. You can have an obsessive attention to detail and calculate your mileage every time you fill your gas tank. You can read the new car ads or Consumer Reports and notice the mileage. Or you can estimate it from how far you travel between fill-ups and how much gas you put in. Most cars travel about 300 miles between fill-ups. If you have to put in 10
gallons of gasoline, then you get 300 mi = 30mpg (miles per gallon) 10 gal

If you have to put in 30 gallons, then you get only (300 mi)/(30 gal) = 10 mpg. Since almost all cars get between 10 and 40 mpg, we’ll use 20 mpg for our estimates.†

Now we can calculate the answer. The total number of gallons used is equal to the total miles driven divided by the miles per gallon: V gas = 105 mi = 5 × 103 gal 20 mpg * (Larry here:) I tried to keep my previous car until it reached the moon, but had to get rid of it at only 225 thousand miles, not the 240 thousand I was aiming for.

† This way, we’ll be within a factor of two of the right answer for almost any vehicle.

92

Chapter 5 Transportation

January 19, 2008

Time: 12:16pm
chapter05.tex

Now we need to convert to metric, since converting to cubic inches or feet or yards would be too complicated.

One gallon equals four quarts, which equals approximately four liters. Thus, 5 × 103 gal ≈ 2 × 104 L. Since there are 1000 L in a cubic meter, this gives a volume of gasoline equal to 20 m3.

Now let’s see how big that is. A 20-m3 volume could be 1 m deep by 4 m wide by 5 m long. At 3 ft per meter (actually 3.3), that’s 3 ft deep by 12 ft wide by 15 ft long.

That is much larger than your car. It’s enough to fill a small above-ground swimming pool!

At a density of 1 ton/m3, your car will burn about 20 tons of fuel in its lifetime. That is ten times the weight of the car itself! Even if you drive a Prius or other car that gets 50 mpg, you will still burn 8 tons of fuel.

Maybe your car lasts longer than 105 miles. Maybe your car is a 10-mpg gas guzzler or a 40-mpg gas sipper. Your answer will still be well within a factor of ten. The fuel your car burns will greatly outweigh the car itself.

93

January 19, 2008

Time: 12:16pm
chapter05.tex

94

January 19, 2008

Time: 12:16pm
chapter05.tex

5.3

Slowly on the highway How much total extra time would Americans spend driving each year if we lowered the highway speed limit from 65 to 55 mph? (Note that we assume that there is some relationship between posted limits and actual speeds on highways.) Give your answer in lifetimes.

*
easier.

arithmetic
the
make
to

65
and

55
of
instead
mph

70
and

60

Use

HINT: highway?

65-mph
on
drive
we
do
those
of
fraction
what
and
highway
the
on
rive
d
we
o
d
miles
those
of
fraction

What

HINT: question.)

previous

See

(
rive?

d
we
o
d
miles
total
many

How

HINT: 95

January 19, 2008

Time: 12:16pm
chapter05.tex

ANSWER: We need to estimate the time we spend driving on 65-mph highways. To do this, we need to estimate the distance we drive on 65-mph highways.

From the answer to an earlier question, we know we drive 2 × 1012 (that’s two trillion) miles per year.

About half of that is spent driving on the highway.

Urban highways do not have 65-mph limits (as if you could ever drive that fast on the Santa Monica Freeway or New York’s West Side Highway). We’ll estimate that only about half of highway driving is done on 65-mph roads. Thus, we drive d 65 mph = 2 × 1012 mi × 1 × 1 = 5 × 1011 mi 2

2
on 65-mph roads.

Now we need to figure out (1) how much time it would take to drive at a 65-mph speed limit, (2) how much time it would take to drive at a 55-mph speed limit, and (3) the difference. Instead of using 65 and 55 mph, we will use 70 and 60 mph. We do this for two reasons: (1) No one knows what the real (i.e., the enforced) speed limit is so most traffic exceeds the limit at least somewhat, and (2) it is much easier to do arithmetic with round numbers.

At 70 mph, driving takes a total time of t 65 mph = 5 × 1011 mi = 7 × 109 hr 70 mph If the speed limit is lowered to 55 mph and people drive at 60 mph, then t 55 mph = 5 × 1011 mi = 8 × 109 hr 60 mph Thus, we will spend an extra time t extra = t 55 mph− t 65 mph = 8×109 hr−7×109 hr = 109 hr That is an extra one billion hours behind the wheel.

Yikes!

96

Chapter 5 Transportation

January 19, 2008

Time: 12:16pm chapter05.tex There are only 24×365 ≈ 104 hours in a year. Thus, the extra 109 hours is an extra 105 years.

At 100 years in a lifetime, that is an extra 1000
lifetimes.

Holy cow! That is a lot more lifetimes than we lose to sharks or lightning or secondhand smoke or arsenic in drinking water or . . . .

Of course, there is a huge difference between one person losing her entire life or millions of people losing a few hours each. (After all, 109 hr divided among 3 × 108 Americans is only 3 hr each.) Note that we have not asked how many lives would be saved by lowering the speed limit. Since traffic deaths depend on actual speed driven, speed differences among drivers, and many other factors, it is very hard to predict the change in fatalities. There was no obvious change in the high fatality rate when the nationwide 55-mph limit was repealed and different regression studies reach different conclusions. Some claim that raising the limit saved lives.

97

January 19, 2008

Time: 12:16pm chapter05.tex 98

January 19, 2008

Time: 12:16pm chapter05.tex 5.4

Rickshaws and automobiles What are the relative costs of fuel (per kilometer or per mile) of New York City bicycle rickshaws (human-pedaled taxis) and of automobiles?

*
gasoline?

of
gallon
a
on
travel
car
a
does
far

How

HINT: bicyclist?

the
feed)

(i.e.,
fuel
to
cost
it
does

What

HINT: ay?

d
one
in
travel
rickshaw
bicycle
a
does
far

How

HINT: 99

January 19, 2008

Time: 12:16pm chapter05.tex ANSWER: Let’s start with automobiles. If your car gets 20 miles per gallon and gas costs $3.00 per gallon, then it will cost you $3.00/gal

= $0.15/mi = $0.10/km 20 mi/gallon Cars are cheap to run!*

Now let’s figure out how much the human costs per kilometer. To do that, we need to estimate the costs and the kilometers (or miles). Bicycling speed with passengers will be between 5 (very slow) and 10 (fast) mph. Thus, during an 8-hour day, the rickshaw can travel between 40 and 80 miles. Since it is not always in motion, we’ll use the lower estimate of 40 miles (or 60 km). We can also consider the performance of experienced bicyclists. They can travel 100 miles in a day. Pulling passengers and dealing with NYC
traffic would certainly slow them down to 50 miles (or less).

Working that hard will burn a lot of calories.†

We can estimate the costs of food several ways, from per diem rates for expense accounts, from fraction of annual income, from the costs of fast food, etc. Let’s see what we get.

Expense account per diems are about $40. This is almost certainly too high, since it applies primarily to business folks who expect to eat three meals a day at good to excellent restaurants. Western European or American per capita income is $30,000–40,000 per year or about $100 per day. We spend more than 5%
and a lot less than half of our income on food, so we’ll estimate we spend 10–20% of our income or $10–20
per day on food. Buying three meals a day in fast-food * Since the difference between miles and kilometers is only a factor of 1.6, it really does not matter which one you use.

† We’ll investigate food as an energy source in Chapter 5.

100

Chapter 5 Transportation

January 19, 2008

Time: 12:16pm chapter05.tex restaurants would also cost about $10–20 per day.*

Thus, the fuel costs for the rickshaw pedaler will be $15/day = $0.25/km 60 km/day

Thus, the rickshaw costs only 2–3 times more than the car for fuel. The car uses more energy per mile, but gasoline is a much cheaper energy source than food.

Note also that the fuel costs for the autmobile depend strictly on distance traveled (i.e., if you double the distance, you double the fuel cost) but that the fuel costs for the rickshaw bicyclist include the 2000 Calories per day just for basic metabolism.

* Since these are New York City rickshaw drivers, we will assume Western diet and food costs.

101

January 19, 2008

Time: 12:16pm chapter05.tex 102

January 19, 2008

Time: 12:16pm chapter05.tex 5.5

Horse exhaust How much waste is generated (per kilometer) by horse-drawn carriages and by automobiles? Give your answer in kg/km.

*
gasoline?

of
gallon
a
on
travel
car
a
does
far

How

HINT: person.

a
as
massive
as
times

10
about
is
horse

A
rink?

d
and
at
e
horse
the
oes
d
much

How

HINT: people.

than
faster
much
that
not
are

Horses
day?

one
in
travel
carriage
horse-drawn a
does
far

How

HINT: (waste).

output
becomes
fuel)

and
food

(
input
the
all
that

Assume

HINT: gaseous.

is
exhaust
car
and
liquid
or
solid
either
is
xhaust
e

Horse

HINT: 103

January 19, 2008

Time: 12:16pm chapter05.tex ANSWER: We’ll need to first estimate how far a horse can pull a carriage in one day and then estimate how much food and water it consumes in the process. Note that we are assuming that the horse does not gain or lose weight and therefore it converts its food and drink into an equal mass of horse exhaust.

How far can an average horse pull an average carriage in one day? Since we are making the horse work for about eight hours, it is certainly not going to gallop, canter, or trot for most of it. Let’s start with the average walking speed of a horse. A horse will certainly walk faster than a person (3–4 mph or about 5–6 kph or about 2 m/s) but not that much faster.

This gives us a range of about 5–10 mph. In 8 hours, it can travel 40–80 miles, so we’ll choose 60 miles (or 100 km).*

How can we figure out the food consumption? We can try to directly estimate the food consumed or we can start with a human and scale up. A horse will certainly eat more than 1 quart (or liter) of grain and less than 100 quarts. We’ll take the geometrical mean and estimate 10 L. That has a mass of about 10 kg (20 lb)†
and will therefore produce 10 kg of waste. Similarly, the horse will drink more than 1 L and less than 100 L so we will estimate that it drinks about 10 L

(also 10 kg).

Now let’s start with a human and scale up. We eat 2–3 lb (1–1.5 kg) of food and drink 1–2 L (or quarts) of liquid per day. A horse is about ten times our size (in volume or mass) and thus probably consumes about ten times more than we do. This gives about the same consumption as the previous estimate.

* That factor of 1.6 is almost irrelevant. Our estimates are just not that exact.

† The density of anything organic is reasonably close to water. The density of water is 1 g/cm3 or 1 kg/L or 1 ton/m3.

104

Chapter 5 Transportation

January 19, 2008

Time: 12:16pm chapter05.tex Thus, the horse will produce about 10 kg each of liquid and solid waste in the course of traveling 60 miles. The total waste will be about 0.3 kg/mi.

The automobile burns about one gallon every 20 miles. A gallon is about 4 L and thus has a mass of about 4 kg. Thus, the car produces 4 kg of waste in 20 miles or 0.2 kg/mi.

A car produces about the same amount of waste per mile as a horse.* The problem is that while the car’s exhaust blows away into the air (and is almost entirely carbon dioxide and water), the horse’s exhaust sticks around (literally). Imagine New York City if it had millions of horses instead of millions of cars. Better yet, don’t.

* Both produce plant food! The car produces carbon dioxide and the horse produces fertilizer.

105

January 19, 2008

Time: 12:16pm chapter05.tex 106

January 19, 2008

Time: 12:16pm chapter05.tex 5.6

Tire tracks How far does a car travel before a one-molecule layer of rubber is worn off the tires?

*
size.

in
m)

10

×
ew
f
a

(i.e.,

10

−
nanometer
a
of
tenths
few
a
only
is
molecule
ubber
r

A

HINT: lifetime?

tire’s
a
during
away
worn
is
this
of
much

How
tire?

new
a
of
tread
the
is
thick

How

HINT: miles)?

in

(
tire
a
of
lifetime
the
is

What

HINT: 107

January 19, 2008

Time: 12:16pm chapter05.tex ANSWER: First we need to figure the lifetime of tires in miles. As usual, there are a few ways to do this.

You can estimate the lifetime of a tire in years and assume the usual 12,000 miles per year. Tires definitely last more than 1 year and less than 10, so estimates of between 3 and 5 years are reasonable. Alternatively, you can read the tire ads, which advertise the tire lifetimes, or remember the lifetime of the last set of tires you bought. Tires typically last 30–60 thousand miles. They typically have between 1/4 and 1/2 in. (i.e., about 1 cm) of tread.

Thus, 1 cm of tread is worn off in about 4 × 104 mi.

We want to know how long it takes to wear off a thickness of one molecule or 5×10−10 m of tread. That distance is d = 4 × 104 mi × 100 cm × 5 × 10−10 m 1 cm 1 m

= 20 × 10−4 mi = 2 × 10−3 mi Now we need to make sense of this result. 10−3 mi is hard to figure out, but 10−3 km is just 1 m. Since a mile is only a little bigger than a km, we have d = 2 × 10−3 mi = 3 × 10−3 km = 3 m Three meters is about 10 feet. That is only one or two complete rotations of the tire.

Thus, you wear off a one-molecule thickness of rubber with every rotation of your tire.

108

Chapter 5 Transportation

January 19, 2008

Time: 12:16pm chapter05.tex 5.7

Working for the car Your car allows you to travel many miles in just a few hours. However, in addition to the hours you spend driving, you have to spend more hours not driving, hours you spend earning money to pay for your car (eg: depreciation, insurance, fuel). This extra time reduces your average car travel speed. For example, if you drive 60 miles in one hour and then spend one more hour earning enough money to pay for the driving, then your average speed is not 60 mph, but 30 mph.

If you add all the time you spend working in order to earn the money to pay for your car to all the time that you spend driving your car, what is your average car travel speed?

*
hour.)

per

$20
to
equivalent
is
year
per

$40,000
of
salary
a
that
so
week]

/
hours

40

×
year

/
weeks

[50
year
a
in
hours
working

2000
are
there
that

(Note
wage?

American
average
the
is

What

HINT: miles?

those
all
rive
d
to
you
take
it
does
hours
many

How

HINT: lifespan?

its
during
car
a
rive
d
you
do
far

How

HINT: costs.

these
of
all
include
should
which
ate,
r
reimbursement mileage

IRS
the
use
can
you

Alternatively, . . . .

parking,
maintenance,
gas,
repairs,
insurance,
price,
purchase

Include
lifespan?

its
over
car
the
of
ost
c
total
the
is

What

HINT: 109

January 19, 2008

Time: 12:16pm chapter05.tex ANSWER: We need to figure out how much time we spend working to pay for the car and how much time we spend driving the car. Let’s put the pedal to the metal and take it from the top.

To figure out the working time, we need to first estimate the life-cycle cost of the car. There are at least two ways to estimate the cost of a car. We can take the IRS mileage reimbursement rate of $0.445 per mile in 2006. That’s the easy way. (As usual, any value between $0.30 and $0.60 is quite reasonable.) Alternatively, we can try to figure out the life-cycle cost of a car. Here we go.

Let’s buy an average new car. No Hummers, no Jaguars, no Minis. Then let’s keep the car until it gets old or about ten years. We will have to pay for many things in those ten years: the car, insurance, gas, repairs, fuzzy dice, and parking.

The best selling vehicle in America in 2006 was the Ford F-150 pickup truck and the best selling car was the Toyota Camry. Both vehicles cost between $20,000
and $25,000 (depending on options and ability to bargain).

Insurance will cost about $1000 per year or $10,000
for 10 years. It will be a lot more for an 18-year-old male driving a Corvette with three speeding tickets and a DUI in the Bronx. It will probably be much less for a 45-year-old woman with a perfect driving record in Salem, Oregon.

Repairs will probably average about $1000 per year.

Repairs will cost much less in the first few years and much more in the last few years.

In ten years, we will drive about 120,000 miles (or about fives times around the Earth). At the Camry’s 20
miles per gallon, we will burn 120,000 = 6000 gallons 20
of gasoline. At $3 per gallon, that will cost $18,000.

The difference in mileage between the Camry and the 110

Chapter 5 Transportation

January 19, 2008

Time: 12:16pm chapter05.tex F-150 will amount to “only” a few thousand dollars of gas and is thus too small for this book to worry about.

If you live in Manhattan (New York, not Kansas), you can add $3000 per year for parking. That’s your problem. Most of us can leave that out.

Thus, the total costs of driving 1.2 × 105 mi in 10 years is (to one significant figure): Category Cost ($)

Vehicle cost 20,000

Insurance

10,000

Repairs

10,000

Gasoline

18,000

Fuzzy dice

5

Total

60,000

If we use the IRS mileage number of about $0.50 per mile, we also get a total cost of $60,000 for 120,000 miles.

Note that the total (life-cycle) cost of the car is three times the purchase price.

Now we need to figure out (1) how much time we spend driving the 120,000 mi and (2) how much time we spend earning the $60,000.

Most people typically drive on a mix of highways (55–65 mph) and local streets (25–45 mph plus traffic lights and stop signs). If we average highway and local (or Montana and New York City), we’ll get a speed of 30–40 mph. This means that we will spend a total of t drive = 1.2 × 105 mi = 3000 hr 40 mph behind the wheel to travel 120,000 miles. If you average 30 mph, you will spend 4000 hours.

Per capita American income is $40,000 per year. Of course, we don’t spend all of our time working (it only 111

January 19, 2008

Time: 12:16pm chapter05.tex seems that way). In one year, we work t work = 40 hr × 50 weeks = 2000 hr week year
year

Since it takes 2000 hours to earn $40,000, we earn an average of $20 per hour.* At $20 per hour, it will take t earn = $60, 000 = 3000 hr $20/hr

Wow! We spend about the same amount of time earning the money to pay for our car as we do driving it. Whodathunkit?

Now we can calculate our average driving speed.

The total time we devote to driving (and we are certainly devoted to our cars) is 3000 hours driving plus 3000 hours earning, which equals 6000 hours. Thus, our average automotive speed is s = 120, 000 mil = 20 mph 6000 hr

And this does not even count the time spent looking for a parking space or waiting for the mechanic.

* Not counting taxes.

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Energy and Work

Chapter 6

*

January 23, 2008

Time: 11:50am
chapter06.tex

6.1 EnergyofHeight Gravity sucks! If you drop something, it will fall. The gravitational acceleration at the surface of the Earth is g = 10 m/s2.* This means that a falling object will increase its speed by 10 m/s (about 20 mph or 36 kph) every second. If you fell for 5 s, you would hit the ground at 50 m/s or 110 mph or 180 kph. Ouch!

g is also the gravitational force (measured in newtons (N)) exerted on an object of mass 1 kg (1 kg is about the mass of 2 lb) at the surface of the Earth.

Thus, a 1-kg block experiences a gravitational force (i.e., has a weight) of 10 N on Earth (less on the moon and more on the “surface” of Jupiter). In general, the gravitational force on an object can be expressed as F = mg , where m is the mass in kilograms.†

It takes work to lift an object against the force of Earth’s gravity. The energy needed to do this is called the potential energy

PE = mg h
where h is the height in meters. This makes sense. If you increase the mass of the lifted object, or the height you lift it to, or the gravitational pull, then you will need more energy to lift the object. The metric unit of energy is the joule, abbreviated J.

* To be precise, which this book is not, g varies from 9.78 to 9.83 m/s2 depending on latitude.

† This difference between mass and weight can be confusing. The object has the same mass everywhere, but the force needed to lift it depends on the planetary gravity. To increase the potential confusion, in US customary units, pounds can refer to either weight or mass.

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6.1.1

Mountain climbing

How much do you change your potential energy climbing a medium-sized mountain? How does this compare to the 6 × 105 J in a can of soda?

*
ft).

(30,000
m

10
is

Earth,
on
mountain
tallest
the
verest,

E

Mt.

HINT:

4
kilograms.

in
mass
approximate
your
get
to
two
by
pounds
in
weight
your

Divide

HINT:

115

January 23, 2008

Time: 11:50am
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ANSWER: Since potential energy is PE = mg h, we need to estimate our mass, the gravitational acceleration, and the height of the mountain. Assuming that we do this on Earth, g = 10 m/s2.

Humans (at least the ones likely to climb mountains) weigh between 100 and 200 lb. This means that we have a mass between 50 and 100 kg. As usual, we will choose 100 kg because it is a round number* and makes the arithmetic easier.

A medium-sized mountain is taller than a building and shorter than Mt. Everest. The tallest building has about 100 floors or stories with about 10 ft per floor and thus is about 1000 ft tall. Since 1 m ≈ 3 ft, this is about 300 m. Thus, a medium-sized mountain is somewhere between 3 × 102 and 1 × 104 m. We’ll take the average of the coefficients (the average of 3 and 1
is 2) and the average of the exponents (the average of 2 and 4 is 3) to get a height of h = 2 × 103 m (or 6000 ft). This is the height of the tallest mountain on the US east coast. If you chose a shorter or taller mountain, that’s fine too.

Now we can calculate the change in potential energy when you climb that mountain: PE = mg h

= 100 kg × 10 m/s2 × 2 × 103 m

= 2 × 106 J

Gee, two million joules! Is a joule big or small? Is that a lot?

To figure that out, we need to compare to other measures of energy. A 12-oz (330-mL) can of soda contains 6 × 105 J of food energy.† Thus, by climbing * And if I massed a 100 kg, I would be round too.

† If you live in the US, the soda’s energy content is given in Calories.

We’ll discuss those in the next chapter. If you live elsewhere, it’s given directly in joules.

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that mountain, we would gain potential energy equal to the food energy contained in N sodas =

2 × 106 J

= 3 sodas

6 × 105 J/soda

That’s not much! Climbing a 2-km (6000-ft) mountain takes a lot more work than that!

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118

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6.1.2

Flattening the Alps How much energy could we get from flattening the Rocky Mountains or the Alps (i.e., how much potential energy is stored in a mountain range)?

*

).

tons/m

10

(
iron
than
less
and

3

)

ton/m

(1
water
than
more
is
rock
of
density
average

The

HINT:

3
rock?

of
density
average
the
is

What

HINT:

it?

is
wide
how
and
range
mountain
the
is
long

How

HINT:

Rockies?

the
or

Alps
the
of
height
average
the
is

What

HINT:

119

January 23, 2008

Time: 11:50am
chapter06.tex

ANSWER: We need to estimate the average height of the Alps or the Rockies and their total mass. To get the mass we will need to estimate the volume and the density. To get the volume we will estimate the average length, width, and height. The highest mountain in either mountain range is about 20,000 ft (6000 m) so we will estimate that the average height including all the peaks and valleys is less than half of that or 2×103 m.*

The Alps stretch about 103 (1000) km from east to west (it is certainly more than 102 km and less than 104 km) and about 200 km from north to south.

(If you did this in miles, that’s fine. Since miles and kilometers are so close, 1 mi = 1.6 km, they are almost interchangeable in this book.) Now we need to convert these distances to meters. Fortunately, it is easier than converting miles to feet. Since 1 km = 103 m, 103 km = 106 m and 2×102 km = 2×105 m. Thus, the volume of the Alps will be V Alps = l × w × h = 106 m × 2 × 105 m × 2 × 103 m = 4 × 1014 m3

The Rockies are a lot longer since they run from southern US to northern Canada. This is about 3000
miles or 5 × 103 km. That would increase the volume by a factor of five from 4 × 1014 m3 to 2 × 1015 m3.

Now we estimate the density in order to convert volume to mass. Mountains are made of rock. Rock sinks and thus is denser than water (1 ton/m3). Rock is less dense than iron (10 ton/m3). Thus, we will use a * If you look at a relief map of the Alps, you will see that most of the Alps are actually only at about 1500 m. Thus, our crude estimate is not too bad.

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Time: 11:50am
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density of d = 3 tons/m3 or 3 × 103 kg/m3. This gives a total mass of M Alps = V × d = 4 × 1014 m3 × 3 × 103 kg/m3

= 1 × 1018 kg
and the Rockies will have five times as much. That’s a lot of mass.

Now we can calculate the potential energy. Note that the average height of the mountain range is 2 ×

103 m. However, the mass extends from sea level up to the average height. Thus, for calculating the potential energy, we need to use half of that average height.*

PEAlps = mg h = 1 × 1018 kg × 10 m/s2 × 1 × 103 m = 1 × 1022 J
and the Rockies will have five times as much or PERockies = 5 × 1022 J (about 1017 cans of soda). Now THAT is a LOT of energy! The energy to raise the mountains came from the motion of the continents (and originally from the heat generated in the core of the Earth).

Note that the height of the mountain range enters our equations twice, once explicitly as the height and once as part of the volume. This means that if we overestimate the height by a factor of two, we will overestimate the potential energy by a factor of four.

Fortunately, we are only trying to get within a factor of ten.

* It’s OK if you left out this factor of two. We are only trying to get within a factor of ten of the correct answer.

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6.1.3

Raising a building How much potential energy
does a 100-story building
have?

*
structure?

steel
is

volume
building’s
the
of
fraction

What

HINT:

. )

tons/m

10

≈

(density

3
steel
of
mostly
made
is
building

The

HINT:

building?

the
of
width
and
length
the
is

What

HINT:

story?

ach
e
is
tall

How

HINT:

building?

the
is
tall

How

HINT:

123

January 23, 2008

Time: 11:50am
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ANSWER: As in the previous problem, we need to estimate the height of the object and its mass.

Each story of a building is about 10 to 12 feet.

Ceilings are about 8 feet (i.e., there is about 8 feet of air between the floor and the ceiling) and there is space used for ductwork and cabling and for structure (holding up the building). This 10 to 12 feet is equal to 3 to 4 meters. Thus, the 100 story building is 100
stories × 3 m/story = 300 m tall. The average height of the material in the building will be half of that (since not all the material is located at the top of the building) or h = 150 m.

Now we need to estimate the mass of the building.

The building’s mass is in its vertical structure and its horizontal floors. The structural supports are all steel (since it is much stronger than the same weight of concrete). The structural supports will occupy much more than 1% and much less than 100% of the volume of the building. We will take the geometric mean and estimate that they occupy about 10% of the volume of the building (if it was much less then we could build much taller buildings and if it was much more then there would be little usable space in the building).

Therefore, we need to estimate the volume of the building. We already have the height. The area of a typical 100-story building will be somewhere between a football field (50 yards by 100 yards = 5 × 103 yd2 =

5 × 103 m2) and a private house (103 ft2 or 100 m2).

We’ll use 103 m2 for our building (this is 30 m by 30 m or 100 ft by 100 ft). Thus, the volume is V bldg = h × A = 3 × 102 m × 103 m2

= 3 × 105 m3

We’ll assume that 10% of this is steel with a density d = 10 tons/m3 = 104 kg/m3. Thus, the mass of the 124

Chapter 6 Energy and Work

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Time: 11:50am
chapter06.tex
building is
m = V steel × d = V bldg × 10% × d = 3 × 105 m3 × 0.1 × 104 kg/m3

= 3 × 108 kg

Note that the 110-story Sears Tower has a mass of M = 2 × 108 kg (so we are gol-durn close!).

Now the potential energy is easy: PE = mg h = 3 × 108 kg × 10 m/s2 × 1.5 × 102 m = 5 × 1012 J

That’s equivalent to the energy contained in 10 million cans of soda. Now that is a lot of energy. It helps explain why it is much easier to collapse a building than it is to build it.

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6.2 EnergyofMotion It takes work to make an object change its speed (e.g., from 0 to 60 mph) even ignoring pesky details such as friction and air resistance. The kinetic energy (energy of motion) of an object (in joules) is KE = 1 m v2

2
where m is the mass in kilograms and v is the velocity (or speed) in meters per second.* (It is easy to convert from meters per second to miles per hour since 1 m/s ≈ 2 mph.)

If you drop a water balloon from a building, it will start with a large amount of potential energy (PE), convert PE to KE as it falls, and then convert KE to a large splash as it hits.

* Technically, velocity includes both magnitude and direction, while speed includes just the magnitude. Since we will only be using them to calculate kinetic energy where the direction does not matter, we will use both terms interchangeably.

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6.2.1

At your service What is the kinetic energy in joules of a served tennis ball?

*

. h
p
m

2

≈
s /
m

1
d
n
a
b l 2

≈
kg

1
that

Remember

HINT:

. . .

mph,

100
mph,

0 1
mph,

1
have?

ball
tennis
the
does
speed

What

HINT:

hundred?

one
ten,

One,
kilogram)?

a

(or
pound
a
weigh
to
needed
are
balls
tennis
many

How

HINT:

ball?

tennis
a
of
weight
the
is

What

HINT:

127

January 23, 2008

Time: 11:50am
chapter06.tex

ANSWER: Since KE = 1 m v2, we need to estimate the 2
mass and the velocity. There are definitely more than one and less than 100 tennis balls in a pound so we will estimate that there are 10 tennis balls per pound.

This means that there are 20 tennis balls in a kilogram (since 1 kg ≈ 2 lb). Thus, the mass is m =

1 kg

= 5 × 10−2 kg/ball

20 balls

Now we need to estimate the velocity. We can place certain limits on the speed. Since the tennis ball does not make a sonic boom, we know it is slower than the speed of sound (300 m/s). The tennis ball travels faster than a car (30 to 60 mph or 15 to 30 m/s). Let’s see if we can get more information. We can compare it to baseball. A pitched fastball (in baseball) travels at 100 mph (50 m/s). A served tennis ball should go faster than that because the racket gives some mechanical advantage. Alternatively, we can use the size of the tennis court. The ball travels about 60 ft (20 m) from the service line to where it hits the court. It must take more than 0.2 s to do this (or we could never return a serve) and it certainly takes less than a second. This gives a range of speeds from 20 to 100 m/s. We’ll use 60 m/s (or 120 mph).

Now we can calculate the kinetic energy: KE = 1 m v2 = 0.5 × 5 × 10−2 kg × (60 m/s)2

2

= 100 J

100 J is the energy output of a 100-W light bulb for 1 s (since 1 watt = 1 joule/second).

Not much!

Now let’s compare to reality. The fastest tennis serve was 164 mph (73 m/s) so 60 m/s is quite reasonable for an excellent player. The mass of a tennis ball is 57 g, so our estimate of 50 g is also very close.

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6.2.2

Kinetic trucking What is the kinetic energy (in joules) of a large truck at highway speed?

*
ton.

1
about
of
mass
a
has
car
small

A

HINT:

mass?

its
is

What

HINT:

. s /
m

1

≈
mph

2
that

Remember

HINT:

speed?

its
is

What

HINT:

129

January 23, 2008

Time: 11:50am
chapter06.tex

ANSWER: We need to estimate the speed and mass of the truck. US highway speed limits are typically between 55 and 65 mph. We’ll use a speed of 60 mph ≈ 30 m/s. (Remember that we need to use metric units in our equations.) A small car weighs about 1 ton. A large truck will weigh a lot more than one car but less than 100 cars. This would give us a weight of 10 tons (the geometric mean of 1 and 100). Alternatively, we can estimate the mass from bridge weight-limit signs.

Bridge weight limits, if posted, tend to be about 10 or 20 tons. This implies that large trucks weigh more than that (since otherwise the warning signs would not be needed). This gives a weight of about 40 tons. We’ll use the average of 10 tons and 40 tons, or 20 tons. Since 1 ton = 103 kg, 20 tons = 20 × 103 kg = 2 × 104 kg. Now we calculate that PE = 1 m v2 = 0.5 × 2 × 104 kg × (30 m/s)2

2

= 1 × 107 J

This is 106 (one million) times less than the potential energy of a large building. Since a typical can of soda contains 105 J, this is the energy contained in 100 12-ounce (330-mL) cans of soda. We’ll investigate automotive energy sources (batteries and gasoline) more in the next chapter.

Note that a small car with a mass of about 1 ton will have a lot less kinetic energy. We can calculate it directly, or we can use the fact that the car has a mass that is twenty times less than the truck and therefore will have a kinetic energy that is twenty times less.

Thus, the car’s kinetic energy will be KEcar = 1 ×

107 J/20 = 5 × 105 J. This is the energy contained in five cans of soda.

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6.2.3

Racing continents What is the kinetic energy of a drifting continent?

(Ignore the rotation and other motion of the Earth.

We are only interested in the motion of the continent with respect to the rest of the Earth.)

*

).

kg/m

(10
iron
and

)

kg/m

(10

3

4

3

3
water
between
somewhere
is
rock
of
density

The

HINT:

s.

10

×
π

=
year

One

HINT:

7
point.

widest
its
at
wide
km

5000
is

Ocean

Atlantic

The

HINT:

Europe.

from
away
moving

America

North
by
formed
be
to

Ocean

Atlantic
the
for
years

10
took

It

HINT:

8
continent?

the
of
speed
the
is

What

HINT:

km.

50
about
only
is
continent
the
of
depth

The

HINT:

km.

10

×

4
is

Earth
the
of
circumference

The

HINT:

4
continent?

the
of

volume
the
is

What

HINT:

131

January 23, 2008

Time: 11:50am
chapter06.tex

ANSWER: As usual, we need to estimate the mass and the speed. Let’s start with the mass. We will estimate the volume (i.e., length, width, and height [or depth]) and the density using North America. We estimated the width of the contiguous US back in question 3.9
as 1/8 of the Earth’s circumference or 5 × 106 m.

Assuming that North America is square (insert your bad joke here; we’re temporarily out of stock), the length is also 5 × 106 m.

The depth of the crust is a little more difficult. It is certainly more than 1 km (many mines go deeper than that) and less than 103 km (since we know that the tectonic plates are a small fraction of the radius of the Earth). We’ll take the geometric mean of 1 and 103 km which is 30 km or 3 × 104 m.* Thus, the volume of the North American plate is V NA = l w h = 5 × 106 m × 5 × 106 m × 3 × 104 m = 8 × 1017 m3

Now we can multiply that by the density of rock to get the mass. The density must be more than that of water ( d = 1 × 103 kg/m3) and less than that of iron ( d = 10×103 kg/m3) so we will use d = 3×103 kg/m3: m NA = d × V = 3 × 103 kg/m3 × 8 × 1017 m3

= 2 × 1021 kg

Now we need the velocity. If you read about California earthquakes, you might know that the North American plate is moving at about 1 to 2 cm per year. Alternatively, we can use a longer time scale. The Atlantic Ocean has been widening for about 108 years as North America has been moving away from Europe.

At present, the maximum width of the Atlantic Ocean

* The actual thickness of the continental crust is between 20 and 80 km.

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is about 5000 km = 5 × 106 m. Thus, the velocity* of North America is
vNA = d = 5 × 106 m
t

108 yr

= 5 × 10−2 m/yr ×

1 yr
π × 107 s

= 2 × 10−9 m/s

This estimate of 5 cm per year is a few times larger than the actual speed. It is still not very fast.

The kinetic energy is now

KE = 1 m v2 = 0.5 × 2 × 1021 kg × (2 × 10−9 m/s)2

2

= 4 × 103 J

That is a lot more than the tennis ball but a lot less than the truck! It is much less than one can of soda.

To explain why continents are so hard to stop, we would need to estimate their momentum. Perhaps in the next book . . .

* We use V for volume and v for velocity. Scientists rapidly run out of variables and have to reuse letters.

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6.2.4

“To boldly go ...”

How much energy is needed to get a spaceship from Earth to Alpha Centauri (the nearest star, about 4 light-years away) before the passengers die of old age?

How many tons of fuel will this
take (assuming 4 × 109 J/ton
of TNT or rocket fuel)?

*
velocity?

that
at
have
spaceship
the
would
energy
kinetic
much

How

HINT:

passengers?

living
with
there
get
to
need
ship
the
would
speed
light
of
fraction
what
years,

4
in

Centauri

Alpha
to
journey
the
makes
light

If

HINT:

have?

to
need
spaceship
the
would
velocity

What

HINT:

s).

10

×
π
or

(
year
one
in
travels
light
distance

7
the
is
light-year

A
m/s.

10

×

3

=
c
is
light
of
speed

The

HINT:

8
tons.

10
is
carrier
aircraft

An

HINT:

5
want?

you
o
d
ship
a
large

How

HINT:

135

January 23, 2008

Time: 11:50am
chapter06.tex

ANSWER: The rocket fuel will provide the kinetic energy for the rocket ship. To figure out the kinetic energy, we need to estimate the mass and the velocity.

Thus, we first need to figure out how fast the spaceship will need to travel. The spaceship needs to travel the 4 light-years from Earth to Alpha Centauri in 40 years or less.* Since it takes us 40 years to travel the distance that light travels in four, our velocity is one-tenth of the speed of light. Thus, v = 1 c = 0.1 × 3 × 108 m/s

10

= 3 × 107 m/s

Now we need to estimate the mass of the rocketship.

A modern aircraft carrier displaces 105 tons. Colum-bus sailed to the New World in the Santa Maria, which weighed only about 100 tons.† It’s hard to imagine cramming the life-support systems needed for a 40-year voyage in a tiny ship. Let’s use a mass of 104 tons or 107 kg. Since the kinetic energy is directly proportional to the mass, it’s easy to try different masses. Anyway, let’s calculate the fuel we need.

The spaceship will have a kinetic energy of KE = 1 m v2 = 0.5 × 107 kg × (3 × 107 m/s)2

2

= 5 × 1021 J

That’s a lot of energy. We could get that much from flattening the Alps, but that would supply the energy for only one small vessel.

* Why 40 years? We could have chosen any number between 20 and 60 years. There are two reasons: (1) it worked for Moses and (2) it simplifies the arithmetic because it is exactly ten times four.

† But his voyage lasted only a few months and he did not have to bring all his oxygen.

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Now let’s look at the fuel needed. At 4 × 109 J/ton of fuel, our 104-ton spaceship will need fuel with a total mass m fuel =

5 × 1021 J

= 1 × 1012 ton

4 × 109 J/ton

Yikes! It does not matter what size spaceship you chose because the fuel will weigh 100 million (108) times more than the spaceship. Now we need to include the extra fuel needed to accelerate the fuel. We did not even take into account the energy needed to decelerate the spaceship at its destination or the inefficiency of rocket engines.

There is no way we can do this. We cannot fly a spaceship to the stars in a human lifetime using conventional chemical fuels.

Someone needs to start working on dilithium crys-tals or antimatter drives.

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6.3 Work

“Work” is not just a generic term. In physics it means transferring energy by pushing on something (applying a force) and making it move. When you kick a football (either the spherical or nonspherical kind), you are applying a force over a distance. When you shoot a bullet from a gun, the expanding gases in the gun barrel exert a force over a distance. When you apply the brakes in your car, the ground applies a force on your car over a distance. The energy transferred by this force is W = Fd
where F is the force in newtons (10 N is the weight of 1 kg and 104 N is the weight of 1 ton [on Earth]) and d is the distance over which the force is applied (in meters). Note that the force must be in the same (or opposite) direction as the distance traveled. If you push in the same direction that the object travels, you increase its energy and speed (e.g., kicking a football). If you push in the opposite direction, then you decrease the object’s energy and speed (e.g., apply the brakes).* Thus, work = change in kinetic energy

* If you push sideways (perpendicular to the object’s motion), then you do no work. You change the object’s direction but not its energy or speed (e.g., the Earth pulling on the Moon).

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6.3.1

Crash!

In the interests of researching this book, John offers to crash his car into a hard barrier at highway speed.

What force would be exerted on his body as he stopped? (Assume that he is wearing his seat belt and his air bag deploys properly.)

*
ft).

2
and

1

(between
m

0.5
about
crumples
car

The

HINT:

crash.

the
uring
d
crumples
car
his
of
nd
e
front
the
that
amount
the
is
involved
istance
d

The

HINT:

car.

his
as
rate
same
the
at
decelerates
he
that
so
seat
his
in
him
hold
bag
air
and
belt
seat

The

HINT:

speed.

and
mass
his

Estimate

HINT:

crash?

the
before
energy
kinetic

John’s
is

What

HINT:

crash.

the
uring
d
travels
he
distance
the
and
energy
kinetic
in
change
his
estimate
to
need
will

You

HINT:

139

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ANSWER: For John to change his velocity from highway speed to a complete stop, a force needs to be exerted on him. This is Newton’s first law: Your speed and direction of motion will not change unless an outside force acts on you. In this case, the force acting on John is exerted by the seat belt and air bag that compel him to slow down as the car slows down. We will estimate the force exerted on him by the seat belt and air bag. To do this, we will estimate his initial kinetic energy (since the work done by the seat belt must change that value to zero) and the distance over which the force acts.

We’ll assume that he is driving at 60 mph or 90 kph (30 m/s) and has a mass of 100 kg (220 lb).* In that case, his kinetic energy is KE = 1 m v2 = 0.5 × 100 kg × (30 m/s)2

2

= 5 × 104 J

Thus, the work done in stopping him must be 5×104 J.

Fortunately, the front end of his car is well designed and crumples. While the front bumper stops immediately, the passenger compartment slows and stops more gradually as the front end crumples. A typical front end will crumple less than 1 m (3 ft) and more than 0.1 m (10 cm or 4 in.). We’ll take the average and use 0.5 m.† This means that the force exerted is F crash = W = 5 × 104 J
d

0.5 m

= 105 N

* This mass is in the cause of science. He will lose the extra weight when the problem is finished.

† In this case, there is little difference between the geometric mean of 0.3 m and the average of 0.5 m.

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Let’s see how much that is. Since 10 N is the gravitational force exerted on a 1-kg object, 105 N is the gravitational force exerted on a 104-kg object. Thus, the force exerted on John during the collision is equivalent to one hundred 100-kg (200-lb) people standing on your chest. That is 10 tons! Yikes! Driving into a wall at 60 mph (even if you are wearing your seat belt) is not a good idea!

Car dashboards are padded and cars have crumple zones to increase the stopping distance in a crash in order to decrease the stopping forces involved. The reason that you should always wear your seat belt is that the seat belt holds you in your seat, so that you stop gradually as the car stops. If you are not wearing your seatbelt, your head will stop abruptly as it hits the windshield. Since the stopping distance will be much less, the forces involved will be much greater. This is generally not a good thing.

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6.3.2

Spider-Man and the subway car

In the movie Spider-Man 2, Spider-Man stops a runaway New York City six-car subway train by attaching his webs to nearby buildings and pulling really hard for 10 or 20 city blocks. How much force does he have to exert to stop the subway train? Give your answer in newtons and in tons (1 ton = 104 N).

How does this compare to the force that you can exert?

*
km.

1
about
is
distance
stopping

The

HINT:

KE.)

=

W
or
nergy
e
kinetic
in
change

=
work
that

(Remember
train?

the
stop
to
needed
is
work
much

How

HINT:

velocity?

and
mass
its
are

What

HINT:

speed?

normal
at
train
subway
a
of
nergy
e
kinetic
the
is

What

HINT:

143

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ANSWER: This is very similar to the previous question.

Since the work done by Spider-Man to stop the train is equal to the train’s initial kinetic energy, we need to estimate the mass and velocity of the train. We will then need to estimate the stopping distance in order to calculate the force exerted.

A subway car is about the same size and weight as a semi-trailer (18-wheeler) truck. This is between 10
and 40 tons. We’ll use 20 tons (or 2 × 104 kg). There are six cars on a train so that the mass of the train is 6 × 2 × 104 kg = 105 kg. They certainly go faster than 20 mph and slower than 100 mph. Since it is not that far between subway stops, subways travel at only about 40 mph (20 m/s). Thus, the kinetic energy of a subway train is KE = 1 m v2 = 0.5 × 105 kg × (20 m/s)2

2

= 2 × 107 J

Now we need to figure out the stopping distance.

There are 20 blocks per mile in Manhattan. Thus, 10 or 20 blocks is about 1 km or 103 m. (It’s certainly more than 100 m and less than 10 km.) Thus, Spider-Man needs to exert a force F = KE = 2 × 107 J
d

103 m

= 2 × 104 N

A force of 2 × 104 N is the weight of 2000 kg or 2 tons. For a superhero who can lift cars, this is quite possible (although definitely not easy). A human could definitely not do it.

Wow! Hollywood got the physics correct, in a superhero movie no less! Hurray!

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Hydrocarbons and Carbohydrates

Chapter 7

*

7.1 ChemicalEnergy

Unless we are capable of photosynthesis, we get most of our energy from chemical reactions: from eating food and from burning hydrocarbon fuels. In a typical chemical reaction, one electron is exchanged between two atoms. The energy of this exchange is about January 23, 2008

Time: 11:54am
chapter07.tex

1.5 electron volts or 1.5 eV.* If you want more precision than that, ask a chemist or look it up. To convert this to a useful number, we need to know two things: 1. The conversion from electron volts to joules: 1 eV ≈ 2 × 10−19 J

2. The number of molecules involved in the reaction

To determine the second, we need to introduce a little chemistry. We will be concerned primarily with hydrocarbons and therefore will limit ourselves to the reactions C + O → CO2 (carbon plus oxygen reacts to form carbon dioxide) and H + O → H2O (hydrogen plus oxygen reacts to form water). All the oxygen for these reactions comes from the atmosphere.

In this book, there are only three hydrocarbons: coal (pure carbon), natural gas (methane or CH4), and everything in between (including gasoline), which we will call CH2.† One mole of carbon has a mass of 12 g (the atomic weight of carbon) and contains N A =

6×1023 atoms. Burning that will result in N A chemical reactions. One mole of methane has a mass of 16 g (the atomic weight of the carbon plus four hydrogens) and contains N A = 6 × 1023 molecules. Burning that will result in 3 × N A reactions (one CO2 and two H2O).

* We know this because batteries convert chemical energy to electrical energy. Common batteries provide an electrical potential of 1.5 V. Therefore, each single electron flowing through the battery gains an energy of 1.5 electron volts and each coulomb of electricity (a coulomb is a LOT of electrons, 1 C = 6 × 1018 e) flowing through the battery gains an energy of 1.5 coulomb volts (or 1.5 joules).

† This is because many hydrocarbons consist of long carbon chains with two hydrogen atoms per carbon atom. We can think of these as being made up of repeated CH2 units. In practice, the ratio of hydrogen to carbon varies depending on the exact chemical involved. We are not chemists; this is not our problem.

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7.1.1

Energy in gasoline

How much chemical energy (in joules) can be released by burning 1 kg (about 1 L or 1/4 gal) of gasoline?

What is its energy density (in J/kg)?

*

2

2
molecule).

O

H
one
and

CO
one
make
we

2(since

CH
of
molecule
one
burning
from
eV

3
get
can

We

HINT:

2
g.

14
of
mass
a
has

CH
of
mole

One

HINT:

2
gasoline?

of
kg

1
in
are

CH
of
moles
many

How

HINT:

2.

CH
approximately
is
gasoline
that

Assume

HINT:

147

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ANSWER: Since we know that the energy we get for each chemical reaction is 1.5 eV, we need to estimate the number of chemical reactions that occur when we burn 1 kg of gasoline. To do this we need to estimate the chemical composition.

We will assume that the hydrogen to carbon ratio in gasoline is two (since it is more than zero [pure carbon] and less than four [pure methane]). Thus, we will assume that gasoline is made of CH2 molecules.*

The atomic masses of carbon and hydrogen are 12
and 1, respectively, so that the molecular mass of CH2
is 14. This means that one mole of CH2 has a mass of 14 g or 1.4 × 10−2 kg. Thus, 1 kg of gasoline contains N =

1 kg

= 70 moles

1.4 × 10−2 kg/mole

Each of these CH2 “molecules” will give us two reactions: the carbon atom will oxidize and form CO2
and the two hydrogen atoms will oxidize and form H2O. Thus, each CH2 molecule will provide 3 eV. The total energy released by burning (oxidizing) 1 kg of gasoline will be E = 70 moles × 6 × 1023 reactions ×

3 eV
kg
mole
reaction

×

1 J

= 2 × 107 J/kg

6 × 1018 eV

Thus, we estimate that 1 kg of gasoline will release 2 × 107 J when burned.

Looking this up on the web [15], we find that gasoline has an energy density of about 4.5 × 107 J/kg so we are only off by a factor of two. Not bad, considering the approximations we made.

* This ignores the fact that the carbon atoms are in long-chain molecules (octane has 8 carbon atoms in a chain) and thus ignores the energy needed to break the carbon–carbon bonds.

We’ll let the chemists worry about that.

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Note also that gasoline has a density of about 3/4
that of water. This is definitely close enough to one for this book. However, if you need to be precise,* you should use a volume energy density of 3 × 107 J/L.

Note that 1 kg of TNT contains only 4×106 J, which is only 10% of gasoline. However, the TNT can release that energy MUCH more rapidly.

* Which this book is not.

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7.1.2

Battery energy

How much chemical energy is stored
in a common D-battery?

*
ashlight?

fl
that
in
last
battery
a
will
long

How

HINT:

night-light.

4-W
a
with
or
bulb
light

100-W
a
with
flashlight

(non-LED)

D-battery-powered
a
of
output
light
the

Compare

HINT:

seconds.

in
lifetime
the
times
watts)

in

(
output
power
the
equals

J)

in

(
used
nergy
e

The

HINT:

151

January 23, 2008

Time: 11:54am
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ANSWER: To estimate the chemical energy contained in a battery, we need to estimate its power output and its lifetime. This is because power (measured in watts) is the energy used each second. Thus, a 100-W lightbulb consumes 100 J of electrical energy every second.

The easiest way to estimate the power output of a battery is to compare the light output of a flashlight powered by the battery with other lights of known power consumption. Since different types of light bulb (e.g.; incandescent, fluorescent, LED) have very different efficiencies (i.e., very different amounts of light produced for the same amount of energy consumed), we need to make sure we compare light from the same type of bulb. We’ll use the common incandescent bulbs for our comparison. (Those are the ones that emit light by heating the filament to thousands of degrees and hence get very hot.) A flashlight definitely gives off much less light than standard lamp bulbs (100-W, 60-W, or 25-W). It gives off about the same amount of light (within a factor of 10) as a 4-W night-light, although it is difficult to compare precisely because a flashlight is usually directional (i.e., it uses a mirror to focus the beam in one direction) and a night-light is usually omni-directional and partly shielded.* Thus, we will estimate that the flashlight consumes 4 W (i.e., 4 J/s).

Now we need to estimate the lifetime of the battery.

The flashlight can stay lit for more than an hour but less than a day (24 hr). We will take the geometric mean and estimate that the lifetime is 5 hours. Now * You also need to compare the two light sources under similar ambient light conditions (since the light sensitivity of the human eye changes by orders of magnitude from bright sunlight to darkness).

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we can calculate the chemical energy stored in the battery:

E D−battery = power × time

= 4 W × 5 hr × 3.6 × 103 s/hr

= 7 × 104 J

Thus, one D-battery contains 7 × 104 J of chemical energy.

Let’s compare our estimates to reality. A Duracell alkaline D-battery has a specified capacity of 15,000
mA-hr (milliamp-hours) or 15 A-hr. The power output equals the current (in amperes) times the voltage.

Thus, we have

E spec = 15 A-hr × 1.5 V × 3.6 × 103 s/hr

= 8.1 × 104 J

Now we need to ask, is this a lot of energy? It is about the same amount as the chemical energy contained in a can of soda. It is also about 1/1000 of the energy in a kilogram (or liter) of gasoline. Since 10−3 L = 1 mL = 1 cm3, it is the energy contained in a cubic centimeter (about the volume of the last joint your smallest finger) of gasoline. Not much.

We’ll explore this further in the next question.

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7.1.3

Battery energy density

What is the energy density of a D-battery (in joules per kilogram)? How does this compare to the energy density of gasoline?

*

1000?

00?

1

10?

1?

pound?

a
or
kilogram
a
in
there
are

D-batteries
many

How

HINT:

155

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ANSWER: To estimate the energy density, we need to estimate the energy and mass of one battery. We estimated the energy of one D-battery in the previous problem. Now we just need to estimate the mass. It is frequently easier to estimate how many objects it takes to have the weight of 1 kg rather than to estimate the weight of a single object. One D-battery clearly weighs less than a kilogram and 100 D-batteries clearly weigh more. We estimate that there are about 10 D-batteries per kilogram. (In reality, a four-pack of D-batteries has a shipping weight of one pound. This would imply exactly nine batteries in a kilogram.) Now we can calculate the energy density of alkaline D-batteries: energy density

= energy × batteries

D-cell
battery
kg

= 8 × 104 J × 10 D − cell

D − cell
kg

= 8 × 105 J/kg

Remember that gasoline has an energy density of 4 × 107 J/kg. This is a factor of fifty (50) worse. Wow!

In fact, reusable batteries are much worse than that. Reusable batteries are designed for, among other things, energy density and cycle life (how many times you can recharge it) [16]. The energy densities of rechargeable batteries are several times worse than the nonrechargeable ones. They range from 1 × 105 J/kg for lead-acid batteries to 6 × 105 J/kg for the highest energy-density lithium-ion batteries. The batteries can be recharged only 200 to 500 times (depending on type of battery).

Note that the cycle life can also be a serious limitation. If you refuel your car once per week, that is 156

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50 times per year. This means that a 200-cycle battery would last only four years as the energy storage for a car. If you are buying an electric car, you do not want to have to replace the large, heavy, expensive battery every four years.

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7.1.4

Batteries vs. gas tanks

How many tons of batteries will you
need to contain the same amount of
energy as the gasoline in your gas
tank? How much does a full tank
of gasoline weigh?

*
contain?

tank
gas
your
oes
d
nergy
e
much
how

L,

/ J

10

×

3

At

HINT:

7

.)

L

4

=
gal

1
that

(Remember
empty?

almost
is
tank
the
when
car
your
efuel
r
to
take
it
oes
d
gas
of
gallons
many

How

HINT:

159

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ANSWER: Automobile (passenger car and SUV) gas tanks range from 10 to 30 gallons depending on the size and mileage of the car. A compact car takes about 10 gallons, a minivan takes about 20, and a Hummer H2 takes 32. Let’s use 20 gallons as a reasonable average. Twenty gallons of gas contains energy E = 20 gal × 4 L/gal × 3 × 107 J/L

= 2 × 109 J
or 2 GJ. Twenty gallons is about 80 L and therefore has a mass of about 80 kg or 160 lb.

Now let’s see how large a rechargeable battery we need to store that much energy. Let’s use the best lithium-batteries available in 2006, with an energy density of 6 × 105 J/kg. We will need a battery with mass M =

2 × 109 J

= 3 × 103 kg

6 × 105 J/kg
or three tons.

Now, this is not quite fair, even by the imprecise standards of this book. When we convert chemical energy to mechanical energy by burning fuel in an engine or a generator, the efficiency is only about one-third. About two-thirds of the chemical energy is lost to heat. Since we have already generated the electrical energy (typically by burning fossil fuel), we need only one-third as much electrical energy as gasoline energy.

Therefore, we need only(!) one ton of batteries.

Note that this still does not give you a heating system in your car. Car heaters use some of the two-thirds of the fuel’s energy that is converted to heat in order to warm the passenger compartment. Battery-powered electric cars will need to have special air conditioners (called heat pumps) that can be run backward to heat the passenger compartment (thereby cooling the outside).

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Since car weights range from one ton (compact car) to two tons (minivan) to three tons (Hummer H2), adding one ton of batteries is significant.

This battery weight is not surprising. Battery technology currently limits many consumer products, from laptop computers to cell phones to cordless power tools. We need to increase the energy density of batteries (with no degradation of lifetime or safety or environmental effects) by a factor of at least five.*

Well, what are you waiting for? Get to work!

* We’ll discuss energy transfer speeds later in this chapter.

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7.2 FoodIsEnergy

We get energy (perhaps too much energy) from the food we eat.* The amount of energy in food is listed on the nutrition statement on the back of the package.

Europeans find this conveniently measured in joules; Americans need to convert from calories to joules.

One food Calorie (C) equals 103 physics calories (c) equals 4 × 103 joules (J).†

* This energy comes from chemical reactions of the food molecules with oxygen in our body (oxidation). Rapid oxidation is called burning. The energy content of food is measured by burning the food and measuring the energy released.

† Calories were originally used to measure heat and joules were originally used to measure kinetic and potential energy. It was only later that we learned that they are just different forms of energy. A physics calorie is the amount of energy needed to raise the temperature of 1 g of water 1 ◦C.

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7.2.1

Eat here, get gas

How much energy does a typical well-fed human consume in one year (in joules)?

How much energy does a
typical well-fed car
consume in one
year (in joules)?

*
tank?

gas
the
efill
r
you
do
often

How

HINT:

year?

a
in
use
car
your
does
gas
much

How

HINT:

. L

4

≈ l a g 1
nergy.

e
chemical
of

J

10

×

3
ontains
c
gasoline
of
liter

One

HINT:

7
ay.

d
per

Calories

500

2
and

2000
between
on
based
are

US
the
in
guidelines

Nutrition

HINT:

day?

every
consume
you
do

Calories
food
many

How

HINT:

163

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ANSWER: We consume between 2000 and 3000 food Calories per day. This means that in one year we consume E = energy per day × days per year

= 2.5×103 C/day × 4×103 J/C × 4×102 days/year

= 4 × 109 J/year

This is an average. Athletes consume more, couch potatoes should consume less.

Now let’s consider the hungry automobile. The energy used equals the gasoline used times the energy density of gasoline. From a previous question, we know the energy density of gasoline is 3 × 107 J/L. So how much gas do we use?

In a previous chapter we estimated that the average American drives 104 miles each year and gets about 20 miles per gallon. Using these numbers, we can calculate the average car’s yearly gasoline consumption V = 104 mi/yr × 4L/gal 20 mi/gal

= 500 gal/yr × 4 L/gal

= 2 × 103 L/yr

Thus, the average car uses 500 gallons or 2000 liters per year.

Alternatively, you can estimate your own gasoline consumption by multiplying how often you refuel your car with how many gallons you put in each time.

For example, if you refuel your car every week and put in 12 gallons each time, your car’s yearly gasoline consumption is V = 12 gal/refuel × 1 refuel/week × 50 weeks/yr

= 600 gal/yr × 4 L/gal = 2.4 × 103 L/yr 164

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Gasoline has an energy density of 3 × 107 J/L. Thus, in one year the average car uses energy E = 2 × 103 L/yr × 3 × 107 J/L

= 6 × 1010 J/yr

That is 60 billion joules. Rather a lot!

Thus, your car consumes about 15 times more energy than you do. This is not surprising, since most humans cannot push one ton along the road at 30 mph.

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7.2.2

Farmland for ethanol

How much more farmland would America need to farm to grow corn for enough ethanol to completely replace the gasoline used in all of our cars? Give your answer as a multiple of the amount of farmland we use today. Note that this question assumes, first, that cars can run on 100% ethanol fuel, second, that we use only the human-edible parts of the plant for
ethanol, and, third,
that we get a lot
more energy from
burning the ethanol
than it takes to
produce it.

*
person?

a
than
consume
car
a
does
energy
more
much

How
driver?

its
by
consumed
energy
the
to
car
a
by
consumed
energy
the
of
ratio
the
is

What

HINT:

167

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ANSWER: All the energy consumed (i.e., eaten) by humans comes from growing crops or from animals fed by those crops. Almost all of the energy consumed (i.e., burned) by cars comes from fossil fuels. If we use crops to fuel cars, we will need more farmland.

We can assume that the amount of farmland we need is proportional to the amount of energy we need from crops. The typical American or European consumes 4 × 109 J/yr (4 GJ/yr) of energy. The typical American car consumes 6 × 1010 J/yr (60 GJ/yr), or about 15 times as much.

However, we need to include the effects of eating meat. Much of the American and European diet comes from meat, not from vegetables. The conversion efficiency (the ratio of animal feed to animal weight) varies from two for chickens and fish through four for pork to seven for beef [17]. Thus, on average, every calorie that comes from meat represents four calories that come from grain and other animal feed.

If half of our food calories come from meat, then we really consume 2 GJ/yr of plants directly and another 4×2 = 8 GJ/yr of plants indirectly. Thus, we consume 10 GJ/yr of plant energy.

There is about one car for every two Americans, so the energy consumed by one American plus her share of the car is E = 10 GJ + 1 60 GJ

2

= 40 GJ

Thus, instead of depending on farmland to provide 10 GJ/yr for each American, we will now need farmland to provide 40 GJ/yr! Yikes! This is a huge increase.

To feed both cars and people from farmland, we would have to vastly increase the farmland under cultivation with a consequent loss of wilderness and animal habitat.

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It’s not even clear how much gasoline we would save by doing this. That depends on how much fossil fuel it takes to produce one gallon of ethanol from corn.

Experts are still wrangling about this. Some even claim that it takes more fossil fuel energy to produce the ethanol than the ethanol contains.

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7.3 Power!

In physics, “power” is just the rate at which we use energy. We measure power in watts (W). 1 watt =

1 joule/second. Your 100-W lightbulb uses 100 joules of energy every second. Thus, in one year, your 100-W
bulb could use as much as

E = 100 W × π × 107 s/yr = 3 × 109 J/yr Wow! Your parents were right when they told you to turn off the lights.

Rather than measuring energy in joules, power companies persist in using kilowatt-hours which is the energy consumed when you use 1 kW for 1 hour.

This is

1 kW-hr = 1000 W × 1 hr × 60 min × 60 s 1 hr

1 min

= 3.6 × 106 J

Thus, your 100-W light bulb uses about 103 kW-hr per year.

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7.3.1

Hot humans

What is the power (in heat) output
of a human (in W or J/s)?

*
day.

/ s

10
approximately
are

There

HINT:

5
heat.

as
off
given
is
nergy
e
that

All

HINT:

ay?

d
one
in
consume
we
o
d
nergy
e
much

How

HINT:

out.

energy

=
in

Energy

HINT:

171

January 23, 2008

Time: 11:54am
chapter07.tex

ANSWER: To answer this problem, we need to estimate our energy consumption over some reasonable period.

This is typically given per day. We consume about 2500
food calories per day. Since this is already in units of power (i.e., energy/time) we just need to convert the units: P = 2.5 × 103 Cal/day × 4 × 103 J × 1 day 1 Cal

105 s

= 107 J

105 s

= 100 J/s = 100 W

Alternatively, we could have used the answer to problem 7.2.1 where we estimated that we each consume 4 × 109 J/yr. Converting this to watts, we get P = 4 × 109 J/yr =
π ×

100 W

107 s/yr

Thus, humans have the same heat output as a 100-W light bulb. Fifteen of us are equivalent to a 1500-W space heater.

Architects and engineers must take this heat output into account when designing heating and cooling systems for theaters, airplanes, and other structures containing large numbers of people.

172

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7.3.2

Fill ’er up with gasoline

At what rate (in watts) is energy
transferred to your car’s gas tank
when you fill the tank?

*
energy/time.

=

Power

HINT:

gas.

the
pumping
spend
you
time
the
only

Include
tank?

the
fill
to
take
it
oes
d
time
much
how

About

HINT:

nergy.

e
chemical
of

J

10

×

3
contains
gasoline
of
liter

Each

HINT:

7
tank?

the
in
put
you
do
gas
much

How

HINT:

173

January 23, 2008

Time: 11:54am
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ANSWER: When you fill your car’s gas tank, you are transferring energy. Since power is energy divided by time, we need to estimate the energy transferred by the gasoline and the time it takes to fill the tank. To get the energy transferred, we first need the amount of gasoline transferred when you fill your car’s tank. We don’t know what car you drive, so we’ll use an average car.

A car’s gas tank can hold between 10 and 30 gallons.

Twenty gallons of gas contains

E = 20 gal × 4 L/gal × 3 × 107 J/L

= 2 × 109 J
of chemical energy.

Now we need to estimate the time it takes to transfer this energy. We are not counting the time it takes to drive to the gas station or the time it takes to pay for the gas, only the time it takes to pump the gas. It takes more than 1 minute and less than 10 minutes to pump the gas. We’ll choose 3 minutes as a reasonable average.

(If you need to be more precise, use a stopwatch and time it yourself.)

Thus, the power transfer at a gas station is P =

2 × 109 J

3 min × 60 s/min

= 2 × 109 J

2 × 102 s

= 107 W

= 10 MW

Wow! We transfer chemical energy to our car at a rate of 10 megawatts! That is pretty fast!

174

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7.3.3

Fill ’er up with electricity

At what rate can you transfer electrical energy to your electric car? Assume
that you are plugging the car
in overnight at home.

*
home?

your
in
have
you
do
breakers)

circuit
or
fuses
i.e.,

(
circuits
many

How

HINT:

deliver.

can
circuit
house

20-amp
typical
a
that
power
maximum
the
about
kW,

1.5
about
use
heaters

Space

HINT:

oven?

microwave
a
or
heater
space
a
by
used
power
the
is

What

HINT:

provide?

can
house
your
in
circuit
one
power
electrical
maximum
the
is

What

HINT:

uses?

house
your
power
electrical
maximum
the
is

What

HINT:

175

January 23, 2008

Time: 11:54am
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ANSWER: You drive your electric car home and plug it in. The energy transfer rate will be limited by the power that your electric circuit can transfer. If you have a special charging station built for the car, then the power will be limited to the power that your house can transfer. Let’s consider the power limitations of American houses.

We can find this out in a few ways. We could go around our house and plug in as many space heaters as possible without tripping our circuit breakers. That is too much work and too expensive. A 1.5-kW space heater uses about the maximum possible power for one electrical circuit. One circuit typically supplies several outlets and lights. You can count the number of independent electrical circuits by counting the number of circuit breakers in your electrical panel. Most houses have about ten or twenty. This means that your home has a limit of about P max = 20 × 1.5 kW = 30 kW

This will overestimate the limit since electrical wiring is not designed to have all circuits drawing their maximum current simultaneously.

Another way to do this is to look at the main circuit breaker that limits the total electrical current coming into our house. A typical main circuit breaker for a medium-sized house might limit the electrical current to about 125 amps. Since power = voltage × current, this means that the maximum power available is P = V × I = 110 V × 125 A = 104 W

This gives a limit of 10 kW.

Thus, by plugging in an electric car, we can recharge it at a rate of at most 10 kW, about one thousand (!) times slower than refueling it with gasoline. Note that this only works with a special dedicated circuit.

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Regular wall outlets would recharge the battery ten times slower still.

This is OK if we are always willing to recharge our electric cars overnight. This would be seriously inconvenient if we needed to refuel our electric car in the middle of a long trip.

Note that this is why automakers are working on developing a plug-in hydrid car. The batteries alone would give the car a 40-mile range, enough for most daily driving. The gas tank and gas engine would be available for the occasional longer trip (or faster acceleration or using the heater or air conditioner).

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178

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06:59pm
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The Earth, the Moon,
and Lots of Gerbils

Now it’s time to look at more cosmic questions, concerning meteors, moons, planets, stars, and gerbils.

Chapter 8

*

January 29, 2008

Time:

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Time:

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8.1

“And yet it moves” (e pur si muove) What is the orbital speed of the Earth around the Sun?

What is its kinetic energy?

*

).

m

/
kg

10

×

8 (
iron
than
less
and

3

3

)

m

/
kg

(10
water
than
more
is

Earth
the
of
density

The

HINT:

3

3
m.

10

×

6
is

Earth
the
of
adius
r

The

HINT:

6

Earth?

the
of
mass
the
is

What

HINT:

2

.

v m

=

KE

2

1
or
speed
the
of
square
the
times
mass
the
to
proportional
is
motion,
of
energy
the
energy,
kinetic
that

Remember

HINT:

m.

10

×

5 . 1
is
orbit

Earth’s
the
of
adius
r

The

HINT:

11
orbit?

Earth’s
the
of
circumference
the
is

What

HINT:

Sun?

the
around
orbit
e
n
o
e t e l p
m
o c
o t
h t r a

E
e
h t
e k a t t i s e
o
d
e
m
i t
h c u
m
w
o

H

HINT:

181

January 29, 2008

Time:

06:59pm
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ANSWER: To estimate the speed, we need the distance traveled and the time elapsed. The Earth takes one year to travel entirely around the Sun. The distance traveled is the circumference of the circle. The radius of that circle is the distance from the Earth to the Sun, or R =

1.5 × 1011 m. If you used R = 93 million miles, that’s fine too (93 million miles = 150 million kilometers =

1.5 × 1011 m). Thus, the Earth’s speed is v = distance traveled
time

= 2π × 1.5 × 1011 m = 2π × 1.5 × 1011 m 1 year
π × 107 s

= 3 × 104 m/s

Thus, the Earth has a speed of 60 thousand miles per hour (since 1 m/s ≈ 2 mph). Rather a lot.

To estimate the kinetic energy, we also need the mass of the Earth. We could Google that, but we would not learn anything from the exercise. Let’s estimate it. There are several ways to do this. If we know the formula relating the mass of the Earth and the gravitational acceleration at its surface, we could use that*. Instead we’ll estimate the mass from the Earth’s volume and its density. We know the radius of the Earth, R = 6 × 103 km or 6 × 106 m. If you remember the formula for the volume of a sphere, you can use V = (4/3)π R 3. If you don’t remember that formula, you can use the fact that a sphere is about half the size of a cube with sides equal to the diameter of the sphere ( V < (2 R)3 = 8 R 3). This gives a volume of V = 4 π R 3 = 4(6 × 106 m)3 = 1021 m3

3

* The physicists among us can use g = G M/ R 2, where the gravitational acceleration g = 10 m/s2, Newton’s constant G ≈ 7 × 10−11 N-m2/kg2, and R is the Earth’s radius.

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(where we made the approximation that π = 3, appropriate for this book and for miscellaneous state legislatures).

We know the Earth is denser than water (103 kg/m3) and less dense than iron (8 × 103 kg/m3).

We’ll use a density of 3 × 103 kg/m3. This gives a mass of

M Earth = dV = 3×103 kg/m3 ×1021 m3 = 3×1024 kg The actual mass of the Earth is twice as much, or 6 × 1024 kg. This indicates that the Earth’s density is about 6, or much closer to iron than we estimated.

Now we can calculate the kinetic energy of the Earth orbiting the Sun:

KEEarth = 1 m v2 = 1 × 6 × 1024 kg × (3×104 m/s)2

2

2

= 3 × 1033 J

Now that is a lot of energy!

183

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184

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Time:

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8.2

Duck!

What is the kinetic energy (in joules and in megatons of TNT [1 kg of

TNT contains 4 × 106 J of
chemical energy])

of a 1-km meteorite
when it hits the

Earth?

*
orbit.

its
in

Earth
the
as
speed
same
the
about
has
meteorite

The

HINT:

Earth.

the
is

So

Sun.

the
orbiting
is
meteorite

The

HINT:

meteorite?

the
of
speed
the
is

What

HINT:

meteorite.

the
of
density
the

Estimate

HINT:

meteorite?

the
of
mass
the
is

What

HINT:

185

January 29, 2008

Time:

06:59pm
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ANSWER: To estimate the kinetic energy of the meteorite, we need to estimate its mass and its speed. We will estimate the meteorite’s density and calculate its mass from its density and volume. Since the problem does not specify whether the meteorite is spherical or cubic or whether the size refers to the radius or the diameter, we will make assumptions that make life easier for us. In this case, we will assume that the meteorite is a cube. This gives a volume V =

(1 km)3 = 1 km3. A sphere of radius 1 km will have a volume four times larger and a sphere of diameter 1 km will have a volume half as much.

Now we need to estimate the density. Again the density is somewhere between those of water and iron.

We know that some meteorites are iron and some are rock. We will use the density of iron so the impact is more spectacular. (You can divide that by two if your meteorite is rocky.) This means that the mass of the meteorite is 3
m = V d = 1 km3 × 103 m

× 8 × 103 kg/m3

1 km

= 8 × 1012 kg
where we needed the factor of [(103 m)/(1 km)]3 =

109 m3/km3 to convert from km3 to m3. Be very careful with factors like that. If you put them in backward or upside-down, you will make a mistake of 1018. That is a huge error, even in this book.

We now estimate the speed of the meteorite. The speed of the Earth in its orbit is 3×104 m/s. This speed comes from the Earth’s orbit in the Sun’s gravitational field. The meteorite is also orbiting the Sun. Therefore, it will also have a speed reasonably close to the Earth’s (i.e., within a factor of two). The meteor could hit the Earth head on (so the relative speed would be 6 × 104 m/s) or it could bump the Earth gently from 186

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Time:

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behind (so the relative speed would be only 1×104 m/s or less). We’ll just use 3×104 m/s for the impact speed.

The kinetic energy of the meteorite as it hits the Earth is thus

K E meteorite = 1 m v2 = 1 × 8×1012 kg×(3×104 m/s)2

2

2

= 4 × 1021 J

This is much less than the kinetic energy of the Earth orbiting the Sun so the meteorite will not change our orbit. However, at 4 × 109 J/ton of TNT, this is the energy contained in 1012 tons or 106 megatons of TNT.

This is rather a lot.

According to [18], a 700-m-diameter meteorite will have a yield of 104 to 105 megatons and would destroy an area the size of a moderate state (e.g., Virginia).

A meteorite this size is expected to hit the Earth every 105 years.

The dinosaur killer meteorite is estimated to have been 10 km in size. Small wonder that it caused massive extinctions.

Don’t forget to duck.

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8.3

Super-sized Sun

What is the radius
of the Sun?

What is the Sun’s
average density?

Note that the mass
of the Sun is about
one million times
the mass of the Earth,
or 2 × 1030 kg.

*
m.

10

×

5 . 1
or

11
miles
million

93
is

Sun
the
to

Earth
the
from
istance
d

The

HINT:

distance.

its
to
width
finger’s
your
of
ratio
the
to
qual
e
is
istance
d
its
to
width

Sun’s
the
of
atio
r

The

HINT:

arm.

outstretched
your
on
finger
a
with

Sun
the
of
image
the
cover
can

You

HINT:

189

January 29, 2008

Time:

06:59pm
chapter08.tex

ANSWER: Here we use our knowledge of our own bodies and of the distance to the Sun to determine the radius of the Sun.

Sun
diameter

Finger

????

width

(1 cm)

Finger

Sun

Arm length (1 m)

11

Distance to Sun (1.5x10 m)

If we hold a finger out at arm’s length, it will cover the image of the Sun. This means that the finger and the Sun both subtend the same angle. This means that the size of the finger divided by the distance to the finger equals the size of the Sun divided by the distance to the Sun. See the figure (which is certainly not to scale). The width of my finger is about 1 cm = 10−2 m.

The length of my arm is about 1 m. The distance to the Sun is 1.5 × 1011 m. Thus, Sun size

=
finger size
distance to Sun
distance to finger

Sun size =
finger size

×distance to Sun
distance to finger

= 10−2 m × 1.5 × 1011 m

1 m

= 1.5 × 109 m

Note that this is the distance from one side of the Sun to the other, or its diameter. This is actually correct to within 10%! Humans really are the measure of the universe!

Now we can calculate the density of the Sun using M Sun = 2 × 1030 kg. Rather than looking up the mass of the Sun, you might remember from elementary 190

Chapter 8 The Earth, the Moon, and Lots of Gerbils January 29, 2008

Time:

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school that the mass of the Sun is about a million times greater than the mass of the Earth.* The density is the mass divided by the volume: d = M Sun = M Sun V

4 π R 3

3

= 2 × 1030 kg

4(7 × 108 m)3

= 2 × 1030 kg

1.4 × 1027 m3

= 1.4 × 103 kg/m3

Wow! This is only slightly more than the density of water. The low density of the outer gas envelope must compensate for the very high density of the center of the Sun.

* We could also calculate the mass of the Sun from the period of the Earth’s orbit and the gravitational acceleration of the Earth caused by the Sun, but that is beyond the scope of this book.

191

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192

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Time:

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8.4

Sun power

What is the power output of the Sun (in W)?

*
m.

10

×

5 . 1

=

R

11
adius
r
of
sphere
a
over
uniform
is
density
power

This

HINT:

second.

every
meter
square
each
through
passing

J

1400
is

That
orbit.

arth

E
at
nergy
e
solar
of
m

/

W

1400
about
is

There

HINT:

2

193

January 29, 2008

Time:

06:59pm
chapter08.tex

ANSWER: The solar power density at Earth orbit is 1400 W/m2, or 1400 J passing through each square meter every second. This means that all points that are the same distance from the Sun experience the same solar power density. Since we are in outer space, we do not have to worry about clouds or nighttime diminishing the solar energy. The distance from the Sun to the Earth is R = 1.5 × 108 km or 1.5 × 1011 m.

Thus, we need to find the total area of all points that are a distance R from the Sun. These points form the surface of a sphere centered on the Sun. The area of this sphere is A = 4π R 2 = 12 × (1.5 × 1011 m)2

= 2.5 × 1023 m2

The total power output of the Sun is thus the power density times the area:

P = solar power density × sphere area

= 1.4 × 103 W/m2 × 2.5 × 1023 m2

= 4 × 1026 W

The actual answer is 3.6 × 1026 W so our answers agree when rounded to one digit.

What relevance does this have to our lives? The total solar energy output determines when we will face the ultimate energy crisis. Forget about high oil prices, forget about running out of oil, forget about oil shale and tar sands. Those are only minor problems. The ultimate limit on human energy consumption will be the total energy output of the Sun.

In one year, the Sun emits energy

E = 4 × 1026 W × π × 107 s/yr = 1034 J/yr In 2003 humans used approximately 4 × 1020 J of energy [19]. This means that we will ultimately be able 194

Chapter 8 The Earth, the Moon, and Lots of Gerbils January 29, 2008

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to increase our energy output by a factor of 1034/4 ×

1020 = 2 × 1013 (20 trillion). This sounds like a lot.

Heck, it is a lot.

The US Department of Energy projects that energy use will continue to increase by about 2% per year for the next 25 years. That does not seem like much. Let’s see how long it will take us to reach that if our energy use keeps increasing by a mere 2% per year. In one year, our energy usage will increase by a factor of 1.02, in two years by a factor of (1.02)2 = 1.04, in three years by (1.02)3 = 1.06, and in 100 years by (1.02)100 ≈ 7.

This is not getting us close. Let’s take a short cut. If 1.02 n = 2 × 1013, then n = log(2 × 1013)/ log(1.02) =

1550.* Thus, 1550 years from now (in the year 3550), energy use will have increased by a factor of F = (1.02)1550 = 2 × 1013
and we will be using every single watt the Sun puts out!

The real energy crisis is coming! Doomsday strikes in 3550!

What this exercise really shows is the absurdity of extrapolating exponential growth trends too far into the future.

* Sorry about using logs here. Just remember: A good logarithm enables lumberjacks to be natural musicians.

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8.5

Gerbils 1, Sun 0

If the Sun were made out of gerbils, then the Earth would be incinerated. Compare the power output per mass of the Sun and a small mammal.

*
human.)

a
than
insulation
thermal
less
has
it
because
greater
is
gerbil
a
for
ratio

(The
ten.

of
factor
a
about
by
only
iffer
d
human
a
for
and
gerbil
a
for
mass
to
output
power
of
ratio
the
warmblooded,
are
mammals
all

Since

HINT:

mass.

gerbil’s
a
to
output
heat)

i.e.,

(
power
gerbil’s
a
of
ratio
the

Compute

HINT:

mass.

Sun’s
the
to
output
power

Sun’s
the
of
ratio
the

Compute

HINT:

197

January 29, 2008

Time:

06:59pm
chapter08.tex

ANSWER: We can estimate this one of two ways. We can estimate the power output divided by the mass for the Sun and for gerbils or we can estimate the total power output from a mass of gerbils equal to that of the Sun. Let’s start with the ratios. The Sun is easy because we have already estimated its power output ( P Sun = 4 × 1026 W) and mass ( M Sun = 2 × 1030 kg).

Thus, the Sun’s power density is

PSun = P = 4 × 1026 W = 2 × 10−4 W/kg.

M

2 × 1030 kg

The power density of a gerbil is more difficult. We have already estimated the power output of a human.

Since both gerbils and humans are mammals, we can expect that the metabolic rates are somewhat similar.

However, since humans are much bigger, they have much more thermal insulation and therefore radiate comparatively less heat. The linear scale of a human is about 1 m (it is certainly more than 10 cm [4 in.] and less than 10 m [30 ft]). The linear scale of a gerbil is about 10 cm (it is more than 1 cm and less than 1 m).

Thus, humans have about ten times the linear size and thus about ten times thicker insulation. This means that gerbils emit about ten times more heat (per unit mass) than humans.

We will calculate the power density for a human and adjust it upward for a gerbil. A human has a mass of about 100 kg and a power output of about 100 W

(see section 7.3.1). This means that the human power density is about

Phuman = P = 100 W = 1 W/kg

M

100 kg

The power density of a gerbil will be even greater.

Thus, a gerbil’s power density will be at least 104
times greater than the Sun’s. This means that if the Sun were made of gerbils, it would emit 104 times more 198

Chapter 8 The Earth, the Moon, and Lots of Gerbils January 29, 2008

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power. This would increase the surface temperature of the Earth by a factor of ten,* from 300 K (25 ◦C or 70 ◦F) to 3000 K (3000 ◦C or 5000 ◦F). This would get mighty uncomfortable (although not as uncomfortable as all those poor gerbils).

It is quite unexpected that the Sun, which is powered by nuclear reactions, is less powerful (pound for pound or kilogram for kilogram) than gerbils, which are powered by chemical reactions.

This is, of course, a very silly problem. The Sun carries all of its fuel for billions of years; gerbils do not.

If we include all the food, water, and oxygen needed by the gerbils for ten billion years, then the Sun would be way ahead. A human eats about two pounds (1 kg) of food per day. This means that in one year we eat several times our body weight. In a billion years, we would eat several billion times our body weight. This more than makes up for the mere factor of 104.

* Since power is proportional to temperature to the fourth, P ∝ T 4.

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Time: 12:54pm
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8.6

Chemical Sun

If the Sun were powered only by
chemical reactions, for how
long could it continue to
burn at its current power
output?

*
kg.

/ J

10

×

4
about
contains

Gasoline

HINT:

7
three.

of
factor
a
by
result
the
changes
only

It
oxygen.

the

Ignore

HINT:

contain?

it
would
energy
chemical
much
how
gasoline,
of
made
were

Sun
the

If

HINT:

n

Su
kg.

10

×

2

=

M
is

Sun
the
of
mass

The

HINT:

30

W.

10

×

4
is

Sun
the
of
output
power

The

HINT:

26

201

January 29, 2008

Time:

06:59pm
chapter08.tex

ANSWER: The Sun can contain at most M = 2 ×

1030 kg of chemical fuel. We can estimate the maximum energy contained either by assuming that the Sun is made out of gasoline (which we have already calculated) and oxygen or by investigating the optimal fuel. Let’s start with gasoline because it is easier. The energy density of gasoline (CH2 in this book) is about 4×107 J/kg. If we ignore the oxygen for now, this gives a total energy content of the Sun E Sun = mass × energy density

= 2 × 1030 kg × 4 × 107 J/kg

= 8 × 1037 J

Since the Sun emits energy at a rate of P = 4×1026 J/s, this means that this amount of energy will last a time T Sun = energy content = 8 × 1037 J
power output

4 × 1026 J/s

= 2 × 1011 s × 1 year
π × 107 s

= 104 years

That is not a very long time. Human agriculture started about 104 years ago. The time is even shorter if we replace about 2/3 of the gasoline with the oxygen needed to burn the gasoline.

We can probably get more energy by oxidizing either pure hydrogen or some other fuel. However, since we cannot do a factor of ten better, it is not worth the effort to calculate it.

This fact bothered 19th century physicists tremendously. William Thomson, who was raised to the British peerage as Lord Kelvin for his great discoveries in thermodynamics, wrote about this in 1862 [20].

He found that the most energetic chemical reaction known could power the Sun for only 3000 years. He 202

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further calculated that gravitational potential energy, the energy generated by objects falling into the Sun, could have provided energy for at most 107 years. This calculation assumed that the Sun itself was assembled from smaller objects falling in.

This conclusion contradicted the then-recent geo-logical discoveries of the ages of rocks. Kelvin therefore concluded his article thus: As for the future, we may say, with equal certainty, that inhabitants of the earth can not continue to enjoy the light and heat essential to their life for many million years longer unless sources now unknown to us are prepared in the great storehouse of creation.

The answer to Kelvin’s conundrum is, of course, nuclear energy. We will discuss that in the next chapter.

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8.7

Nearby supernova

If a star 30 light-years distant went supernova (and distributed most of its mass in all directions uniformly), how much of its mass (in kg) would hit the Earth? A light-year is the distance light travels in one year so that 1 l-y = 3 × 108 m/s × π × 107 s =

1016 m. If the planet Krypton was orbiting the star when it went supernova, how much of its mass (in kg) would hit the Earth?

*
kg.

10

×

6
is

Earth
the
of
mass

The

HINT:

24
question.

the
of

part
previous
the
in
planet
a
of
mass
the
with
star
the
of
mass
the

Replace

HINT: kg.

10

×

2
s i
n
u

S
e
h t f o
s s a
m
e
h

T

HINT: 30

Sun.

the
than
massive
more
times
ten
about
are
supernova
go
that

Stars

HINT: sphere.

30-light-year
that
of
area
the
with

Earth
the
of
area
the

Compare

HINT: light-years.

30
of
adius
r
a
have
would
sphere
that
system,
solar
our
passed
mass
the

As
sphere.

xpanding
e
an
of
surface
the
over
uniformly
itself
istribute
d
would
star
the
of
mass

The

HINT: 205

January 29, 2008

Time:

06:59pm
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ANSWER: Stars get their energy from nuclear fusion.

They start by fusing hydrogen into helium. When they run out of hydrogen, they fuse helium into heavier nuclei. Eventually, if the star is large enough ( M star > 8 M Sun), it will fuse silicon into iron. Once the core of the star becomes iron, fusion no longer produces energy (since iron is about the most tightly bound nucleus). The core then cools and collapses inward. As it collapses, it converts gravitational potential energy to kinetic energy (just like dropping a water balloon from a tall building, only MUCH more so). Some of the energy generated by this collapse creates a giant explosion that blows the outer part of the star rapidly and violently outward.* The remnant of the core then forms either a neutron star or a black hole (depending on the initial mass of the star).

As time passes, the mass of the ejected outer part of the star expands outward in a spherical shell. At some point, part of this shell will pass through our solar system. When it does, the radius of the shell will be 30
light-years. One light-year is the distance light travels in one year: 1 l-y = v t = 3 × 108 m/s × π × 107 s =

1016 m. Therefore, the surface area of the shell at that distance will be A shell = 4π R 2 = 12 × (30 l-y)2

= 12 × (30 × 1016 m)2

= 1036 m2

Since the mass of the ejected material is about ten times the mass of the Sun ( M Sun = 2×1030 kg), we can * You can try this yourself at home on a smaller scale. Place a tennis ball on top of a basketball and drop them together. (It takes a little coordination to drop them so that the tennis ball is still on top of the basketball when they hit the ground.) The basketball plays the part of the falling core. The tennis ball is ejected upward rapidly.

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estimate the density of the ejecta as it passes through the solar system: d = M = 2 × 1031 kg = 2 × 10−5 kg/m2

A 1036 m2

Now all we need is the area of the Earth in order to figure out how much mass hits the Earth. We could use the surface area (4π R 2) but none of the mass will hit the far side. To be geometrically correct, we should use the cross-sectional area (π R 2). Using the wrong area will change your answer by only a factor of four.

A Earth = π R 2 = π(6 × 106 m)2 = 1014 m2. Thus, the total mass of the supernova ejecta that hits the Earth would be m = area × density = 1014 m2 × 2 × 10−5 kg/m2

= 2 × 109 kg

That is a lot, but there is only 20 micrograms (a few cubic millimeters of dust) spread over each square meter of surface.

If we now consider how much mass from the planet Krypton would hit the Earth, we need to replace the mass of the star with the mass of the planet. The mass of the Earth is about 106 times less than the Sun so we will assume that the mass of Krypton is 106 times less than its star. This means that if 2 × 109 kg of star dust hits the Earth, then 10−6 × 2 × 109 kg = 2 × 103 kg of kryptonite reaches the Earth. That is two tons of kryptonite.

Watch out, Superman!

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8.8

Melting ice caps How much would the ocean surface rise if the ice caps melted?

*
ice.

of
kg

1
as

volume
same
the
about
has
water
of
kg

1

HINT:

thick.

km

2
about
is
cap
ice

The

HINT:

cap?

ice
the
is
thick

How

HINT:

latitude.

south

66◦
at
circle

Antarctic
the
within
entirely
lies

Antarctica

HINT:

???

. . .

00,

1

10,

1,
globe?

the
cover
to
needed
be
would

Antarctica
of
copies
many

How
cover?

Antarctica
does
area
surface

Earth’s
the
of
fraction

What

HINT:

it.

Ignore
projection.

Mercator
a
on
big
looks
only

Greenland

HINT:

water.

of
weight
its
isplacing
d
already
is
it
because
level
water
the
raise
not
oes
d
it
melts,
cap)

ice
polar
north
the

(e.g.,
ice
floating

When
level.

water
the
aise
r
they
melt,
land
on
caps
ice

When

HINT:

209

January 29, 2008

Time:

06:59pm
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ANSWER: To solve this problem, we need to estimate the volume of ice resting on the land and then figure out how high it would be if we melted the ice and spread it over the ocean. To do this, we can either estimate the volume of the ice cap and divide by the surface area of the Earth (since the Earth is 3/4 ocean), or we can estimate the height of the ice cap and scale that by the relative areas of the ice cap and the earth.

Since only the land-based ice cap matters, we will ignore the Arctic ice cap. Greenland is much smaller than Antarctica, so we will consider only Antarctica.

Let’s first estimate the average height of the ice cap. It must be much more than 100 m, since scientists have drilled several-kilometer ice cores that go back 700,000
years. It must be much less than 104 m (10 km) since that is the height of Mt. Everest. We’ll take the average height to be 103 m, the geometric mean of 102 and 104 m.

Now we need the area of Antarctica. Actually, we just need the relative area of Antarctica and the Earth.

If the area of Antarctica is one-tenth the area of the Earth, then the 103 m tall ice cap will raise the ocean level by 1 × 103 m = 100 m. If you look at a globe, 10
you can see that Antarctica is contained well within the Antarctic Circle at 66◦ south latitude. We’ll assume that Antarctica fills the entire circle beyond 70◦ S. Let’s see how many of these circles will go around the world from south to north and back again. Antarctica itself goes from 70◦ S, to the South Pole at 90◦ S, and then to 70◦ S, so that it spans 40◦ of latitude. The next copy will go from 70◦ S to 30◦ S. The third will go from 30◦ S
to 10◦ N. You get the idea.

There are 360◦ of latitude as you go around the globe from the South Pole to the North Pole and back again. Therefore, you can place 360◦

40◦

= 9 copies of

Antarctica on that north–south circle. You can place another 9 copies of Antarctica along an east–west circle 210

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at the equator. If the surface of the Earth was a square, you could place 9 × 9 = 80 copies of Antarctica on it. The actual number will be a bit less than that. We’ll estimate that the surface area of the Earth is 50 times greater than the surface area of Antarctica.

Alternatively, you could estimate the area of Antarctica. Assume that Antarctica is a circle. The radius of Antarctica is the distance from the South Pole to 70◦ S, which is 20◦

360◦ of the circumference of the Earth or R ant = 20◦

360◦ × 2π R Earth = 2 × 103 km. This gives an area A ant = π R 2 = 107 km2. If you compare this to ant the surface area of the Earth, you will also get a factor of 50.

Now we will take the ice covering each square meter of Antarctica, melt it, and spread it out so that it covers 50 square meters of water. If there are 103 m of ice covering Antarctica, then the extra water will raise the ocean height by h = 103 m = 20 m

50

This would be, shall we say, rather unfortunate for all coastal cities.

The actual expected sea-level rise would be about 80 m [21]. Note that the consequences of an 80-m rise are not that much worse than a 20-m rise. Either would be catastrophic.

To quote Bill Cosby, “How long can you tread water?”

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Energy and the Environment

One measure of the advance of civilization is the amount by which we use external sources of energy to multiply our own puny efforts. We have advanced from animal power, through water and wind power, to electrical power generated mostly by fossil fuel.

Chapter 9

*

January 29, 2008

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9.1

Power to the people

How much electrical power does one American or European family

(or household) use

(on average)?

*
kW-hr.

per

$0.10
paid
you
assume
and
bill
energy
monthly
typical
your
take

Alternatively,

HINT:

on.

is
one
ach
e
day
the
of
fraction
the
including
bulbs,
light
and
appliances
ifferent
d
by
used
power
the
up

Add

HINT:

hours.

ten
in
bulb
light

00-W

1
one
or
hour
one
in
bulbs
light

00-W

1
ten
by
used
energy
of
amount
the
is

(kW-hr)

kilowatt-hour

A
second.

every

J

100
uses
bulb
light

100-W

A
second.

per
joules)

(in
used
energy
of
amount
the
is
watts)

in

(

Power
nergy.

e
use
you
uickly
q
how
of
measure
a
is
power
that

Remember

HINT:

215

January 29, 2008

Time:

06:45pm
chapter09.tex

ANSWER: There are two ways to estimate the average power we use. We can work from the bottom up by adding up the contributions from individual appliances or we can work from the top down, using our electric bill to estimate the total electric energy we use in a month.

Let’s start from the bottom up. Consider the appliances that are on for a significant fraction of the day. Since we are writing this book in August, we will consider air conditioners, stoves and ovens, refrigerators, and light bulbs. We will assume that microwave ovens, water heaters, washing machines, etc. are not on enough to make a big difference.

There are probably five 100-W light bulbs on from 5:00 to 11:00 PM every day. The refrigerator certainly uses more power than a light bulb (100 W) and less than a space-heater (1500 W) so we’ll take the geometric mean of 400 W. You can listen to your refrigerator and hear it switch on and off. It is on about one-fourth of the time (that is the time that the compressor is running and it is actively cooling food). The stove and oven probably use more power than a space heater (3×103 W) for about an hour a day. The air conditioner is harder to estimate. A room air conditioner can be plugged into a room outlet and therefore uses about the same power as a space heater (1500 W). If you have six rooms, then your total air conditioning (AC) uses about 10 kW. Central air conditioning should be more efficient than individual room air conditioners and probably uses about half that power. Now we want the average AC usage from April to October.* At 2:00 PM

* We are assuming a location in about the middle of the US or Europe, where the heating season runs from November to March and the cooling season runs from April to October. We are also assuming that the heating and cooling costs are equal. Your results will vary depend on where you live, but they should be within a factor of ten.

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on the hottest day of the year, your AC is probably running constantly. At night in October, your AC is probably unused. During those six months, the AC certainly runs less than 100% and more than 1% of the time. We will take the geometric mean and estimate 10% (or 2.4
hours per day). Let’s see what we have so far: Peak power Time used Average

Item

(kW)

(hr/day)

power (kW)

Lights

0.5

6

0.1

Refrigerator

0.4

12

0.2

Stove

3.0

1

0.1

Central AC

5.0

2.4

0.5

Total

0.9

The average household uses an average 0.9 kW of electrical power.

Let’s attack the problem from the other end. A typical monthly electric power bill (it’s really an energy bill, since we pay for kilowatt-hours, not for kilowatts) is about $100. At an average price of $0.10 per kW-hr, that means we use P

= $100/(0.10 $/kW-hr)=

103 kW-hr each month. Since there are 24 hours/day ×

30 days/month = 700 hours per month, this means that our average power consumption is P = energy used (in kW-hr) time (in hr)

= 103 kW-hr = 1.4 kW

700 hr

The two methods give the same answer to within a factor of two!

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9.2

Continental power

How much electrical power does the US (or Europe) use? How much electrical energy in one year?

*
year.

one
in
s

10

×
π
re
a

There

HINT:

7
use?

industrial
and
commercial
for
add
to
need
you
do
extra
much

How

HINT:

US?

the
in
there
are
households)

or

(
families
many

How

HINT:

219

January 29, 2008

Time:

06:45pm
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ANSWER: There are two ways to estimate this. We can estimate either the power produced or the power consumed. To estimate the total electrical power consumed by the US or Europe,* we need to add up the total power used by residences, commerce, and industry. We just estimated that one American household uses 1 kW of electrical power. Therefore, we need to estimate the total number of households.

There are 3 × 108 Americans. With two or three Americans per household, there are about 108 households. The total residential power usage is then P res = 103 W/household×108 households = 1011 W

There is probably about as much commercial store and office space as there is residential space, so commercial and residential power use is probably about the same or 1011 W.

Industrial power use is probably between one and ten times as much as residential or commercial use, so we will take the geometric mean and estimate a factor of three or 3 × 1011 W.

Thus, the total electrical power used in the US is P elec = 1011 W+1011 W+3×1011 W = 5×1011 W

Now let’s try to estimate the total power produced.

We read somewhere that there are about 100 nuclear power plants and that they produce about 10% of the nation’s electricity. Each power plant produces about 1 GW (109 W). Thus, the total electrical power produced is about ten times the nuclear power produced or P = 10×number of nuclear plants×power per plant = 10 × 102 plants × 109 W/plant

= 1012 W

* The US and Europe have about the same population and standard of living and so consume about the same amount of electricity (at least at the level of precision of this book).

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Hmmmm. The two estimates differ by only a factor of two.

Now let’s calculate the total electrical energy used by the US in one year. Since energy equals power times time, we have E = P × t = 1012 W × π × 107 s/yr = 3 × 1019 J

That is certainly a lot of energy (although it will not get a rocketship to Alpha Centauri).

Now let’s compare it to reality. According to the CIA World Factbook [22], the US used 3.6 × 1012 kW-hr of electricity in 2003. We need to convert this from kW-hr to J.

E = 3.6 × 1012 kW-hr × 103 W × 60 s × 60 min 1 kW

1 min

1 hr

= 1.3 × 1019 J

We’re off by a factor of two.

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9.3

Solar energy

How much solar energy reaches the

Earth in a one year?

*
m.

10

×

6

≈

R
is

Earth
the
of
adius
r

The

HINT:

6
s/yr.

10

×
π
re
a

There

HINT:

7
area.

surface
the
not

Earth,
the
of
area
cross-sectional
the

Use

HINT:

.

W/m

1400
is
orbit

Earth
at
density
power
solar

The

HINT:

2

223

January 29, 2008

Time:

06:45pm
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ANSWER: Since we know that the solar power density is 1400 W/m2, we need to estimate the time and the area to get the energy received. We cannot just use the surface area of the Earth ( A = 4π R 2) for two reasons: (1) Half of it is dark (i.e., night) and (2) the light intensity decreases at higher latitudes due to the angle of incidence of the light. If we imagine a slice through the middle of the Earth, that circle will be perpendicular to the Sun and all the sunlight will hit it from directly overhead. We will use the area of this circle (the same one we used for the supernova debris).

Thus, the area we want is

A = π R 2 = π × (6 × 106m)2 = 1014m2

Now we can calculate the energy

E = 1.4 × 103 W/m2 × 1014 m2 × π × 107 s/yr = 4 × 1024 J/yr

This is more energy than we could gain by flattening the Rockies or the Alps.

As we discussed in the answer to question 8.4, humans used 4 × 1020 J of energy in 2003. Thus, we used a fraction f = 4 × 1020 J/yr = 10−4

4 × 1024 J/yr
of the available solar energy hitting the Earth. Of course, most of the energy we used came from ancient solar energy as stored in fossil fuels.

This means that we could expand our energy use by a factor of 100 and still only use 1% of the available solar energy. However, at a 2% annual increase in energy use, we will reach that point in a mere 230 years.

224

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9.4

Land for solar energy

How much land (in km2) would be needed to supply the US electrical energy needs with solar energy? What fraction
of the US land area
would be needed?

*

3.9.

uestion
q

See
before.

this
did

We

HINT:

states.

48
lower
the
only

Include

US?

the
of
area
the
is

What

HINT:

nergy?

e
lectrical
e
to
nergy
e
solar
converting
for
panels
solar
of
efficiency
the
is

What

HINT:

night.

and
clouds
include
to
forget

Don’t

HINT:

.

W/m

1400
is
orbit

Earth

2
at
atmosphere
the
above
density
power
solar

The

HINT:

225

January 29, 2008

Time:

06:45pm
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ANSWER: We will start with the total solar power density outside the atmosphere and then account for the effects of the atmosphere, clouds, night, and conversion efficiency.

The solar constant is 1.4 × 103 W/m2. About half of that reaches the ground (at noon on a clear day).

Including the effects of clouds and night decreases the total flux by about a factor of ten to 140 W/m2. The efficiency of solar panels is about 10%, so they will generate an average of about 14 W/m2 of electrical power.

From problem 9.2 we know that the US uses an average of 5 × 1011 W. Thus, we would need an area A solar cell =
electrical power used
electrical power per area

=

5×1011 W

1.4×101 W/m2

= 4 × 1010 m2

Now (1 km)2 = (103 m)2 = 106 m2. Thus, the area needed is A = 4 × 104 km2. This is a square that is 200 km (150 mi) on a side.

This seems like a lot of land, but the US is a very big country. Let’s compare this area to the area of the US. We already estimated the area of the contiguous US back in question 3.9 as A = 107 km2.

Therefore, the fraction of US land area needed if we used solar energy cells (photovoltaics) to provide all of our electrical energy would be f = solar cell area = 4 × 104 km2 = 4 × 10−3

US land area

107 km2

We would need to use 0.4% of all US land area for solar arrays.

This would be amazingly expensive for three reasons. First, solar cells are very expensive. They cost 226

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about $10 per watt installed (in 2006), so 5 × 1011 W
would cost 5 trillion dollars. Second, the best sites for solar power (i.e., deserts) are frequently far from population centers and would require expensive long-distance power transmission. Third, we would need an amazing number of batteries to charge during the day and provide power at night.

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9.5

Tilting at windmills

How much electric power can a wind turbine (a modern windmill) generate? (Consider the kinetic energy of the air passing through the area swept by the turbine
blades.)

*
energy?

electrical
to
convert
turbine
wind
the
will
energy
wind
the
of
fraction

What

HINT:

have?

air
that
does
energy
kinetic
much

How

HINT:

.

kg/m

1
about
of
density
a
has

Air

HINT:

3

. s /
m

1

≈
h
p
m

2
that

Remember
speed.

wind
sustained
easonable
r
a

Pick

HINT:

blades?

the
by
swept
area
the
through
passes
kg)

(in
air
much
how
second,
one

In

HINT:

building.

ten-story
a
as
tall
as
are
turbines
wind

Modern

HINT:

229

January 29, 2008

Time:

06:45pm
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ANSWER: Wind turbines convert wind kinetic energy into electrical energy. Therefore, to estimate the power that a wind turbine can generate, we first need to estimate the available kinetic energy. To estimate that, we need the air mass and the wind speed. The air mass passing through the area swept by the blades will be determined by the area swept by the blades and by the wind speed. We will first estimate the wind speed.

Wind turbines are placed in windy locations. Typical sustained wind speeds will be between 20 and 30 mph or between 10 and 15 m/s. (Less than that would not be considered windy, more than that is rather unlikely.) We’ll use 10 m/s because it is a nice round number. We’ll see how the final answer depends on wind speed later.

Modern wind turbines are as tall as a ten-story building. (Some are even taller.) At about 4 m or 12 ft per story, the wind turbine would be 40 m tall. This means that the length of the blades is 40 m and the area of the circle swept by the blades is A = π r 2 =

3 × (40 m)2 = 5 × 103 m2.

At a wind speed of 10 m/s, air travels 10 m in one second. Thus, in one second, a volume of air V =

10 m×5×103 m2 = 5×104 m3 passes through the area swept by the turbine blades. At a density of 1 kg/m3, this air has a mass m = 5 × 104 kg. Thus, 50 tons of air passes through the blade area every second!

Now we can calculate the kinetic energy of one second’s amount of air: KEair = 1 m v2 = 1 × 5 × 104 kg × (10 m/s)2

2

2

= 3 × 106 J

Thus, the power of the wind passing through the turbine blade area is 3 × 106 W or 3 MW. That’s a lot of available power.

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Now we need to estimate the efficiency of the wind turbine in extracting the kinetic energy from the air and converting it to electrical energy. That efficiency has to be more than 1% (since that is a very low efficiency) and less than 100% (since that would bring the wind to a complete stop). We will take the geometric mean and estimate an efficiency of 10%. This means that the available power from a wind turbine at a constant wind speed of 20 mph (10 m/s) would be 10%
of 3 × 106 W, which equals 3 × 105 W or 300 kW.

Now let’s consider the effects of wind speed. If we double the wind speed then we double the speed in the kinetic energy equation, which quadruples the kinetic energy (since the speed is squared). However, it also doubles the mass passing through the turbine blade area, so doubling the wind speed increases the available kinetic energy by a factor of 23 = 8. Thus, our 40-m turbine would produce 300 kW at a wind speed of 10 m/s and eight times more, or 2.4 MW, at a wind speed of 20 m/s (40 mph).

Now its time for the dreaded reality check. According to the Danish Wind Industry Association [23], a modern 92-m blade diameter wind turbine with a maximum tolerable wind speed of 25 m/s would have a nominal output of 2.75 MW and an average power of almost 1 MW. Since our hypothetical turbine would produce 2.4 MW at 20 m/s, our estimate is rather good.

Unfortunately, wind energy is not reliable enough to provide base electrical power. Yet another reason for better batteries.

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9.6

The power of coal

How much fuel does a 1 GW coal-fired electrical power plant require? Express your answer in kilograms per year and in 100-ton railroad cars per day.

Note that the plant uses 3 GW of thermal power to produce 1 GW of electrical power.

*

2
details.

more
for

7.1
section

See

.

CO
form
to

Carbon
of
atom
one
burning
from
eV

1.5
get
can

We

HINT:

g.

12
of
mass
a
has

Carbon
of
mole

One

HINT:

coal?

of
kg

1
in
are

Carbon
of
moles
many

How

HINT:

carbon.

pure
approximately
is

Coal

HINT:

year?

one
in
use
plant
power
the
does
energy
thermal
much

How

HINT:

233

January 29, 2008

Time:

06:45pm
chapter09.tex

ANSWER: To figure out how much coal a 1-GW electric power plant needs per year, we need to estimate the thermal energy needed in one year and the energy density of coal. Oil, gas, coal, and nuclear power plants use the heat from their fuel to boil water and then use the steam to turn a turbine and produce electricity. Only about one-third of the thermal energy is converted to electrical energy; the other two-thirds is emitted as heat. Thus, in one year, a 1-GW electrical power plant will need to burn enough fuel to produce E year = 1 GW × 3 × π × 107s/yr = 1017 J/yr This thermal energy comes from the chemical energy of the coal. Burning one atom of carbon will produce 1.5 eV of energy. One mole of carbon has a mass of 12 g = 1.2 × 10−2 kg. Thus there are 1/1.2 ×

10−2 = 80 moles of carbon in a kilogram. There are 6 × 1023 atoms in a mole. Thus, one kilogram of coal contains chemical energy equal to E coal = 1.5 eV × 6 × 1023 atoms atom
mole

×2 × 10−19 J × 80 moles
eV
kg

= 107 J/kg

Alternatively, we could have taken the results for gasoline (CH2) and divided by two since we get about as much energy from burning the H2 as from the C and the H2 has much less mass than the C. This would give an energy density of 1 × 4.5 × 107 J/kg =

2

2 × 107 J/kg.

When we compare with reality, we find that coal has an energy density of between 10 and 30 MJ/kg [15], so we are within a factor of three.

234

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Time:

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Now, we can calculate the coal needed to operate the power plant for a year: M coal =
energy needed

= 1017 J/yr
coal energy density

2 × 107 J/kg

= 5 × 109 kg/yr

This is 5 million tons of coal. That is a lot of coal.

Let’s try to get a feel for how much that is. We cannot really picture a ton of coal, let alone 5 million tons. One railroad car can carry about 100 tons of coal. One very long railroad train will have 100 cars.

Thus, to carry 5 × 106 tons of coal, we will need 5×104 railroad cars, organized into five hundred 100-car trains. This means that our power plant needs more than one 100-car coal train every single day.

Wow. That really is a lot of coal.

235

January 29, 2008

Time:

06:45pm
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236

January 29, 2008

Time:

06:45pm
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9.7

The power of nuclei

There are two general types of nuclear reactions, fission and fusion. We use fission reactions on Earth to produce nuclear energy and nuclear explosions. The Sun and stars use fusion reactions to produce solar energy. Fissioning (splitting) a heavy element such as uranium or plutonium produces about 200 MeV

(mega-electron volts) of energy. Fusing four hydrogen nuclei (protons) into one helium nucleus produces 28 MeV of energy. Nuclear reactions can convert between 0.1 and 1% of the mass of the atoms to energy.

How much fuel does a 1-GWe nuclear power plant require in one year?

*
energy.

of

MeV

00

2
about
produces
fission

Each

HINT:

inert.

relatively
s i
h c i h
w

U
is
est
r

The

U.

5%
about
only
is
fuel

The

HINT:

238

235
g/mole.

235
of
mass
a
has
which

U
is

Uranium
of
isotope
fissionable

The

HINT:

235
question).

previous
the
of
plant
coal-fired
the
like
just

(
power
electrical
of
e

GW

1
produce
to
power
thermal
of

GW

3
needs
plant
power
nuclear

The

HINT:

237

January 29, 2008

Time:

06:45pm
chapter09.tex

ANSWER: The nuclear power plant needs the same thermal energy as a coal plant to produce its 1 GW of electrical energy for a year. Thus, from the previous question, we know it needs E = 1017 J/yr.

This thermal energy comes from the fission of 235U. Fissioning one atom of 235U will produce about 200 MeV (2 × 108 eV) of energy. One mole of 235U has a mass of 235 g or about 0.24 kg. This means that there are 4 moles/kg. Thus, 235U has an energy density of E 235U = 2 × 108 eV × 6 × 1023 atoms atom 1 mole

×2 × 10−19 J × 4 moles = 8 × 1013 J/kg 1 eV
kg

That is 8 million times more energy density than coal.

However, we need to reduce this because only 5% of the fuel is 235U. The other 95% is the relatively inert 238U. This means that the effective energy density is only 5% of 8 × 1013 J/kg or E U = 4 × 1012 J/kg.

This means that in one year, a 1-GW nuclear power plant will need M U =
energy needed

=

1017 J/yr
uranium energy density

4 × 1012 J/kg

= 2 × 104kg/yr
or 20 tons of 5% enniched uranium. Since the density of uranium is about 2 × 104 kg/m3, this mass of uranium would fit in a cube that is 1 m (3 ft) on each side. Thus, the fuel for a nuclear power plant for an entire year would fit under your dining room table.*

* We do not suggest storing the fuel there for two reasons: (1) You would bang your shins on it repeatedly, and (2) if the fuel was concentrated into such a small volume with no control rods, it would go critical, the chain reaction would be out of control, and you would have a highly radioactive meltdown in your dining room. But at least your food would not get cold.

238

Chapter 9 Energy and the Environment

January 29, 2008

Time:

06:45pm
chapter09.tex

9.8

Hard surfaces

What fraction of the US land
area is impervious

(i.e., roofed or paved)?

*

9.4.

uestion
q

See

US?

the
of
area
the
is

What

HINT:

Americans.

10

×

3
re
a

There

HINT:

8
have?

building
ach
e
oes
d
oad
r
much

How

HINT:

ones?

residential
to
compare
building
commercial
of
areas
the
o
d

How

HINT:

home?

average
an
on
roof
the
of
area
the
is

What

HINT:

homes.

single-family
in
live

Americans

Most

HINT:

239

January 29, 2008

Time:

06:45pm
chapter09.tex

ANSWER: We will estimate the size of the typical American home, estimate the amount of road per home, and add some more space to account for commercial development.

A typical two-story home has about 2000 square feet of living space and thus 1000 square feet of roof area. Since 1 m2 = (3 ft)2 ≈ 10 ft2, the typical home has about 100 m2 of roof area.

Now we need to include roads. Our typical house is about 10 m wide by 10 m deep. The distance between the front doors of adjacent houses will certainly be greater than 10 m (i.e., no space between houses) and less than 100 m (i.e., a football field between houses).

We’ll take the geometric mean of 10 and 100 m and use a house spacing of 30 m (100 ft). Therefore, each house will have 30 m of road.

The width of this road is shared between houses on both sides of the street. A typical suburban road is more than 1 car (3 m) wide and less than 10 cars (30 m) wide so we will use a width of 10 m (30 ft). This means that the typical house will have a 30-m-long stretch of road that is 5 m wide with an area of 150 m2.

Thus, each house will have 100 m2 of roof and 150 m2
of road for a total of 250 m2.

We estimated earlier that the 3 × 108 Americans form about 108 households. Thus, residential roofs and roads will cover an area A = 108 × 2.5 × 102 m2 =

2.5 × 1010 m2.

Stores, offices, and factories will certainly occupy more than one-tenth and less than ten times the amount of residential space so we will estimate that commercial roofs and roads contribute another 2.5 × 1010 m2. Therefore, the total roofed and paved area in the US is A = 5 × 1010 m2 or 5 × 104 km2.

Alternatively, we can estimate this by considering (1) the population density of our cities and (2) the roof and road density of our cities. A typical US city 240

Chapter 9 Energy and the Environment January 29, 2008

Time:

06:45pm
chapter09.tex
has a density of 2000 people per square mile or 1000
people/km2. At this density, if everyone lived in cities, the cities would cover an area A = 3 × 108 people = 3 × 105 km2

103 people/km2

You can look at Google maps for any suburban region and estimate that this city is about 25% paved and roofed (more than 10% and less than 50%). Therefore, the total paved area is about 105 km2. This estimate agrees (within a factor of two) with the other method.

In reality, according to a 2004 study done by the US

National Atmospheric and Oceanographic Administration (NOAA) [24], 1.1 × 105 km2 are impervious.

This is slightly higher than either of our estimates.

We already estimated the land area of the US in question 9.4 to be 8 × 106 km2. Therefore, the fraction of the US that is paved or roofed is F = roofed and paved area US land area

= 8 × 104 km2 = 10−2

8 × 106 km2
or about 1%. This is about twice the amount of land that we estimated we would need to provide enough solar energy for all of our electrical needs. This puts that number in perspective. To provide enough solar energy for the US, we would have to have an area of solar panels equal to half the area of all the roofs and roads in the entire country. Yikes!

241

January 29, 2008

Time:

06:45pm
chapter09.tex

242

January 23, 2008

Time: 12:20pm
chapter10.tex

The Atmosphere

We rely on the atmosphere for airplanes, kites, wind-blown hair, and, oh yeah, breathing.

Let’s find whose breath you just breathed in, how much oxygen plants provide, and how much carbon dioxide cars emit.

Chapter 10

*

January 23, 2008

Time: 12:20pm
chapter10.tex

244

January 23, 2008

Time: 12:20pm
chapter10.tex

10.1

Into thin air

What is the mass of the atmosphere?

*
cube.

a
as

Earth
the
treat
just
can
you
sphere,
a
of
area
surface
the
forget
you

If

HINT:

m.

10

×

6
is

Earth
the
of
adius
r

The

HINT:

6

Earth?

the
of
area
surface
the
is

What

HINT:

kg.

10
of
weight
the
is

N

10

HINT:

4

5
surface.

Earth’s
the
of
meter
square
every
on

N

10
or
inch
square

5
every
on
force
of
lb

15
exerts
atmosphere
the
of
weight

The

HINT:

245

January 23, 2008

Time: 12:20pm
chapter10.tex

ANSWER: Atmospheric pressure is measured in too many different ways. It is 15 pounds per square inch or 105 newtons per square meter or (shudder) 760 mm (30 in.) of mercury or 10 m of water. These last two measures just indicate that the weight of the air over a certain area is the same as the weight of an amount of mercury sufficient to cover the same area to a depth of 760 mm (or enough water to cover the same area to a depth of 10 m).

We can use any of these measures to calculate the mass of the atmosphere. To do that, we just need to multiply the area of the Earth’s surface (in appropriate units) by the mass per square whatever. For example, if we use 15 pounds per square inch, we would need to calculate the area of the Earth in square inches (ick!).

Alternatively, if we use 760 mm of mercury, we would have to calculate the weight of enough mercury to cover the entire Earth’s surface (including the oceans) to a depth of 760 mm (or enough water to cover it to a depth of 10 m).

We will use the pressure of 105 N/m2 so we don’t have to calculate square inches. 105 N is the weight of 104 kg (since F gravity = mg , where g = 10 m/s2).

Now we need the surface area of the Earth. We happen to remember that the surface area of a sphere is A = 4π R 2, but not everyone does. You can also pretend that the Earth is a cube. In that case, the area of each of the six faces is (2 R)2 = 4 R 2 so that the total area is A = 24 R 2. This is close enough for this book and saves you from having to memorize pesky formulas. We’ll use the correct equation if you don’t mind: A Earth = 4π R 2 = 12 × (6 × 106 m)2 = 4 × 1014 m2

Thus, the mass of all the air is

M air = 104 kg/m2 × 4 × 1014 m2 = 4 × 1018 kg That is 4 billion billion kilograms or more than the mass of the Alps!

246

Chapter 10 The Atmosphere

January 23, 2008

Time: 12:20pm chapter10.tex 10.2

Ancient air How many molecules of Alexander the Great’s last breath do you inhale with each breath?

*
atmosphere?

the
in
there
are
breaths many

How

HINT: volume.

of

L

20
occupies and
molecules 10

×

6
has
gas
of
mole

One

HINT: 23
breath?

a
in
are
molecules many

How

HINT: breath?

a
of

volume
the
is

What

HINT: distributed.

randomly are
molecules the
so
years

2000
last
the
in
thoroughly mixed
has
atmosphere the
that

Assume

HINT: 247

January 23, 2008

Time: 12:20pm chapter10.tex ANSWER: To answer this question, we need to estimate the number of molecules in Alexander’s last breath and the fraction of the atmosphere that you just breathed in a few seconds ago. The second part amounts to estimating the number of breaths in the atmosphere.

The volume of a breath is about 1 L (1 quart). It is certainly more than a cup (1/4 L) and less than a gallon (4 L). We have no idea how many molecules are in a liter, but we do know the number of molecules in a mole. A mole contains 6 ×1023 molecules and takes up 20 L (as a gas at standard temperature and pressure).

Thus, a 1-L breath contains 1/20 of a mole and hence contains 3 × 1022 molecules.

Now we need the number of breaths in the atmosphere. We can estimate this using either the volume of a breath (in which case we need the volume of the atmosphere) or the mass of a breath (in which case we can use the mass of the atmosphere that we just estimated). Since we’re lazy, we’ll use the mass.*

The average density of air is about 1000 times less than that of water. Since 1 L of water has a mass of 1 kg, 1 L of air would have a mass of 1 g. You can also calculate the mass of air from its molecular weight.

Our 1-L breath is 1/20 of a mole. Air is composed of N2 (molecular weight = 2 × 14 = 28) and O2

(molecular weight = 2 × 16 = 32) molecules and so the average molecular weight is 30. One mole of air would have a mass of 30 g. Therefore, air has a density d = (30 g/mole)/(20 L/mole) = 1.5 g/L. (The difference between 1 g and 1.5 g is due to roundoff error and is negligible for this book.) * This also lets us avoid the question of how high the atmosphere is.

Unlike the oceans, the atmosphere thins out as you ascend but never really ends. The boundary of outer space is somewhere between 100 and 300 km (60 to 200 mi). The maximum height at which there is enough air for humans to breathe is a bit less than the height of Mt. Everest (10 km).

248

Chapter 10 The Atmosphere

January 23, 2008

Time: 12:20pm chapter10.tex The total mass of the atmosphere (from the previous question) is m air = 4 × 1018 kg = 4 × 1021 g. Since one breath has a mass of 1 g, with each breath you inhale one part in 4 × 1021 of the atmosphere. Thus, the number of Alexander the Great’s (AG) molecules that you just inhaled is N AG = AG s molecules in the atmosphere × fraction of the atmosphere inhaled = 3 × 1022 molecules ×

1

4 × 1021

= 8 molecules Thus, you just inhaled about eight molecules from Alexander the Great’s (and Confucius’ and Ramses’

and . . .) last breath (and second-to-last and . . .).

We hope you treated those molecules with reverence and care.*

* Alas, we know from quantum mechanics and thermodynamics that all oxygen molecules are alike and cannot be tagged like classical macroscopic particles [25]. Thus, this question is ultimately meaningless. The following analogy can be helpful. If you pay for a $20 purchase with a $20 bill and then six months later we give you $20 (for being such a loyal reader) then there is a chance that the same bill returned to you (since each has a unique serial number). If, however, you pay for your purchase with a credit card and then six months later we transfer $20 to your bank account, it is completely meaningless to ask whether it is the same bill. However, it is still fascinating to know that are about as many molecules in one breath as there are breaths in the atmosphere.

249

January 23, 2008

Time: 12:20pm chapter10.tex 250

January 23, 2008

Time: 12:20pm chapter10.tex 10.3

Suck it up How much time would it take human respiration to use up 10% of all atmospheric O2 (oxygen), ignoring all other contributions?

*
atmosphere?

entire
the
out)

(and
in
breathe
to
people
the
all
for
take
it
would
time
much

How

HINT: Earth.

the
on
people

10

×

6
re
a

There

HINT: 9
breathe?

we
do
frequently

How

HINT: breath?

one
of
mass
the
is

What

HINT: alive.

lse
e
someone
keep
to
xhale
e
you

2
air
the
in

O
nough
e
is

There
works.

espiration
r

Artificial

HINT: used?

is
oxygen
inhaled
the
of
much

How

HINT: 2

2

O.

H
or

CO
as
xhaled
e
is
oxygen
used
he

T
food.

our

2
oxidizing
by
energy
get
to

)

O

(
air
the
from
oxygen
use

We

HINT: 251

January 23, 2008

Time: 12:20pm chapter10.tex ANSWER: We need to estimate the amount of oxygen used up in each breath. From there, we can go on to estimate the number of breaths it will take to use up 10% of the oxygen in the atmosphere. We know that cardiopulmonary resuscitation (CPR) can keep someone alive until the ambulance arrives. This means that the air we force into someone’s lungs by breathing out contains enough oxygen to sustain life (even for someone sick enough to need CPR). This means that we use relatively little of the oxygen in each breath. We estimate that we use 10% of the oxygen that we inhale (a lot more than 1% and a lot less than 100%).

This means that we now need to estimate the amount of time it will take humans to breathe in (and out) the entire atmosphere once (since that will use up 10% of the oxygen). Note that we have completely avoided needing to know the abundance of oxygen in the atmosphere.*

In the previous question we estimated that each breath has a volume of about 1 L and a mass of about 1 g and that there are 4 × 1021 breaths in the atmosphere. Therefore, we just need to estimate how much time will be needed for humanity to take those 4 × 1021 breaths. Well, there are 6 × 109 of us. This means that the number of breaths you need to take is N breaths = 4 × 1021 breaths = 7 × 1011

6 × 109

We take about one breath every few seconds. Let’s estimate four seconds per breath. This means that it will take each of us a time t = 7 × 1011 breaths × 4 s/breath = 3 × 1012 s × 1 year =
π ×

105 years

107 s

That is 100 thousand years.

* If you’re curious, it’s about 20%. If you’re not curious, why are you reading this footnote?

252

Chapter 10 The Atmosphere

January 23, 2008

Time: 12:20pm chapter10.tex OK. Individually, we won’t survive long enough to accomplish this. However, all the other animals are helping us. Since human (or at least hominid) life on this planet dates back millions of years and animal life on this planet dates back hundreds of millions of years, it’s a good thing that plants are continually replenishing the oxygen supply.

253

January 23, 2008

Time: 12:20pm chapter10.tex 254

January 23, 2008

Time: 12:20pm chapter10.tex 10.4

CO2 from coal How much carbon dioxide (in kg) does a 1-GWe coal-fired power plant release into the atmosphere each year? What fraction of the mass of the atmosphere is this?

*
weight.

atomic
same
the
about
have
oxygen
and

Carbon

HINT: C?

of
kilogram

2
each
from
produced
are

CO
of
kilograms
many

How

HINT: 9.6.

uestion
q

See

HINT: 255

January 23, 2008

Time: 12:20pm chapter10.tex ANSWER: The amount of carbon dioxide released into the atmosphere will be directly proportional to the amount of carbon burned by the power plant. If all of the carbon is completely burned, then it will all be emitted from the smokestack in the form of carbon dioxide. Fortunately, we know how much coal the 1-GWe coal-fired power plant uses in one year. We estimated that in question 9.6 to be 5 × 109 kg. Each carbon atom is oxidized and becomes a molecule of carbon dioxide. Carbon and oxygen are relatively light elements and have about the same atomic weight.*

Therefore, each molecule of carbon dioxide (CO2) has about three times the mass of a carbon atom.

Therefore, the total mass of CO2 emitted into the atmosphere in one year by a 1-GWe coal-fired power plant is M CO = 3 × 5 × 109 kg C = 2 × 1010 kg 2

That seems to be a lot of CO2. Let’s compare it to the mass of the atmosphere. We calculated the mass of the atmosphere in question 10.1 to be M air = 4 × 1018 kg.

This plant will add f = 2 × 1010 kg 4 × 1018 kg = 5 × 10−9
or five parts per billion CO2 to the atmosphere every year. If you remember those golf balls circling the equator in the first chapter, that would amount to five CO2 golf balls out of the billion needed to circle the globe. While this does not seem like much, there are a lot of power plants.

A natural gas-fired power plant, fueled by methane (CH4), will produce about three times less carbon * If you insist on being precise, the atomic weight of carbon is 12
and the atomic weight of oxygen is 16.

256

Chapter 10 The Atmosphere

January 23, 2008

Time: 12:20pm
chapter10.tex
dioxide per GW than a coal-fired plant because it oxidizes the hydrogen and the carbon: the carbon will oxidize to carbon dioxide (CO2) and the four hydrogen atoms will oxidize to two water molecules (H2O). With three reactions, it will produce three times as much energy per carbon atom.

Note that carbon dioxide is a colorless gas. The “smoke” you see emerging from a smokestack will be some combination of partially combusted fuel, impu-rities (e.g., ash), and steam (condensing water vapor).

Note also that carbon dioxide is naturally present in the atmosphere at a level of a few hundred parts per million. Plants breathe in CO2 and breathe out oxygen.

Thus, carbon dioxide is not a standard pollutant (like smog) that harms people directly. Any damage it might do is indirect.

257

January 23, 2008

Time: 12:20pm
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258

January 23, 2008

Time: 12:20pm
chapter10.tex

10.5

A healthy glow How much high-level nuclear (i.e., highly radioactive) waste does a 1-GWe nuclear power plant produce in a year?

*
waste.

radioactive
highly
the
of
most
is
fuel
used

The

HINT: 9.7.

uestion
q

See

HINT: 259

January 23, 2008

Time: 12:20pm
chapter10.tex

ANSWER: The nuclear fuel in the reactor is composed primarily of two isotopes of uranium, 235U and 238U.

The 235U fissions into smaller nuclei, many of which are highly radioactive. The daughter nuclei and the 238U in the reactor are exposed to a tremendous amount of radiation, which also produces many highly radioactive by-products. Thus, all of the fuel becomes highly radioactive waste. That is most of the high-level nuclear waste produced by the power plant. Thus, the mass of the highly radioactive waste is between one and ten times the mass of the fuel. We’ll take the geometric mean and use a factor of three.

Fortunately, we know how much nuclear fuel the 1-GWe nuclear power plant uses in one year. We estimated that in question 9.7 to be 2 × 104 kg. Thus, a nuclear power plant produces about M nuclear waste = 3 × 2 × 104 kg = 6 × 104 kg or about 60 tons of highly radioactive nuclear waste per year.

This waste is extremely dangerous. On the other hand, since it so compact (and solid) it should be much easier to dispose of safely.

260

Chapter 10 The Atmosphere

January 23, 2008

Time: 12:20pm chapter10.tex 10.6

CO2 from cars How much carbon dioxide (in kg) does one car emit into the atmosphere each year?

All American cars?

*

Americans.

two
every
for
car
one
about
is

There

HINT: have?

ach
e
we
o
d
cars
many

How

HINT: Americans.

10

×

3
re
a

There

HINT: 8

2.

CO
make
to
oxygen
of
kilograms
more

3
almost
with
combines
gasoline
the
in
carbon
of
kilogram

Each

HINT: kg.

4
about
of
mass
a
has
and

L

4
contains
gas
of
gallon

Each

HINT: get?

it
does
mile
per
gallons
many
how
and
rive
d
car
each
does
far

How

HINT: year?

a
in
burn
car
one
oes
d
gasoline
much

How

HINT: 261

January 23, 2008

Time: 12:20pm chapter10.tex ANSWER: We already estimated that the average American car is driven about 104 miles per year and gets about 20 miles per gallon. This means that it consumes about (104 miles per year)/(20 mpg) = 500
gallons per year. At 4 liters per gallon, this is 2000 liters per year.

Now we need to convert from volume to mass. A liter of water has a mass of 1 kg. Gasoline is a bit lighter than water, but not enough for us to worry about.

Thus, your car burns about 2000 kg of gasoline per year. That is 2 tons of gasoline, more than the mass of most cars!

Gasoline is a hydrocarbon with about two hydrogen atoms for each carbon atom. Since carbon has an atomic mass of 12 and hydrogen has an atomic mass of 1, we can (and will) ignore the hydrogen. The two oxygens that combine with each carbon to make CO2
have a total atomic mass of 32. Thus, the mass of the CO2 (44) is almost four times the mass of the carbon (12). Thus, your car will emit 4 × 2000 kg = 8 × 103 kg of carbon dioxide every year.

However, there is more than one car in the US.

As we discussed in question 5.1, there are 3 × 108

Americans and about 0.5 car per person. Thus, all American cars emit a total of M CO = 8 × 103 kg/car-year × 1.5 × 108cars 2

= 1 × 1012 kg/yr That is 1 billion (109) tons.

Now let’s compare to reality. According to the US

Department of Energy [26], total CO2 emissions in the US in 2004 were 6 billion tons. Transportation, which includes cars, trucks, railroads, and airplanes, accounted for 2 billion tons of CO2. That is only twice as much as we estimated for cars alone.

262

Chapter 10 The Atmosphere

January 23, 2008

Time: 12:20pm
chapter10.tex

That seems to be a lot of CO2. Let’s again compare it to the mass of the atmosphere. Driving cars will add f = 1 × 1012 kg 4 × 1018 kg

= 2 × 10−7
or 200 parts per billion CO2 to the atmosphere every year. If you remember those golf balls circling the equator in the first chapter, that would amount to 200

CO2 golf balls out of the billion needed to circle the globe. This is 1000 times less than natural level of CO2 in the atmosphere of about 200 parts per million (200 ppm or 2 × 10−4).

If we want to reduce our share of transportation greenhouse gas emissions, we can either drive less or drive a car with better gas mileage. Driving less means either car pooling* or living closer to work (which restricts our house and neighborhood choices).

Driving a more efficient car means either paying more for a hybrid (but not all hybrids get good gas mileage) or driving a smaller car. If you drive an electric car, the electricity still has to be generated somehow, generally by burning fossil fuels. Alternatively, we can pass laws that make other people pay the costs of reducing CO2
emissions. The choice is ours.

* Which carries the risk of developing car pool tunnel syndrome.

263

January 23, 2008

Time: 12:20pm
chapter10.tex

264

January 23, 2008

Time: 12:20pm
chapter10.tex

10.7

Turning gas into trees How much carbon dioxide (in kg) does a 1-km2 new*
forest absorb per year? Trees absorb carbon dioxide, emit oxygen (thank you, trees!), and use the carbon to make more tree. Assume that the forest is growing in a temperate zone that gets plenty of rainfall (e.g., New Jersey†
or Germany).

* So we do not need to account for trees dying and rotting.

† New Jersey may be the “landfill of opportunity,” but we assure you that it has some lovely forests too.

*
floor?

forest
the
on
make
they
would
layer
a
thick
how
them,
compacted
and
years

0

2
after
trees
the
all
down
cut
you

If

HINT: years.

0

2
after
trees
the
all
of
mass
total
the

Estimate

HINT: carbon.

mostly
is
trees
the
of
mass

The

HINT: years.

0

2
over

Average

HINT: 265

January 23, 2008

Time: 12:20pm
chapter10.tex

ANSWER: This is a great problem, because there are four very different ways to solve it. We can estimate (1) the number and size of 20-year-old trees, (2) how thick a layer they would make if we cut them all down and compacted them, (3) the solar energy used by the forest and how much CO2 would be converted to carbon with that energy, or (4) the water used by the trees and how much cellulose would be made from the hydrogen in the water. Methods (1) and (2) will work much better if we average over 20 years.

The easiest method is to estimate how thick a layer the trees would make if after 20 years we cut the forest down and compacted it. A 20-year-old tree is less than a meter thick. Thus, the layer of biomass will certainly be more than 1 cm (10−2 m or 1/2 in.) and less than 1 m, so we will take the geometric mean and use a thickness of 0.1 m (4 in.). Thus, the volume of biomass will be V = (103 m)2 × 0.1 m = 105 m3. Since wood has about the same density as water (103 kg/m3), this gives a total mass of M = 103 kg/m3 ×105 m3 = 108 kg.

Since almost of all of this mass is carbon and since CO2 has about four times the mass of C, we estimate that 1 km2 of new forest removes 4 × 108 kg of CO2 in 20 years or 2 × 107 kg/year of carbon dioxide.

Now let’s try the solar energy method. The solar power available at Earth orbit is 103 W/m2. Between night and clouds and the atmosphere, about 10% of that energy is available to the trees. After a billion years of evolution, trees must be more than 1% efficient at using solar energy (but definitely less than 100%) so we’ll assume 10% efficiency.* Some of that energy probably goes to maintaining the tree’s metabolism so we’ll estimate that 50% of the available energy is used to make more tree. Thus, after accounting for night * This is the same efficiency as our best photovoltaic solar cells today.

266

Chapter 10 The Atmosphere

January 23, 2008

Time: 12:20pm
chapter10.tex
and clouds (10%), energy conversion efficiency (10%), and energy used for making more tree (50%), only 5 × 10−3 of the 103 W/m2 or 5 W/m2 is used to make more tree. This gives an available energy E = 5W/m2 × π × 107 s/yr × 106 m2/km2

× 6 × 1018 eV/J = 1033 eV/yr-km2

We converted this from joules to electron-volts because extracting the carbon from CO2 requires 1.5 eV

(our standard chemical reaction energy). Thus, we can extract the carbon from 6 × 1032 CO2 molecules per year. The mass of all this CO2 is M CO = 6 × 1032 molecules/yr × 44 g/mole CO

2

2

6 × 1023 molecules/mole = 4 × 1010 g/yr = 4 × 107 kg/yr This is a factor of about two more than our previous estimate. Not too bad for such different methods.

Now let’s compare to reality. According to the Encyclopedia of Energy [27], the average insolation ranges from 100 to 200 W/m2 and the solar energy capture efficiency ranges from 0.2 to 5%. The highest production levels are from a tree plantation in the Congo, which produces 36 tons/hectare/year or 4 × 106 kg/km2/yr. Our estimates are about a factor of ten higher even than this maximal production.

267

January 23, 2008

Time: 12:20pm
chapter10.tex

268

January 23, 2008

Time: 12:20pm
chapter10.tex

10.8

Turning trees into gas Humans have cleared a lot of forest for agriculture in the last 8000 years. Assuming that all the cleared forest has been burned or otherwise oxidized, how many tons of CO2
has deforestation added to the atmosphere?

How much is
this in ppm

(parts per
million)?

*

269

January 23, 2008

Time: 12:20pm
chapter10.tex

ANSWER: We need to estimate the area of forested land cleared and the amount of biomass per area.

Let’s start with the area. Land is 25% of the Earth’s surface. We’ll estimate that half of all land area was originally forested and half of that has been cleared.

Thus, humans have cleared forest from 1/4 of the land area of Earth. This estimate is certainly within a factor of four of reality since we have certainly cleared less than 100% of the land area and more than 1/16 or 6%.

The surface area of the Earth (see question 10.1) is A Earth = 4π R 2 = 4π(6 × 103 km)2 = 4 × 108 km2

The land area is one-quarter of that or 108 km2. Humans have cleared the forests from one-quarter of that or 2 × 107 m2.

Now we need the amount of biomass. We know from the previous question that new forests produce less than 4 × 106 kg/km2/yr of biomass. However, after a forest becomes mature, it becomes a steady-state system where the amount of wood that rots (i.e., oxidizes) equals the amount of new wood produced. It takes at least 10 years and less than 100 years for a forest to become mature.* We’ll take the geometric mean and use 30 years. Thus, clearing forest for agriculture releases 30 × 4 × 106 kg/km2=108 kg/km2 of biomass.

Since CO2 has four times the mass of carbon, this is 4 × 108 kg/km2 of CO2. Before we continue, let’s compare this to reality. Typical forest biomass is about 107 kg carbon/km2 [28]. Thus, our estimate of 108 is a factor of 10 too high.

Using the real biomass density, clearing forests released M CO = 2 × 107 km2 × 4 × 107 kg/km2

2

= 8 × 1014 kg

* This definition of maturity refers to the carbon cycle, not to the species of trees in the forest.

270

Chapter 10 The Atmosphere

January 23, 2008

Time: 12:20pm chapter10.tex of carbon dioxide. That is 104 times more than American cars emit each year.

To put it another way, the total mass of the atmosphere is only 4 × 1018 kg. Thus, this is about 200 parts per million (ppm) of the atmosphere. The current atmospheric concentration of CO2 is about 350 ppm, so that this is a significant input.

Wow! Maybe clearing forests over the last ten thousand years has averted another ice age [29]. Or maybe not.

271

January 23, 2008

Time: 12:20pm chapter10.tex 272

January 21, 2008

Time: 09:23am chapter11.tex Risk Life is dangerous. But some things are far more dangerous than others. Do we really need to worry about shark attacks? Should we be required to put infants in car seats on airplanes? After figuring out these questions, you should be able to see through the scare tactics used against us by politicians and other salesmen.

Chapter 11

*

January 21, 2008

Time: 09:23am chapter11.tex 274

January 21, 2008

Time: 09:23am chapter11.tex 11.1

Gambling on the road What is the risk (in the US) of dying per mile traveled in an automobile? What fraction of American deaths are caused by automobiles?

*
year.

each
e i d
l l i w
s u
f o

0

0 1 / 1
then
years,

100
is
expectancy
life
our

If

HINT: expectancy?

life
our
is

What

HINT: .

10

×

3
about
is

US
the
of
population

The

HINT: 8
year.

each
roads
the
on
killed
are

Americans

10

×

4

About

HINT: 4

5.1.

uestion
q

See
year?

per
drive

Americans
o
d
miles
total
many

How

HINT: 275

January 21, 2008

Time: 09:23am chapter11.tex ANSWER: We need to estimate the total number of miles traveled and the total number of deaths on the road. In question 5.1 we estimated the total distance that Americans drive each year to be about 2×1012 mi.

About 4 × 104 Americans die each year in car crashes.

Thus, the risk of death is R = 4 × 104 deaths/yr = 2 × 10−8 deaths/mi 2 × 1012 mi/yr Wow, there are only two deaths per 100 million miles.

That sounds pretty safe.

Let’s look at this number a different way. Let’s figure out the probability that you, the reader, will die in a car crash. The probability that any one of us will be killed in a car crash is the same as the fraction of Americans who die in car crashes. That is, if one in fifty Americans die in car crashes, then your chance of dying in a car crash is also one in fifty.

Now we need to estimate the number of Americans that die each year. There are 3 × 108 of us. The average American life span is about 75 years (although we plan to live much longer than that). Thus, one in every 75 Americans dies every year. This means that the number of deaths every year is N d = 3 × 108 Americans = 4 × 106 deaths/yr 75 yr Thus, the total lifetime probability of dying in a car crash is P crash = 4 × 104 car deaths/yr = 0.01

4 × 106 deaths/yr or 1%. Thus, 1% of us will die in a car crash. (This is the total probability starting at age zero. Since you, gentle reader, are probably slightly older than that, your probability of dying in a car crash is somewhat reduced. Whew!) 276

Chapter 11 Risk

January 21, 2008

Time: 09:23am chapter11.tex 11.2

The plane truth What is the risk in America of dying per mile traveled in a large airplane?

How does this compare to the risks of driving?

*
crash?

per
ie
d
people
many

How

HINT: crash?

planes
large
do
often

How

HINT: flight?

each
is
far

How

HINT: introduction.

the

See
take?

we
do
ights
fl
airplane
many

How

HINT: year?

each
fly

Americans
o
d
miles
total
many

How

HINT: 277

January 21, 2008

Time: 09:23am chapter11.tex ANSWER: We need to estimate the total miles traveled and the total deaths per year. We estimated in the introduction that Americans take 7 × 108 flights per year. The average distance per flight is probably more than 300 miles (it is almost easier to drive) and less than 3000 miles (the distance from NY to LA), so we will use the geometric mean of 1000 miles. Thus, Americans fly a total distance of d = 7 × 108 flights ×103 miles/flight = 7 × 1011 mi. This is 1/3 as far as we drive.

Large planes do not crash every year. Whew! The crash frequency is less than once a year and more than once a decade. We’ll take the geometric mean of √
once every three years (since 3 =

1 × 10). When a plane crashes, it kills about 100 people. Thus, large plane crashes kill about 30 people per year. This means that the probability of dying in a large plane crash per mile is R = 30 deaths/yr = 4 × 10−11 deaths/mi 7 × 1011 mi/yr That is almost 1000 times safer per mile than driving.

Note that expressing airplane fatalities in terms of deaths per mile is somewhat misleading since almost all crashes occur during takeoff and landing. A 3000-mile plane flight is almost exactly as dangerous as a 300-mile plane flight.

This means that some airline safety regulations could be misguided. For example, if we required infants to ride in infant seats on airplanes, it could cost lives. This would make plane travel more expensive, since parents would have to pay for a plane ticket for their infant. Thus, fewer families would fly and more would drive. Because driving is so much more dangerous than flying, many more people would be killed on the roads than would be saved by using infant seats on the airplane.

278

Chapter 11 Risk

January 21, 2008

Time: 09:23am chapter11.tex 11.3

Life’s a beach Compare the risks of getting killed by a shark at the beach and of driving to the beach.

*
mile.

per
people

10

×

2
kills

Driving HINT: 8

−
beach?

the
to
drive
people
do
far

How

HINT: there?

are
ays
d
summer
typical many

How

HINT: ay?

d
summer
typical a
on
beach
the
at
are
people
many

How

HINT: nationwide.

reported is
death
every
that
assume
can
you
newsworthy, so
are
attacks Shark

HINT: US?

the
in
year
ach
e
sharks
by
killed
are
people
many

How

HINT: 279

January 21, 2008

Time: 09:23am chapter11.tex ANSWER: Shark attacks are scary. Shark attacks are newsworthy. Just about every person killed by a shark will make the national news. We read about shark deaths at the beach every few years. This indicates that there is about one fatality per year. (To humans, that is.

We kill lots more sharks per year than that.) In reality, the average for the US is about 0.5 per year.

Now we need to estimate the number of beach goers per year. About 10% of the US population lives within 20 miles of the coast (it’s certainly between 1 and 100%). About 10% of them go to the beach on any given summer day. This means that in July and August, 3×108×0.1×0.1 = 3×106 Americans are at the beach on any given day. Since there are 60 days in July and August, this means that there are 60×3×106 = 2×108
beach visits each year. This is a bit low but reasonably close to reality since California alone has 108 beach visits per year [30].

Each beach goer probably drives about 10 miles each way to get to the beach (more than one and less than 100). The total distance driven by beach goers is thus about d = 20 mi × 2 × 108 visits = 4 × 109 mi.

This will cause an average of N deaths = 4×109 mi×2×10−8 deaths/mi = 80 deaths Thus, driving to the beach is about 100 times more dangerous than shark attacks. Despite this, or perhaps because of this, the shark attacks get far more media coverage and publicity.*

* Driving accidents are just so, you know, pedestrian.

280

Chapter 11 Risk

January 21, 2008

Time: 09:23am chapter11.tex 11.4

Up in smoke On average, how much does each cigarette smoked by a heavy smoker shorten his or her life expectancy?

*
lose?

smoker
average
the
does
life
of
years
many

How

HINT: smoke?

smokers
heavy
do
cigarettes
many

How

HINT: 281

January 21, 2008

Time: 09:23am chapter11.tex ANSWER: We need to estimate how many cigarettes a heavy smoker smokes and how many years of life he or she will lose. We will do this for the average smoker.

Your Uncle Joe may have smoked like a chimney until he was shot by a jealous lover at age 97 while skydiving, but he got lucky. Someone will get lucky and win the lottery, but the average lottery gambler loses money.

Some people live long lives despite smoking, but the average smoker dies early.

Smoking kills primarily through lung cancer and heart disease. These are late-onset diseases. They typically start killing at age 50 or so. The average smoker must lose much more than one year of life* and less than 30 years (since smoking starts to kill at about age 50 and life expectancy is less than 80). Taking the geometric mean of 1 and 30, smokers die five years earlier than nonsmokers.

If you started smoking at age 18 (when it became legal to buy cigarettes) and continued until you died at age 70, at one pack of cigarettes per day, you would have smoked N = 50 years × 400 days/year × 20 cigarettes/day = 4 × 105 cigarettes If we make the totally ridiculous assumption that each cigarette makes the same contribution to mortality, then each cigarette will cost you t =

5 years

= 10−5 years 4 × 105 cigarettes = 10−5 years × π × 107 s/yr = 300 s = 5 min Thus, each cigarette would cost you about the amount of time it takes to smoke it.

* If a smoker only lost one year of life, we would not make so much of a fuss about smoking.

282

Chapter 11 Risk

January 21, 2008

Time: 09:23am
chapter11.tex

Comparing to reality,* a study in the British Medical Journal [31] found that there is a difference in life expectancy between smokers and nonsmokers of 6.5
years and the average smoker consumes a bit less than a pack per day. They found that each cigarette costs an average of 11 min of life. Maybe we should have published.

Note that the assumption that each cigarette does an equal amount of damage is both unverifiable and probably nonsense. Biological systems frequently have thresholds below which no damage is done. One ton of rocks will crush you. However, the first pound or kilogram of rock will not have any adverse affect at all. Similarly, swallowing two tablets of acetaminophen will help your headache but swallowing the entire bottleful will kill you.

* Or at least to a published study.

283

January 21, 2008

Time: 09:23am
chapter11.tex

284

December 27, 2007

Time: 12:46pm
chapter12.tex

Unanswered Questions

There are many, many more questions that could be asked and answered. Just look at the world around you and apply your newfound skills. Here is a set of unanswered questions to get you started.

Chapter 12

*

December 27, 2007

Time: 12:46pm
chapter12.tex

1. How many 1-gal (4-L) buckets of water are needed to empty Loch Ness (or Lake Erie)?

2. How many cigarettes are smoked annually in the US? If you place them end-to-end, how far will they stretch?

3. How many video rental stores are there in the US (or Europe)?

4. How many people are talking on their cell phones at this instant?

5. How many people are eating lunch at this instant?

6. How fast does human hair grow (in m/s or mph)?

7. How many grains of sand are there in all the beaches of the world?

8. How many blades of grass are there on the average (natural) football field?

9. If all the lottery tickets sold in the US in one year were stacked up, how tall would the stack be?

10. How much total time do Americans spend driving in one year? Express your answer in hours, years, and lifetimes.

11. What is your average bicycle travel speed?

Include the time you spend riding your bicycle and the time you spend earning money to pay for your bicycle.

12. How much gasoline would the US save if the speed limit was lowered from 65 to 55 mph?

How many extra hours of the time would be spent driving for each gallon of gas saved?

13. How much rubber (in kg) is deposited on American roadways every year by automobile tires?

286

Chapter 12 Unanswered Questions

December 27, 2007

Time: 12:46pm
chapter12.tex

14. What is the potential energy of a large raindrop in a cloud?

15. You crash your car into highway crash barrels (large barrels filled with sand) at highway speed.

What force is exerted on your body as you stop?

(Assume that you are wearing your seat belt and your air bag deploys properly.) 16. How much potential energy does the landfill referred to in chapter 3 have after Americans have dumped all their garbage there for 100 years?

17. What is the kinetic energy of a rifle bullet?

18. What force does a catcher exert on a baseball when he catches a fast pitch?

19. How much energy can be released by eating 1 kg of chocolate chip cookies?

20. What is the total kinetic energy of all the vehicles on the road in the United States (or Europe) at this instant?

21. If you drop a water ballon from a 30-story building, with what speed will it hit the ground?

22. How much does a human sweat in a hot climate ( T = 98.6 ◦F = 37 ◦C)? Express your answer in liters/day. (Assume that all of the energy you ingest can be removed only by sweating. Water absorbs about 1000 J/g when it evaporates.) 23. How much food (in kg) does a typical human consume in her lifetime? How does that compare to her mass?

24. What is the kinetic energy of the Moon as it orbits the Earth?

25. What is the energy density of 235U (in J/kg)?

26. If every American switched to driving an electric car we would use more electrical energy.

287

December 27, 2007

Time: 12:46pm
chapter12.tex

How many 1-GW power plants would it take to supply that much electrical energy?

27. How many tons of hydrogen are converted to helium in the Sun every year?

28. What proportion of the Sun’s mass is converted to energy during the 10-billion year lifetime of the Sun? (Assume that its power output is relatively constant over its lifetime.) 29. How many miles of roads does the US (or Europe) have?

30. How much carbon dioxide (in kg) does one person emit into the atmosphere each year? All humans?

31. How much force does the atmosphere exert on the front of your body?

32. How large a helium ballon (in m3) would be needed to lift your car? Helium has about one-tenth the density of air.

33. How much, on average, will each high-fat meal consumed shorten your life?

288

Chapter 12 Unanswered Questions

December 27, 2007

Time: 12:49pm
app-a.tex

Appendix A Needed Numbers and Formulas Although it is extraordinary how much can be deduced from just a little knowledge, a little knowledge is necessary. We do not believe that it is dangerous. Here are the important numbers and formulas we use in this book.

A.1 Useful Numbers US population (2006) = 3 × 108

World population (2006) = 6 × 109

Number of items/mole (Avogadro’s number) = 6×1023

Electron volts/joule (eV/J) = 6 × 1018

1 m/s = 2 mph

1 year = π × 107 s

Size of an atom = 10−10 m

Radius of the Earth = 6 × 106 m Earth–Sun distance = 1.5 × 1011 m 1 Calorie = 4 × 103 J

Chemical reaction energy = 1.5 eV

Mass of one mole of carbon = 12 g Acceleration of gravity at Earth’s surface g = 10 m/s2

A.2 Handy formulas Potential energy, PE = mg h Mass = volume × density

289

December 27, 2007

Time: 12:49pm
app-a.tex

Kinetic energy, KE = 1 m v2

2

Work = force × distance

Work = change in KE

Energy = power(W) × time(s)

A.3 Metric Prefixes Size

Prefix

Abbreviation

109
giga

G

106
mega

M

103
kilo
k

10−2
centi
c

10−3
milli
m

10−6
micro
µ

10−9
nano
n

290

Appendix 1 Needed Numbers and Formulas January 21, 2008

Time: 09:27am
app-b.tex

Appendix B

Pegs to Hang Things On Length in meters (m)

Object

1011

Earth–Sun distance

(1.5 × 1011 m)

107 (104 km)

Earth’s diameter

(8000 mi, or

1.3 × 104 km)

106 (103 km)

Distance from

New Orleans to

Detroit (1600 km)

105 (102 km)

Lake Michigan (length)

104 (10 km)

Mt. Everest (height)

103 (1 km; 0.6 mi)

George Washington

Bridge

102

Football field (length)

101

Tennis court

100

Tall man’s stride

10−1 (10 cm)

Person’s hand (width)

10−2 (1 cm)

Sugar cube

10−3 (1 mm)

Coin (thickness)

10−4

Human hair (thickness)

10−5

Human cell (diameter)

10−6 (1 micron [1 µm])

Soap-bubble film

(thickness)

10−9 (1 nanometer [1 nm])

Small molecule

10−10

Atom

291

January 21, 2008

Time: 09:27am
app-b.tex

Area in square meters (m2)

Typical object

1014

Land area of the Earth

1012

Egypt; Texas

1011

New York State; Iceland

109

Los Angeles;

Virginia Beach

108

Manhattan

106 (1 km2)

City of London

104

Football field

102

Volleyball court

100

Small office desk

10−4 (1 cm2)

Sugar cube (one side only)

10−6 (1 mm2)

Head of a pin

10−8

Pixel on computer display

Density in kilograms per
cubic meter (kg/m3)

Item

1018

Neutron star;
atomic nucleus

109

White dwarf star

104

Lead; iron

103 (1 ton/m3, 1 kg/L,

Water; human body

1 g/cm3)

100

Earth’s atmosphere at
sea level

292

Appendix 2 Pegs to Hang Things On

February 1, 2008

Time: 12:57pm
app-b.tex

Mass in kilograms (kg)

Object

1030

The Sun

1027

Jupiter

1025

Earth

1021

Earth’s oceans

1018

Earth’s atmosphere

1015

World coal reserves

(estimated)

1012

World oil production in 2001

1011

Total mass of human world
population

1010

Great Pyramid of Giza

109

Matter converted into energy by the Sun each second

108

Aircraft carrier

107

RMS Titanic 106

Launch mass of the space
shuttle

105

Largest animal, the blue whale 104

Large elephant

103 (1 ton)

Automobile (small)

102

Lion; large human

101

Microwave oven; large cat

100

1 liter or quart of water

10−1

Human kidney; apple; rat

10−2

Lethal dose of caffeine;
adult mouse; large coin

10−3 (1 g)

Sugar cube

10−4

Caffeine in a cup of coffee

10−6 (1 mg)

Mosquito

10−7

Lethal dose of ricin

10−9 (1 µg)

Sand grain (medium)

10−12 (1 ng)

Human cell

10−27

Neutron; proton; hydrogen
atom

10−30

Electron

293

January 21, 2008

Time: 09:27am
app-b.tex

294

December 27, 2007

Time: 12:52pm
references.tex

Bibliography [1] A. Asaravala. Landing a job can be puzzling.

Wired.com, June 2003. http://www.wired.com/news/
culture/0,1284,59366,00.html.

[2] J. Kador. How to Ace the Brain Teaser Interview.

McGraw-Hill, New York City, 2004.

http://www.jkador.com/brainteaser/

[3] W. Poundstone. How Would You Move Mount Fuji?

Little, Brown, New York, 2003.

[4] C. Swartz. First, the answer. The Physics Teacher 33:488, 1995.

[5] P. Morrison. Fermi questions. American Journal of Physics, 31:626, 1963.

[6] T. Isenhour. Private communication. Originally attributed to H. Jeffries.

[7] G. Lucier and B.-H. Lin. Americans relish cucumbers.

Agricultural Outlook, page 9, 2000. www.ers.usda.gov/
publications/agoutlook/dec2000/ao277d.pdf [8] T. L. Isenhour and L. G. Pedersen. Passing Freshman Chemistry. Harcourt Brace Jovanovich, New York, 1981.

[9] Basic facts: municipal solid waste (US). Technical report, US Environmental Protection Agency, Washington, DC, 2006. http://www.epa.gov/
msw/facts.htm

[10] P. Jillette and Teller. Recycling. Bullsh∗ t! 2004.

[11] Stanford Comprehensive Cancer Center. Stanford bmt body surface area calculator, 2006.

http://bmt.stanford.edu/calculators/bsa.html [12] McGraw-Hill Encyclopedia of Science & Technology.

McGraw-Hill Professional, New York, 2004.

[13] G. Elert. The physics hypertextbook, 2006.

http://hypertextbook.com/facts/1998/

StevenChen.shtml

295

December 27, 2007

Time: 12:52pm
references.tex

[14] N. Juster. The Phantom Tollbooth. Knopf, New York, 1961.

[15] G. Elert. The physics hypertextbook, 2006.

http://hypertextbook.com/physics/matter/energy-chemical/

[16] I. Buchmann. Batteries in a Portable World. Cadex Electronics, Richmond, BC, Canada, 2001.

http://www.batteryuniversity.com/index.htm [17] L. Brown. Plan B: Rescuing a Planet Under Stress and a Civilization in Trouble. W.W. Norton, New York, 2003.

http://www.earth-policy.org/Books/PB/

PBch8_ss4.htm

[18] M. S. Matthews T. Gehrels, and A. M. Schumann, editors. Hazards Due to Comets and Asteroids.

University of Arizona Press, Tucson, AZ, 1994.

http://seds.lpl.arizona.edu/nineplanets/nineplanets/
meteorites.html

[19] International energy outlook 2006. Technical Report DOE/EIA-0484(2006), DOE Energy Information Administration, Washington, DC, 2006.

http://www.eia.doe.gov/oiaf/ieo/world.html [20] W. Thomson. On the age of the sun’s heat. Macmillan’s Magazine 5:288, 1862. Reprinted in Popular Lectures and Addresses, vol. 1, by Sir William Thomson.

Macmillan, London, 1891.

[21] Sea level and climate. Technical Report Fact Sheet 002-00, US Geological Survey, Washington, DC, 2000.

http://pubs.usgs.gov/fs/fs2-00/

[22] The World Factbook 2006. Central Intelligence Agency, Office of Public Affairs, Washington, DC, 2006.

https://www.cia.gov/cia/publications/factbook/
index.html

[23] Danish Wind Industry Association. Wind turbine power calculator. http://www.windpower.org/en/tour/
wres/pow/index.htm

296

Bibliography

December 27, 2007

Time: 12:52pm
references.tex

[24] C. D. Elvidge et al. U.S. constructed area approaches the size of Ohio. Eos, Trans. AGU 85:233, 2004.

http://www.agu.org/pubs/crossref/2004/

2004EO240001.shtml

[25] G. Grimvall. Socrates, Fermi, and the second law of thermodynamics. American Journal of Physics, 72:1145, 2004.

[26] Emissions of greenhouse gases in the United States 2004. Technical Report DOE/EIA-0573(2004), DOE

Energy Information Administration, Washington, DC, 2005. http://www.eia.doe.gov/oiaf/1605/ggrpt/
carbon.html

[27] D. L. Klass. Biomass for renewable energy and fuels. In Encyclopedia of Energy. Elsevier, Amsterdam, 2004.

www.bera1.org/cyclopediaofEnergy.pdf [28] R. A. Houghton. Aboveground forest biomass and the global carbon balance. Global Change Biology 11:945, 2005. www.whrc.org/resources/published_literature/
pdf/HoughtonGCB.05.pdf

[29] W. Ruddiman. Plows, Plagues and Petroleum.

Princeton University Press, Princeton, NJ, 2005.

[30] G. H. Burgess. California beach attendance, surf rescues, and shark attacks 1990–2000. International Shark Attack File, 2005. http://www.flmnh.ufl.edu/
fish/Sharks/statistics/CAbeachattacks.htm [31] M. Shaw, R. Mitchell, and D. Dorling. Time for a smoke? One cigarette reduces your life by 11 minutes.

British Medical Journal 320:53, 2000.

http://bmj.bmjjournals.com/cgi/content/full/320/

7226/53

297

December 27, 2007

Time: 12:52pm
references.tex

298

January 21, 2008

Time: 09:30am
index.tex

Index
airplane flights, 5
from deforestation, 269

Alexander the Great, 247
into forests, 265
area
cars
human, 65
distance, 89
landfill, 47
cats, 39
people, 21
cells, 57
rally, 79
coal, 233, 255
towel, 31
continent, 131
trash, 47
cow, 73

Armageddon, 61
atmosphere, 247

DNA, 83
mass of, 245
driving, 89
using up, 251
automobiles

Earth
average speed, 109
circumference, 25
cost, 99, 109
mass, 39
crash, 139
electricity
distance, 89
power, 175
fuel, 91
energy, 170
time, 95
battery, 151, 159
tires, 107
car, 163
waste, 103
chemical, 145, 147, 160,

162, 201
ball
electrical, 159
golf, 25
electrical, 219
tennis, 127
electron-Volts, 146
battery, 151, 155, 159
ethanol, 167
blood, 61
food, 163
books, 53
gasoline, 147, 159
breathing, 247, 251
gravitational, 114
building, 123

Joules, 114
kinetic, 126, 138, 139, 143, calories, 162

230
carbon dioxide
continent, 131
from cars, 261

Earth, 181
from coal, 255
meteorite, 185

299

January 21, 2008

Time: 09:30am
index.tex
rocket ship, 135
kilowatt-hour, 170
tennis ball, 127

Krypton, 205
truck, 129

Kryptonite, 207
nuclear, 237
potential, 114, 115, 119, 123
landfill, 47
solar, 193
library, 53

Sun, 201, 223, 225
lottery, 4, 41

TNT, 149
work, 138
mass
energy crisis, 195
trash, 43
energy density
mean, geometric, 3
battery, 155
meteorite, 185
gasoline, 147

Million Man March, 79
ethanol, 167
mole, 147
mountain
farmland, 167
climbing, 115
fission, 237
flattening, 119
flying, 5, 51
food, 162
noses, 77
force, 138, 139, 143
nuclear energy, 237
fossil fuel, 167
nuclear waste, 259
fusion, 237
oxygen, 251
gasoline, 147, 159, 163
part per billion, 27
power, 173
paved area, 239
gerbil, 197
piano tuners, 7

Goldilocks, 2
pickles, 29
gravitation, 114
porta-potties, 79
hair, 69
power, 170
heat, 171
electrical, 215, 219
horses, 103
human, 171
hot dog, 73
refueling
human
electricity, 175
area, 65
gasoline, 173
cells, 57

Sun, 193
noses, 77

Watt, 151
power, 171
wind, 229
hydrocarbons, 146
power density
gerbil, 197
ice caps, 209

Sun, 197

300

Index

January 21, 2008

Time: 09:30am
index.tex
power plant
power density, 197, 193
coal fired, 233
radius, 189
nuclear, 237

Superman, 207
supernova, 205
refueling, 173, 175
rickshaws, 99

TNT, 149
risk
trash, 43, 47
shark attack, 279
truck, 129
smoking, 281
rocket ship, 135
roofs, 239
velocity, 126

volume
shark attack, 279
blood, 61
smoking, 281
cow, 73
soccer, 75
human, 57
solar energy, 223, 225
trash, 43
speed, 126

U.S. Capitol, 35

Spider-Man, 143
subway, 143
waste, 103

Sun

Watt, 170
density, 189
wind turbine, 229
energy, 201
work, 138, 139, 143

301

Document Outline

- 0691129495
- prelims
- chapter01
- chapter02
- chapter03
- chapter04
- chapter05
- chapter06
- chapter07
- chapter08
- chapter09
- chapter10
- chapter11
- chapter12
- app-a
- app-b
- references
- index

Table of Contents

- Chapter 3 General Questions
- Chapter 3 General Questions
- Chapter 3 General Questions
- Chapter 3 General Questions
- Chapter 3 General Questions
- Chapter 3 General Questions
- Chapter 3 General Questions
- Chapter 3 General Questions
- Chapter 3 General Questions
- Chapter 3 General Questions
- books
- Chapter 3 General Questions
- Chapter 5 Transportation
- Chapter 5 Transportation
- Chapter 5 Transportation
- Chapter 5 Transportation
- Chapter 5 Transportation
- Chapter 5 Transportation
- Chapter 5 Transportation
- Chapter 5 Transportation
- part
- Chapter 10 The Atmosphere
- Chapter 10 The Atmosphere
- Chapter 10 The Atmosphere
- Chapter 10 The Atmosphere
- Chapter 10 The Atmosphere
- Chapter 10 The Atmosphere
- Chapter 10 The Atmosphere
- (parts per
- Chapter 10 The Atmosphere
- Chapter 11 Risk
- Chapter 11 Risk
- Chapter 11 Risk
- Chapter 11 Risk
- Chapter 12 Unanswered Questions
- Chapter 12 Unanswered Questions