INSTRUCTOR SOLUTIONS MANUAL (vol & ch 01-33) INSTRUCTOR SOLUTIONS MANUAL VOLUMES & VOLUME DOUGLAS C GIANCOLI’S PHYSICS PRINCIPLES WITH APPLICATIONS 7TH EDITION BOB DAVIS TAYLOR UNIVERSITY J ERIK HENDRICKSON UNIVERSITY OF WISCONSIN – EAU CLAIRE San Francisco Boston New York Cape Town Hong Kong London Madrid Mexico City Montreal Munich Paris Singapore Sydney Tokyo Toronto President, Science, Business and Technology: Paul Corey Publisher: Jim Smith Executive Development Editor: Karen Karlin Project Manager: Elisa Mandelbaum Marketing Manager: Will Moore Senior Managing Editor: Corinne Benson Managing Development Editor: Cathy Murphy Production Service: PreMedia Global, Inc ISBN 10: 0-321-74768-2 ISBN 13: 978-0-321-74768-6 Copyright © 2014, 2009, 1998 Pearson Education, Inc 1301 Sansome St., San Francisco, CA 94111 All rights reserved Manufactured in the United States of America This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise To obtain permission(s) to use material from this work, please submit a written request to Pearson Education, Inc., Permissions Department, 1900 E Lake Ave., Glenview, IL 60025 For information regarding permissions, call (847) 486-2635 Many of the designations used by manufacturers and sellers to distinguish their products are claimed as trademarks Where those designations appear in this book, and the publisher was aware of a trademark claim, the designations have been printed in initial caps or all caps www.pearsonhighered.com CONTENTS Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter Chapter 10 Chapter 11 Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18 Chapter 19 Chapter 20 Chapter 21 Chapter 22 Chapter 23 Chapter 24 Chapter 25 Chapter 26 Chapter 27 Chapter 28 Chapter 29 Chapter 30 Chapter 31 Chapter 32 Chapter 33 PREFACE - iv Introduction, Measurement, Estimating - 1-1 Describing Motion: Kinematics in One Dimension 2-1 Kinematics in Two Dimensions; Vectors 3-1 Dynamics: Newton’s Laws of Motion - 4-1 Circular Motion; Gravitation 5-1 Work and Energy - 6-1 Linear Momentum 7-1 Rotational Motion 8-1 Static Equilibrium; Elasticity and Fracture 9-1 Fluids 10-1 Oscillations and Waves -11-1 Sound 12-1 Temperature and Kinetic Theory 13-1 Heat 14-1 The Laws of Thermodynamics -15-1 Electric Charge and Electric Field -16-1 Electric Potential -17-1 Electric Currents -18-1 DC Circuits -19-1 Magnetism 20-1 Electromagnetic Induction and Faraday’s Law 21-1 Electromagnetic Waves -22-1 Light: Geometric Optics 23-1 The Wave Nature of Light 24-1 Optical Instruments 25-1 The Special Theory of Relativity 26-1 Early Quantum Theory and Models of the Atom 27-1 Quantum Mechanics of Atoms -28-1 Molecules and Solids 29-1 Nuclear Physics and Radioactivity 30-1 Nuclear Energy; Effects and Uses of Radiation -31-1 Elementary Particles -32-1 Astrophysics and Cosmology 33-1 © Copyright 2014 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher PREFACE This Instructor’s Solutions Manual provides answers and worked-out solutions to all end of chapter questions and problems from chapters – 15 of Physics: Principles with Applications, 7th Edition, by Douglas C Giancoli At the end of the manual are grids that correlate the 6th edition questions and problems to the 7th edition questions and problems We formulated the solutions so that they are, in most cases, useful both for the student and the instructor Accordingly, some solutions may seem to have more algebra than necessary for the instructor Other solutions may seem to take bigger steps than a student would normally take: e.g simply quoting the solutions from a quadratic equation instead of explicitly solving for them There has been an emphasis on algebraic solutions, with the substitution of values given as a very last step in most cases We feel that this helps to keep the physics of the problem foremost in the solution, rather than the numeric evaluation Much effort has been put into having clear problem statements, reasonable values, pedagogically sound solutions, and accurate answers/solutions for all of the questions and problems Working with us was a team of five additional solvers – Karim Diff (Santa Fe College), Thomas Hemmick (Stony Brook University), Lauren Novatne (Reedley College), Michael Ottinger (Missouri Western State University), and Trina VanAusdal (Salt Lake Community College) Between the seven solvers we had four complete solutions for every question and problem From those solutions we uncovered questions about the wording of the problems, style of the possible solutions, reasonableness of the values and framework of the questions and problems, and then consulted with one another and Doug Giancoli until we reached what we feel is both a good statement and a good solution for each question and problem in the text Many people have been involved in the production of this manual We especially thank Doug Giancoli for his helpful conversations Karen Karlin at Prentice Hall has been helpful, encouraging, and patient as we have turned our thoughts into a manual Michael Ottinger provided solutions for every chapter, and helped in the preparation of the final solutions for some of the questions and problems And the solutions from Karim Diff, Thomas Hemmick, Lauren Novatne, and Trina VanAusdal were often thought-provoking and always appreciated Even with all the assistance we have had, the final responsibility for the content of this manual is ours We would appreciate being notified via e-mail of any errors that are discovered We hope that you will find this presentation of answers and solutions useful Bob Davis (rbdavis@taylor.edu) Upland, IN J Erik Hendrickson (hendrije@uwec.edu) Eau Claire, WI © Copyright 2014 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher INTRODUCTION, MEASUREMENT, ESTIMATING Responses to Questions (a) A particular person’s foot Merits: reproducible Drawbacks: not accessible to the general public; not invariable (size changes with age, time of day, etc.); not indestructible (b) Any person’s foot Merits: accessible Drawbacks: not reproducible (different people have different size feet); not invariable (size changes with age, time of day, etc.); not indestructible Neither of these options would make a good standard The distance in miles is given to one significant figure, and the distance in kilometers is given to five significant figures! The value in kilometers indicates more precision than really exists or than is meaningful The last digit represents a distance on the same order of magnitude as a car’s length! The sign should perhaps read “7.0 mi (11 km),” where each value has the same number of significant figures, or “7 mi (11 km),” where each value has about the same % uncertainty The number of digits you present in your answer should represent the precision with which you know a measurement; it says very little about the accuracy of the measurement For example, if you measure the length of a table to great precision, but with a measuring instrument that is not calibrated correctly, you will not measure accurately Accuracy is a measure of how close a measurement is to the true value If you measure the length of an object, and you report that it is “4,” you haven’t given enough information for your answer to be useful There is a large difference between an object that is meters long and one that is feet long Units are necessary to give meaning to a numerical answer You should report a result of 8.32 cm Your measurement had three significant figures When you multiply by 2, you are really multiplying by the integer 2, which is an exact value The number of significant figures is determined by the measurement The correct number of significant figures is three: sin 30.0° = 0.500 Useful assumptions include the population of the city, the fraction of people who own cars, the average number of visits to a mechanic that each car makes in a year, the average number of weeks a mechanic works in a year, and the average number of cars each mechanic can see in a week © Copyright 2014 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 1-1 1-2 Chapter (a) (b) There are about 800,000 people in San Francisco, as estimated in 2009 by the U.S Census Bureau Assume that half of them have cars If each of these 400,000 cars needs servicing twice a year, then there are 800,000 visits to mechanics in a year If mechanics typically work 50 weeks a year, then about 16,000 cars would need to be seen each week Assume that on average, a mechanic can work on cars per day, or 20 cars a week The final estimate, then, is 800 car mechanics in San Francisco Answers will vary Responses to MisConceptual Questions (d) One common misconception, as indicated by answers (b) and (c), is that digital measurements are inherently very accurate A digital scale is only as accurate as the last digit that it displays (a) The total number of digits present does not determine the accuracy, as the leading zeros in (c) and (d) are only placeholders Rewriting the measurements in scientific notation shows that (d) has two-digit accuracy, (b) and (c) have three-digit accuracy, and (a) has four-digit accuracy Note that since the period is shown, the zeros to the right of the numbers are significant (b) The leading zeros are not significant Rewriting this number in scientific notation shows that it only has two significant digits (b) When you add or subtract numbers, the final answer should contain no more decimal places than the number with the fewest decimal places Since 25.2 has one decimal place, the answer must be rounded to one decimal place, or to 26.6 (b) The word “accuracy” is commonly misused by beginning students If a student repeats a measurement multiple times and obtains the same answer each time, it is often assumed to be accurate In fact, students are frequently given an “ideal” number of times to repeat the experiment for “accuracy.” However, systematic errors may cause each measurement to be inaccurate A poorly working instrument may also limit the accuracy of your measurement (d) This addresses misconceptions about squared units and about which factor should be in the numerator of the conversion This error can be avoided when students treat the units as algebraic symbols that must be cancelled out (e) When making estimates, students frequently believe that their answers are more significant than they actually are This question helps the student realize what an order-of-magnitude estimation is NOT supposed to accomplish (d) This addresses the fact that the generic unit symbol, like [L], does not indicate a specific system of units Solutions to Problems (a) 214 significant figures (b) 81.60 significant figures (c) 7.03 significant figures © Copyright 2014 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Introduction, Measurement, Estimating (d) 0.03 significant figure (e) 0.0086 significant figures (f) 3236 significant figures (g) 8700 significant figures (a) 1.156 = 1.156 × 100 (b) 21.8 = 2.18 × 101 (c) 0.0068 = 6.8 × 10−3 (d) 328.65 = 3.2865 × 102 (e) 0.219 = 2.19 × 10−1 (f) 444 = 4.44 × 102 (a) 8.69 × 104 = 86,900 (b) 9.1 × 103 = 9100 (c) 8.8 × 10−1 = 0.88 (d) 4.76 × 102 = 476 (e) 3.62 × 10−5 = 0.0000362 (a) 14 billion years = 1.4 × 1010 years (b) ⎛ 3.156 × 107 (1.4 × 1010 yr) ⎜ ⎜ yr ⎝ 1-3 s⎞ 17 ⎟⎟ = 4.4 × 10 s ⎠ 0.25 m × 100% = 4.6% 5.48 m % uncertainty = (a) % uncertainty = 0.2 s × 100% = 3.636% ≈ 4% 5.5 s (b) % uncertainty = 0.2 s × 100% = 0.3636% ≈ 0.4% 55 s (c) The time of 5.5 minutes is 330 seconds % uncertainty = 0.2 s × 100% = 0.0606% ≈ 0.06% 330 s © Copyright 2014 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 1-4 Chapter To add values with significant figures, adjust all values to be added so that their exponents are all the same (9.2 × 103 s) + (8.3 × 104 s) + (0.008 × 106 s) = (9.2 × 103 s) + (83 × 103 s) + (8 × 103 s) = (9.2 + 83 + 8) × 103 s = 100.2 × 103 s = 1.00 × 105 s When you add, keep the least accurate value, so keep to the “ones” place in the last set of parentheses When you multiply, the result should have as many digits as the number with the least number of significant digits used in the calculation (3.079 × 102 m)(0.068 × 10−1 m) = 2.094 m ≈ 2.1 m The uncertainty is taken to be 0.01 m % uncertainty = 10 0.01 m 1.57 m × 100% = 0.637% ≈ 1% To find the approximate uncertainty in the volume, calculate the volume for the minimum radius and the volume for the maximum radius Subtract the extreme volumes The uncertainty in the volume is then half of this variation in volume Vspecified = 34 π rspecified = 43 π (0.84 m)3 = 2.483 m3 Vmin = 34 π rmin = 34 π (0.80 m)3 = 2.145 m3 Vmax = 34 π rmax = 43 π (0.88 m)3 = 2.855 m3 ΔV = 12 (Vmax − Vmin ) = 12 (2.855 m3 − 2.145 m3 ) = 0.355 m3 The percent uncertainty is 11 ΔV Vspecified = 0.355 m3 2.483 m3 × 100 = 14.3 ≈ 14% To find the approximate uncertainty in the area, calculate the area for the specified radius, the minimum radius, and the maximum radius Subtract the extreme areas The uncertainty in the area is then half this variation in area The uncertainty in the radius is assumed to be 0.1× 104 cm Aspecified = π rspecified = π (3.1 × 104 cm) = 3.019 × 109 cm 2 Amin = π rmin = π (3.0 × 104 cm) = 2.827 × 109 cm 2 Amax = π rmax = π (3.2 × 104 cm) = 3.217 × 109 cm ΔA = 12 ( Amax − Amin ) = 12 (3.217 × 109 cm − 2.827 × 109 cm ) = 0.195 × 109 cm Thus the area should be quoted as A = (3.0 ± 0.2) × 109 cm 12 (a) 286.6 mm 286.6 × 10−3 m 0.2866 m (b) 85 μ V 85 × 10−6 V 0.000085 V (c) 760 mg 760 × 10−6 kg 0.00076 kg (if last zero is not significant) © Copyright 2014 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Astrophysics and Cosmology 33-3 X-rays, which we could detect on Earth Another way we could “see” a black hole is if it caused gravitational lensing of objects behind it Then we would see stars and galaxies in the “wrong” place as their light is bent as it passes past the black hole on its way to Earth 14 Both the formation of the Earth and the time during which people have lived on Earth are on the far right edge of Fig 33–29, in the era of dark energy 15 Atoms were unable to exist until hundreds of thousands of years after the Big Bang because the temperature of the universe was still too high At those very high temperatures, the free electrons and nuclei were moving so fast and had so much kinetic energy, and they had so many high energy photons colliding with them, that they could never combine together to form atoms Once the universe cooled below 3000 K, this coupling could take place, and atoms were formed 16 (a) (b) All Type Ia supernovae are expected to be of nearly the same luminosity Thus they are a type of “standard candle” for measuring very large distances The distance to a supernova can be determined by comparing the apparent brightness with the intrinsic luminosity and then using Eq 33–1 to find the distance 17 If the average mass density of the universe is above the critical density, then the universe will eventually stop its expansion and contract, collapsing on itself and ending finally in a “big crunch.” This scenario corresponds to a closed universe, or one with positive curvature 18 (a) (b) Gravity between galaxies should be pulling the galaxies back together, slowing the expansion of the universe Astronomers could measure the redshift of light from distant supernovae and deduce the recession velocities of the galaxies in which they lie By obtaining data from a large number of supernovae, they could establish a history of the recessional velocity of the universe and perhaps tell whether the expansion of the universe is slowing down Responses to MisConceptual Questions (c) At the end of the hydrogen-fusing part of the star’s life, it will move toward the upper right of the diagram as it becomes a red giant Low-mass stars, not large enough to end as neutron stars, then move to the lower left as they become white dwarf stars The position of the star on the main sequence is a result of the size (mass) of the star This does not change significantly during its main-sequence lifetime, and therefore the star does not change positions along the main sequence (b) Parallax requires that the star move relative to the background of distant stars This only happens when the distance to the star is relatively small, less than about 100 light-years (b) If the universe were expanding in every direction, then any location in the universe would observe that all other points in the universe are moving away from it with the speed increasing with distance This is observed when galaxy speeds are measured, implying that indeed the universe is expanding The observation does not say anything about whether the expansion will continue forever or eventually stop (c) The Sun is primarily made of hydrogen, not heavy radioactive isotopes As gravity initially compressed the Sun, its core heated sufficiently for the hydrogen atoms in the core to fuse to create helium This is now the primary source of energy in the Sun The Sun is too hot for molecules, such as water, to be formed by the oxidation of hydrogen When fusion began in the © Copyright 2014 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 33-4 Chapter 33 core, the energy release balanced the force of gravity to stop the gravitational collapse Since the Sun is no longer contracting, the gravitational potential energy is no longer decreasing and is not a significant source of energy (b, c, d) Parallax is only useful for finding distances to stars that are less than 100 light-years distant, much less than intergalactic distances The H–R diagram with luminosity and temperature of the star can tell us the absolute brightness of the distant star The relationship between the relative brightness and absolute brightness tells us the distance to the star Certain supernova explosions have a standard luminosity A comparison of the apparent brightness and absolute brightness of the supernova can give the distance to the supernova Finally, since the universe is expanding and the velocity of objects increases with distance, a measure of the redshift in light from distant stars can be used to find the speed of the stars and therefore the distance (b) Due to the speed of light, light takes a finite time to reach Earth from the stars, with light from more distant objects requiring more time to reach Earth Therefore, when an object that is a million light-years away is observed, the observer sees the object as it was a million years ago If the object is 100 million light-years away, the observer sees it as it was 100 million years ago The farther the observed object, the earlier the time that it is being observed (d) At the time of the Big Bang, all of space was compacted to a small region The expansion occurred throughout all space So it occurred near the Earth, near the center of the Milky Way Galaxy, several billion light-years away, near the Andromeda Galaxy, and everywhere else (d) In the first few seconds of the universe, matter did not exist, only energy; the first atoms were not formed until many thousands of years later, and these atoms were primarily hydrogen Stars on the main sequence and novae convert hydrogen into helium They not create the heavier elements These heavy elements are typically created in the huge energy bursts of supernovae (c) The rotational period of a star in a galaxy is related to the mass in the galaxy Astronomers observing the rotational periods found them to be much greater than could be accounted for by the visible matter (stars and dust clouds) Dust clouds are not dark matter, as they can reflect light and are therefore visible The acceleration of the expansion of the universe is the foundation for the theory of dark energy, which is different from dark matter 10 (a) If the Big Bang were a single large explosion that gave the universe an initial large kinetic energy, and since currently the only large-scale force in the universe is the attractive force of gravity, we would expect that the kinetic energy of the universe would be decreasing as that energy becomes gravitational potential energy The expansion, however, is accelerating, so there must be another energy source fueling the acceleration This source has been dubbed “dark energy.” Solutions to Problems Convert the angle to seconds of arc, reciprocate to find the distance in parsecs, and then convert to light-years ⎛ 36000 ⎞ ⎟ = 1.0440 ⎝ 1° ⎠ φ = (2.9 × 10−4 )° ⎜ d (pc) = φ0 = ⎛ 3.26 ly ⎞ = 0.958 pc ⎜ ⎟ = 3.1 ly 1.0440 ⎝ pc ⎠ © Copyright 2014 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Astrophysics and Cosmology Use the angle to calculate the distance in parsecs and then convert to light-years d (pc) = 33-5 ⎛ 3.26 ly ⎞ 1 = = 3.704 pc → 3.704 pc ⎜ ⎟ = 12 ly φ 0.270 ⎝ pc ⎠ The parallax angle is smaller for the farther star Since tan φ = d /D, as the distance D to the star increases, the tangent decreases, so the angle decreases And since for small angles, tan φ ≈ φ , we have that φ ≈ d /D Thus if the distance D is doubled, then the angle φ will be smaller by a factor of The apparent brightness of an object is inversely proportional to the observer’s distance from the L object, given by Eq 33–1, b = To find the relative brightness at one location as compared with 4π d another, take a ratio of the apparent brightness at each location L bJupiter bEarth M M 1.99 × 1030 kg = = = × 10−3 kg/m3 π r3 π (6 × 1010 m)3 V 3 (a) The apparent brightness is the solar constant, 1.3 × 103 W/m (b) Use Eq 33–1 to find the absolute luminosity b= 2 ⎞ ⎛ ⎞ −2 ⎟ =⎜ ⎟ = 3.7 × 10 ⎟ ⎝ ⎠ ⎠ The density is the mass divided by the volume ρ= ⎛ d 4π d Jupiter d2 = = 2Earth = ⎜ Earth L d Jupiter ⎝⎜ d Jupiter 4π d Earth L 4π d → L = 4π d 2b = 4π (1.496 × 1011 m ) (1.3 × 103 W/m ) = 3.7 × 1026 W The angular width is the inverse tangent of the diameter of our Galaxy divided by the distance to the nearest galaxy According to Fig 33–2, our Galaxy is about 100,000 ly in diameter φ = tan −1 Galaxy diameter 1.0 × 105 ly = tan −1 = 4.2 × 10−2 rad ≈ 2.4° Distance to nearest galaxy 2.4 × 106 ly φMoon = tan −1 Moon diameter 3.48 × 106 m = tan −1 = 9.1× 10−3 rad ( ≈ 0.52°) Distance to Moon 3.84 × 10 m The galaxy width is about 4.5 times the Moon’s width The text says that there are about × 1011 stars in the galaxy and about 1011 galaxies, so that means there are about × 1022 stars in the observable universe The density is the mass divided by the volume ρ= M Sun M 1.99 × 1030 kg = = = 1.83 × 109 kg/m3 π R3 π (6.38 ì 106 m)3 V Earth 3 â Copyright 2014 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 33-6 Chapter 33 Since the volumes are the same, the ratio of the densities is the same as the ratio of the masses ρ ρ Earth 10 = M M Earth star 5.98 × 10 24 kg = 3.33 × 105 times larger star ρ white M 1.5(1.99 × 1030 kg) = = 5.354 × 1017 kg/m3 ≈ 5.4 × 1017 kg/m3 π (11× 103 m)3 V ρ neutron = 5.354 × 1017 kg/m3 1.83 × 10 kg/m dwarf 12 1.99 × 1030 kg The density of the neutron star is its mass divided by its volume Use the proton to calculate the density of nuclear matter The radius of the proton is taken from Eq 30–1 ρ neutron = 11 = ρ neutron star = 2.9 × 10 ρ nuclear star 5.354 × 1017 kg/m3 1.673 × 10−27 kg π (1.2 × 10 −15 = 2.3 m)3 The reciprocal of the distance in parsecs is the angle in seconds of arc 1 = = 0.01786 ≈ 0.018′′ d (pc) 56 pc (a) φ0 = (b) ⎛ 1° ⎞ −6 −6 0.017860 ⎜ ⎟ = 4.961× 10 ° ≈ (5.0 × 10 )° 3600 ⎝ ⎠ ( ) Convert the light-years to parsecs and then take the reciprocal of the number of parsecs to find the parallax angle in seconds of arc ⎛ pc ⎞ 65 ly ⎜ ⎟ = 19.94 pc ≈ 20 pc (2 significant figures) ⎝ 3.26 ly ⎠ 13 = φ= = 0.050′′ 19.94 pc Find the distance in light-years That value is also the time for light to reach us ⎛ 3.26 ly ⎞ 85 pc ⎜ ⎟ = 277 ly ≈ 280 ly → It takes light 280 years to reach us ⎝ pc ⎠ 14 It is given that b1 = b2 , r1 = r2 , λP1 = 750 nm, and λP2 = 450 nm Wien’s law (Eq 27–1) says λPT = α , where α is a constant, so λP1T1 = λP2T2 The Stefan-Boltzmann equation (Eq 14–6) says that the power output of a star is given by P = β AT , where β is a constant and A is the radiating area The P in the Stefan-Boltzmann equation is the same as the luminosity L in this Chapter The luminosity L is related to the apparent brightness b by Eq 33–1 λP1T1 = λP2T2 b1 = b2 → → L1 4π d12 T2 λP1 = T1 λP2 = L2 4π d 22 → d 22 d12 = L2 P2 β A2T24 4π r22T24 T24 ⎛ T2 ⎞ = = = = =⎜ ⎟ L1 P1 β A1T14 4π r12T14 T14 ⎝ T1 ⎠ → 2 ⎛λ ⎞ d ⎛ T2 ⎞ ⎛ 750 ⎞ = ⎜ ⎟ = ⎜ P1 ⎟ = ⎜ ⎟ = 2.8 d1 ⎝ T1 ⎠ ⎝ 450 ⎠ ⎝ λP2 ⎠ The star with the peak at 450 nm is 2.8 times farther away than the star with the peak at 750 nm © Copyright 2014 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Astrophysics and Cosmology 15 33-7 Wien’s law (Eq 27–2) says that λPT = α , where α is a constant, so λP1T1 = λP2T2 The StefanBoltzmann equation (Eq 14–6) says that the power output of a star is given by P = β AT , where β is a constant and A is the radiating area The P in the Stefan-Boltzmann equation is the same as the luminosity L in this Chapter The luminosity L is related to the apparent brightness b by Eq 33–1 It is given that b1 /b2 = 0.091, d1 = d , λP1 = 470 nm, and λP2 = 720 nm T2 λP1 = ; b1 = 0.091 b2 T1 λP2 λP1T1 = λP2T2 → 1= d 22 d12 = → L1 4π d12 = 0.091 L2 4π d 22 T24 r22 0.091L2 0.091P2 0.091A2T24 (0.091)4π r22T24 = = = = 091 L1 P1 A1T14 T14 r12 4π r12T14 2 → → ⎛T ⎞ ⎛λ ⎞ ⎛ 470 nm ⎞ r1 = 0.091 ⎜ ⎟ = 0.091 ⎜ P1 ⎟ = 0.091 ⎜ ⎟ = 0.1285 r2 ⎝ 720 nm ⎠ ⎝ T1 ⎠ ⎝ λP2 ⎠ The ratio of the diameters is the same as the ratio of radii, so 16 The Schwarzschild radius is 2GM /c REarth = 17 2GM c = 2(6.67 ×10−11 N ⋅ m /kg )(5.98 ×1024 kg) (3.00 × 10 m/s) c2 = 2(6.67 × 10−11 N ⋅ m /kg )(2 × 1041 kg) (3.00 × 108 m/s) The Schwarzschild radius is given by R = 2GM /c , so M = M = 19 2GM Earth = 8.86 × 10−3 m ≈ 8.9 mm The Schwarzschild radius is given by R = 2GM /c An approximate mass of ordinary matter for our Galaxy is calculated in Example 33–1 A value of twice that mass is also quoted in the text, so the given answer may vary by a factor of R= 18 D1 = 0.13 D2 = × 1014 m Rc The radius is given in Eq 27–14 2G Rc (5.29 × 10−11 m)(3.00 × 108 m/s) = = 3.57 × 10−16 kg −11 2 2G 2(6.67 × 10 N ⋅ m /kg ) The limiting value for the angles in a triangle on a sphere is 540° Imagine drawing an equilateral triangle near the north pole, enclosing the north pole If that triangle were small, then the surface would be approximately flat, and each angle in the triangle would be 60° Then imagine “stretching” each vertex of that triangle down toward the equator, ensuring that the north pole was inside the triangle The angle at each vertex of the triangle would expand, with a limiting value of 180° The three 180° angles in the triangle would sum to 540° © Copyright 2014 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 33-8 20 Chapter 33 (a) (b) 21 For the vertices of the triangle, we choose the north pole and two points on a latitude line on nearly opposite sides of the Earth, as shown on the diagram Let the angle at the north pole be 179° It is not possible to draw a triangle on a sphere with the sum of the angles less than 180° To draw a triangle like that, a hyperbolic surface like a saddle (similar to Fig 33–18) would be needed 180° 179o 90° 90o 90°o 90 We find the time for the light to cross the elevator and then find how far the elevator moves during that time due to its acceleration Δt = Δx g (Δx) (9.80 m/s )(2.4 m) ; Δy = 12 g (Δt ) = = = 3.1× 10−16 m c 2c 2(3.00 × 108 m/s)2 Note that this is smaller than the size of a proton 22 Use Eq 33–4, Hubble’s law υ = H0d → d = 23 H0 = 1850 km/s = 88 Mly = 8.8 × 107 ly 21 km/s/Mly Use Eq 33–4, Hubble’s law υ = H0d → d = 24 υ (a) υ H0 = (0.015)(3.00 × 108 m/s) 2.1× 10 m/s/Mly = 214.3 Mly ≈ 210 Mly = 2.1× 108 ly Use Eq 33–6, applicable for υ c, to solve for the speed of the galaxy z= λobs − λrest υ ≈ c λrest ⎛ 455 nm − 434 nm ⎞ → υ = c⎜ ⎟ = 0.04839c ≈ 0.048c 434 nm ⎝ ⎠ The size of the answer does meet the condition that υ c (b) Use Hubble’s law, Eq 33–4, to solve for the distance υ = H0d → d = 25 υ H0 = 0.04839(3.00 × 108 m/s) = 691 Mly ≈ 6.9 × 108 ly 21000 m/s/Mly We find the velocity from Hubble’s law, Eq 33–4, and the observed wavelength from the Doppler shift, Eq 33–3 (a) υ c = H d (21, 000 m/s/Mly)(7.0 Mly) = = 4.9 × 10−4 c 3.00 × 108 m/s λ = λ0 (b) υ c = + υ /c + 4.9 × 10−4 = (656.3 nm) = 656.62 nm → shift = λ − λ0 = 0.3 nm − υ /c − 4.9 × 10−4 H d (21, 000 m/s/Mly)(70 Mly) = = 4.9 × 10−3 c 3.00 × 10 m/s λ = λ0 + υ /c + 4.9 × 10−3 = (656.3 nm) = 659.52 nm → shift = λ − λ0 = 3.2 nm /c 4.9 ì 103 â Copyright 2014 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Astrophysics and Cosmology 26 33-9 Use Eqs 33–3 and 33–4 to solve for the distance to the galaxy λobs = λrest d= υ H0 (λ − λ ) → υ =c (λ + λ ) ( λ − λ ) = (3.00 ×10 m/s) ⎡⎣(423.4 nm) ( λ + λ ) (2.1×10 m/s/Mly) ⎡⎣(423.4 nm) + υ /c − υ /c = c H0 obs rest obs rest obs rest obs rest − (393.4 nm) ⎤ ⎦ 2⎤ + (393.4 nm) ⎦ = 1.048 × 103 Mly ≈ 1.0 × 109 ly 27 Use Eqs 33–3 and 33–5a to solve for the speed of the galaxy z= λobs − λrest λobs + υ /c = −1 = −1 → − υ /c λrest λrest υ ( z + 1) − c = ( z + 1) + = 1.0602 − 1.0602 + = 0.05820 → υ = 0.058c The approximation of Eq 33–6 gives υ = zc = 0.060c 28 Use Eqs 33–3 and 33–5a to solve for the redshift parameter z= λobs − λrest λobs + υ /c + 0.075 = −1 = −1 = − = 0.078 − υ /c − 0.075 λrest λrest Or we can use the approximation given in Eq 33–6 z ≈ υ /c = 0.075 29 We cannot use the approximation of Eq 33–6 for this circumstance Instead, we use Eq 33–5b combined with Eq 33–3, and then use Eq 33–4 for the distance + υ /c + υ /c + υ /c −1 → z + = =2 → =4 − υ /c − υ /c − υ /c + υ /c = 4(1 − υ /c) → + υ /c = − 4υ /c → 5υ /c = → υ = 0.6c z= d= 30 υ H0 = 0.6(3.00 × 108 m/s) 21× 10 m/s/Mly = 8571 Mly ≈ × 109 ly Use Eqs 33–4 and 33–5a to solve for the speed of the galaxy Δλ = 610 nm − 434 nm = 176 nm 2 ⎛ Δλ ⎞ ⎛ 176 nm ⎞ + 1⎟ −1 ⎜ +1⎟ −1 ⎜ λ +υ /c υ ⎝ ⎠ 0.9755 Δλ ⎝ 434 nm ⎠ z= = −1 → = = = = 0.3278c → υ ≈ 0.33c c ⎛ Δλ ⎞2 λ0 −υ /c 2.9755 ⎛ 176 nm ⎞ +1⎟ +1 + 1⎟ + ⎜ ⎜ ⎝ 434 nm ⎠ ⎝ λ0 ⎠ Use Hubble’s law, Eq 33–4, to solve for the distance υ = H0d → d = υ H0 = υ /c H /c = 0.3278 = 4.68 × 103 Mly ≈ 4.7 × 109 ly (21000 m/s/Mly) 3.00 × 108 m/s © Copyright 2014 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 33-10 31 Chapter 33 For small relative wavelength shifts, we may use Eq 33–6 to find the speed We use Eq 33–4 to find the distance υ c = d= Δλ → υ =c λrest υ H0 = Δλ λrest ; υ = H0d → c Δλ ⎛ 3.00 × 108 m/s ⎞ ⎛ 0.10 cm ⎞ =⎜ ⎟⎜ ⎟ = 68 Mly H λrest ⎜⎝ 21, 000 m/s/Mly ⎟⎠ ⎝ 21 cm ⎠ We also use the more exact relationship of Eq 33–3 λobs + υ /c = − υ /c λrest ⎛ λobs ⎞ ⎛ 21.1 cm ⎞ −1 ⎜ ⎟ −1 ⎜ 21cm ⎠⎟ υ ⎝ λrest ⎠ ⎝ → = = = 4.75 × 10−2 2 c ⎛λ ⎛ 21.1 cm ⎞ obs ⎞ ⎜ ⎟ + ⎜ 21 cm ⎟ + λ ⎝ ⎠ ⎝ rest ⎠ υ = H0d → d = υ H0 = υ /c = H /c 4.75 × 10−2 = 67.86 Mly ≈ 68 Mly ( 21000 m/s/Mly ) 3.00 × 108 m/s 32 Eq 33–3 states λ = λrest + υ /c − υ /c 1/2 + υ /c ⎛ υ⎞ = λrest ⎜1 + ⎟ − υ /c ⎝ c⎠ λ = λrest 1/2 ⎛ υ⎞ ⎜1 − ⎟ ⎝ c⎠ υ ⎞⎛ υ⎞ υ⎞ ⎛ ⎛ ≈ λrest ⎜ + 12 ⎟ ⎜ − − 12 ⎟ = λrest ⎜ + 12 ⎟ c ⎠⎝ c⎠ c⎠ ⎝ ⎝ υ ⎛ ⎛ υ ⎞⎞ ⎛ υ⎞ λ = λrest ⎜1 + ⎜ 12 ⎟ ⎟ = λrest ⎜ + ⎟ = λrest + λrest c c c ⎝ ⎝ ⎠⎠ ⎝ ⎠ ( ) → λ − λrest = Δλ = λrest υ c → Δλ υ = λrest c 33 Wien’s law is Eq 27–2 λPT = 2.90 × 10−3 m ⋅ K → λP = 34 2.90 × 10−3 m ⋅ K 2.90 × 10−3 m ⋅ K = = 1.1× 10−3 m T 2.7 K We use Wien’s law, Eq 27–2 From Fig 33–29, the temperature at that time is about 1010 K λPT = 2.90 × 10−3 m ⋅ K → λP = 2.90 × 10−3 m ⋅ K 2.90 × 10−3 m ⋅ K = = × 10−13 m 10 T 10 K From Fig 22–8, that wavelength is in the gamma ray region of the EM spectrum 35 We use the proton as typical nuclear matter ⎛ −26 kg ⎞ ⎛ nucleon ⎞ ⎟ = nucleons/m ⎜10 ⎟⎜ m3 ⎠ ⎜⎝ 1.67 ì 1027 kg â Copyright 2014 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Astrophysics and Cosmology 36 33-11 If the scale of the universe is inversely proportional to the temperature, then the scale times the temperature should be constant If we call the current scale and know the current temperature to be about K, then the product of scale and temperature should be about Use Fig 33–29 to estimate the temperature at various times For purposes of illustration, we assume that the universe has a current size of about 1010 ly There will be some variation in the answer due to differences in reading the figure (a) At t = 106 yr, the temperature is about 1000 K Thus the scale is found as follows: (Scale)(Temperature) = → Scale = 3 = = × 10−3 Temperature 1000 → Size ≈ (3 × 10−3 )(1010 ly) = × 107 ly (b) At t = s, the temperature is about 1010 K Scale = (c) 3 = = × 10−10 Temperature 1010 → Size ≈ (3 ×10−10 )(1010 ly) = ly At t = 10−6 s, the temperature is about 1012 K Scale = 3 = = × 10−12 Temperature 1012 → Size ≈ (3 × 10−12 )(1010 ly) = × 10−2 ly ≈ × 1014 m (d) At t = 10235 s , the temperature is about 1027 K Scale = 3 = = × 10−27 Temperature 1027 Size ≈ (3 × 10−27 )(1010 ly) = × 10−17 ly ≈ 0.3 m 37 We approximate the temperature–energy relationship by kT = E = mc as suggested in Section 33–7 kT = mc (a) T= → T= mc k mc (500 MeV/c )c (1.60 ×10−13 J/MeV) = = × 1012 K k 1.38 × 10−23 J/K From Fig 33–29, this corresponds to a time of ≈ 10−5 s (b) T= mc (9500 MeV/c )c (1.60 ×10−13 J/MeV) = = 1× 1014 K −23 k 1.38 × 10 J/K From Fig 33–29, this corresponds to a time of ≈ 10−7 s (c) T= mc (100 MeV/c )c (1.60 × 10−13 J/MeV) = = 1× 1012 K −23 k 1.38 × 10 J/K From Fig 33–29, this corresponds to a time of ≈ 10−4 s There will be some variation in the answers due to differences in reading the figure © Copyright 2014 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 33-12 38 Chapter 33 (a) According to the textbook, near Fig 33–33, the visible matter makes up about one-tenth of the total baryonic matter The average baryonic density is therefore 10 times the density of visible matter The data in the Problem are for visible matter only (stars and galaxies) ρ baryon = 10 ρvisible = 10 = 10 (b) M visible π R3 11 (1011 galaxies)(10 stars/galaxy)(2.0 × 1030 kg/star) 4π ⎡(14 × 109 ly)(9.46 × 1015 m/ly) ⎤ ⎣ ⎦ ≈ 2.1× 10−26 kg/m3 Again, according to the text, dark matter is about times more plentiful than baryonic matter ρ dark = 5ρ baryon = 5(2.055 ×10−26 kg/m3 ) ≈ 1.0 ×10−25 kg/m3 39 The angular momentum is the product of the rotational inertia and the angular velocity ( I ω ) initial = ( I ω ) final → 2 ⎞ ⎛ R initial ⎞ ⎛ × 106 m ⎞ ⎟ = ω initial ⎜ = (1 rev/month) ⎜⎜ ⎟ ⎟⎟ ⎜ R final ⎟ ⎟ ⎝ × 10 m ⎠ ⎝ ⎠ ⎠ rev month 1d 1h = 5.625 × 105 rev/month = 5.625 × 105 × × × = 0.217 rev/s month 30 d 24 h 3600 s ⎛ I initial ⎝ I final ω final = ω initial ⎜ ⎛ MR initial ⎞ ⎜ = ω ⎟ initial ⎜ MR ⎠ final ⎝ ≈ 0.2 rev/s 40 The rotational kinetic energy is given by I ω The final angular velocity, from Problem 39, is 5.625 × 105 rev/month KE final KE initial = 1I ω final 1I ω initial final initial = MR ω final MR initialω ⎛ final =⎜ ⎜ ⎝ initial R finalω final ⎞ ⎟ R initialω initial ⎟⎠ 2 ⎛ (8 × 103 m)(5.625 × 105 rev/month) ⎞ 5 ⎟ = 5.625 × 10 ≈ × 10 ⎜ ⎟ (6 × 10 m)(1 rev/month) ⎝ ⎠ =⎜ 41 A: B: C: The temperature increases, the luminosity stays the same, and the size decreases The temperature stays the same, the luminosity decreases, and the size decreases The temperature decreases, the luminosity increases, and the size increases 42 The apparent luminosity is given by Eq 33–1 Use that relationship to derive an expression for the absolute luminosity, and equate the Sun’s luminosity to the star’s luminosity b= L 4π d → L = 4π d 2b Ldistant = LSun star d distant = dSun star 2 bdistant = 4π dSun bSun → 4π d distant star → star lSun ⎛ ly ⎞ = (1.5 × 1011 m) ⎟ = ly 15 −11 ⎜ ldistant 10 ⎝ 9.461× 10 m ⎠ star © Copyright 2014 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Astrophysics and Cosmology 43 The power output is the energy loss divided by the elapsed time P= = 44 33-13 ΔKE = Δt I ω (fraction lost) Δt = 25 MR 2ω (fraction lost) Δt (1.5)(1.99 × 10 kg)(8.0 ×10 m) (2π rad/s) (1× 10−9 ) = 1.7461025 W ≈ 1.7 × 1025 W (1 d)(24 h/d)(3600 s/h) 30 Use Newton’s law of universal gravitation F =G m1m2 r (4 × 1041 kg) = (6.67 × 10−11 N ⋅ m /kg ) ⎡(2 × 106 ly)(9.46 × 1015 m/ly) ⎤ ⎣ ⎦ = 2.98 × 1028 N ≈ × 1028 N 45 We use the Sun’s mass and given density to calculate the size of the Sun ρ= M M = V π r3 Sun 1/3 ⎛ 3M ⎞ rSun = ⎜ ⎟ ⎝ 4πρ ⎠ rSun d Earth-Sun 46 = → 1/3 ⎡ 3(1.99 × 1030 kg) ⎤ =⎢ ⎥ −26 kg/m3 ) ⎥⎦ ⎢⎣ 4π (10 3.62 × 1018 m 11 1.50 × 10 m ≈ × 107 ; 1ly ⎛ ⎞ = 3.62 × 1018 m ⎜ ⎟ = 382 ly ≈ 400 ly 15 ⎝ 9.46 × 10 m ⎠ rSun dgalaxy = 382 ly ≈ × 10−3 100,000 ly The temperature of each star can be found from Wien’s law, Eq 27–2 The peak wavelength is used as a subscript to designate each star’s properties λPT = 2.90 × 10−3 m ⋅ K → T660 = 2.90 × 10−3 m ⋅ K 660 × 10−9 m = 4394 K T480 = 2.90 × 10−3 m ⋅ K 480 × 10−9 m = 6042 K The luminosity of each star can be found from the H–R diagram L660 ≈ × 1025 W L480 ≈ × 1026 W The Stefan-Boltzmann equation says that the power output of a star is given by P = β AT , where β is a constant and A is the radiating area The P in the Stefan-Boltzmann equation is the same as the luminosity L given in Eq 33–1 Form the ratio of the two luminosities 4 L480 β A480T480 4π r480 T480 = = 4 L660 β A660T660 4π r660 T660 → r480 = r660 L480 T660 × 1026 W (4394 K) = = 1.413 L660 T480 × 1025 W (6042 K) The diameters are in the same ratio as the radii d 480 = 1.413 ≈ 1.4 d660 The luminosities are fairly subjective, since they are read from the H–R diagram Different answers may arise due to different readings of the H–R diagram © Copyright 2014 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 33-14 47 Chapter 33 (a) First, calculate the number of parsecs Then use the fact that the number of parsecs is the reciprocal of the angular resolution in seconds of arc ⎛ pc ⎞ 100 ly ⎜ ⎟ = 30.67 pc = φ0 ⎝ 3.26 ly ⎠ → ⎛ ⎞′′ ⎛ 1′ ⎞ ⎛ 1° ⎞ −6 −6 ⎟ ⎜ ⎟⎜ ⎟ = (9.06 × 10 )° ≈ (9 × 10 )° ⎝ 30.67 ⎠ ⎝ 600 ⎠ ⎝ 60′ ⎠ φ =⎜ (b) We use the Rayleigh criterion, Eq 25–7, which relates the angular resolution to the diameter of the optical element We choose a wavelength of 550 nm, in the middle of the visible range θ= 1.22λ D → D= 1.22λ θ = 1.22(550 × 10−9 m) = 4.24 m ≈ m ⎡(9.06 × 10−6 )°⎤ (π rad/180°) ⎣ ⎦ The largest optical telescopes with single mirrors are about m in diameter 48 (a) We approximate the temperature–kinetic energy relationship by kT = KE, as given in Section 33–7 kT = KE → T = (b) 49 (a) k = (14 × 1012 eV)(1.60 × 10−19 J/eV) 1.38 × 10 J/K = 1.6 × 1017 K Find the Q-value for this reaction From Eq 30–2, the Q-value is the mass energy of the reactants minus the mass energy of the products Q= 24 12 Mg 2mC c − mMg c = → [ 2(12.000000 u) − 23.985042 u ] c (931.5 MeV/c ) = 13.93 MeV The total kinetic energy should be equal to the electrical potential energy of the two nuclei when they are just touching The distance between the two nuclei will be twice the nuclear radius, from Eq 30–1 Each nucleus will have half the total kinetic energy r = (1.2 ×10−15 m)( A)1/3 = (1.2 × 10−15 m)(12)1/3 ; PE = qnucleus ; 4πε 2r qnucleus 4πε 2r KE nucleus = PE = 1 = 12 (8.988 ×109 N ⋅ m /C2 ) (c) −23 From Fig 33–29, this is in the hadron era 12 12 6C + 6C (b) KE (6) (1.60 × 10−19 C) 2(1.2 × 10 −15 1/3 m)(12) × eV 1.60 × 10−19 J = 4.71 MeV Approximate the temperature–kinetic energy relationship by kT = KE, as given in Section 33–7 ⎛ 1.60 × 10−19 J ⎞ (4.71× 106 eV) ⎜ ⎟⎟ ⎜ eV KE ⎝ ⎠ = 5.46 × 1010 K = kT = KE → T = 23 k 1.38 ì 10 J/K â Copyright 2014 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Astrophysics and Cosmology 50 (a) 33-15 Find the Q-value for this reaction From Eq 30–2, the Q-value is the mass energy of the reactants minus the mass energy of the products 16 16 8C + 8C Q= 28 14 Si + He 2mC c − mSi c − mHe c → = [ 2(15.994915 u) − 27.976927 u − 4.002603] c (931.5 Mev/c ) = 9.594 MeV (b) The total kinetic energy should be equal to the electrical potential energy of the two nuclei when they are just touching The distance between the two nuclei will be twice the nuclear radius, from Eq 30–1 Each nucleus will have half the total kinetic energy r = (1.2 × 10−15 m)( A)1/3 = (1.2 × 10−15 m)(16)1/3 ; qnucleus 4πε 2r qnucleus 4πε 2r KE nucleus = PE = = 12 (8.988 × 109 N ⋅ m /C ) (c) PE = (8) (1.60 × 10−19 C) 2(1.2 × 10 −15 1/3 m)(16) × eV = 7.61 MeV 1.60 × 10−19 J Approximate the temperature–kinetic energy relationship by kT = KE, as given in Section 33–7 ⎛ 1.60 × 10−19 J ⎞ (7.61× 106 eV) ⎜ ⎟⎟ ⎜ eV KE ⎝ ⎠ = 8.82 × 1010 K kT = KE → T = = k 1.38 × 10−23 J/K 51 We must find a combination of c, G, and U that has the dimensions of time The dimensions of c are ⎡ L3 ⎤ ⎡ ML2 ⎤ ⎡L⎤ , and the dimensions of = are ⎥ ⎢ ⎥ Use dimensional ⎢ T ⎥ , the dimensions of G are ⎢ ⎣ ⎦ ⎣⎢ MT ⎦⎥ ⎣⎢ T ⎦⎥ analysis, as discussed in Chapter β γ α ⎡ L ⎤ ⎡ L ⎤ ⎡ ML ⎤ α + 3β + 2γ t P = cα G β =γ → [T ] = ⎢ ⎥ ⎢ [ M ]γ − β [T ]−α − β −γ ⎥ ⎢ ⎥ = [ L] T T ⎣ ⎦ ⎣⎢ MT ⎦⎥ ⎣⎢ ⎦⎥ α + 3β + 2γ = 0; γ − β = 0; − α − 2β − γ = → α + 5β = 0; α = −1 − 3β → β= −5β = −1 − 3β tP = c 52 −5/2 1/2 1/2 G = = G= c5 1; = γ= 1; α =− → (6.63 × 10−34 J ⋅ s) 2π = 5.38 ×10−44 s (3.00 × 108 m/s)5 (6.67 × 10−11 N ⋅ m /kg ) The radius of the universe is estimated as the speed of light (c) times the age of the universe (T) This radius is used to find the volume of the universe, and the critical density times the volume gives the mass r = cT ; V = 43 π r = 34 π (cT )3 ; m =Vρ = π (cT )3 ρ = 4π ⎡ ⎛ 3.156 × 107 ⎢(3.00 × 10 m/s)(13.8 × 10 yr) ⎜⎜ yr ⎝ ⎣⎢ s ⎞⎤ −26 kg/m3 ) ⎟⎟ ⎥ (10 ⎠ ⎦⎥ = 9.34 × 1052 kg ≈ 1053 kg © Copyright 2014 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 33-16 Chapter 33 Solutions to Search and Learn Problems (a) Assume that the nucleons make up only 2% of the critical mass density, so neutrinos make up 98% of the critical mass density Nucleon mass density = 0.02(10−26 kg/m3 ) Nucleon number density = 0.02(10−26 kg/m3 ) 1.67 ×10 −27 kg/nucleon = 0.12 nucleon/m3 Neutrino number density = 10 (nucleon number density) = 1.2 × 108 neutrino/m3 The neutrino number density times the mass per neutrino must make up 98% of the critical mass density The mass-to- eV/c conversion is in the front of the textbook (mv kg/neutrino)(1.2 × 108 neutrino/m3 ) = 0.98(10−26 kg/m3 ) → mv = (b) 0.98(10−26 kg/m3 ) ⎛ 931.5 × 106 eV/c ⎞ ⎜ ⎟ = 45.8 eV/v ≈ 50 eV/v (1.2 × 108 neutrino/m3 ) ⎜⎝ 1.6605 × 10−27 kg ⎟⎠ Assume that the nucleons make up only 5% of the critical mass density Nucleon mass density = 0.05(10−26 kg/m3 ) Nucleon number density = 0.05(10−26 kg/m3 ) 1.67 ×10 −27 kg/nucleon = 0.30 nucleon/m3 Neutrino number density = 10 (nucleon number density) = 3.0 ×108 neutrino/m3 (mv kg/neutrino)(3.0 × 108 neutrino/m3 ) = 0.95(10−26 kg/m3 ) → mv = 0.95(10−26 kg/m3 ) ⎛ 931.5 ×106 eV/c ⎞ ⎜ ⎟ = 17.8 eV/v ≈ 20 eV/v (3.0 × 108 neutrino/m3 ) ⎜⎝ 1.6605 × 10−27 kg ⎟⎠ Many methods are available • For nearby stars (up to 500 or 100 ly away) we can use parallax In this method we measure the angular distance that a star moves relative to the background of stars as the Earth travels around the Sun Half of the angular displacement is then equal to the ratio of the Earth–Sun distance and the distance between the Earth and that star • The apparent brightness of the brightest stars in galaxies, combined with the inverse square law, can be used to estimate distances to galaxies, assuming they have the same intrinsic luminosity • The H–R diagram can be used for distant stars Determine the surface temperature using its blackbody radiation spectrum and Wien’s law and then estimate its luminosity from the H–R diagram Using its apparent brightness with Eq 33–1 will give its distance • Variable stars, like Cepheid variables, can be used by relating the period to its luminosity The luminosity and apparent brightness can be used to find the distance • The largest distances are measured by measuring the apparent brightness of Type Ia supernovae All supernovae are thought to have nearly the same luminosity, so the apparent brightness can be used to find the distance • The redshift in the spectral lines of very distant galaxies can be used to estimate distances that are farther than 107 to 108 ly © Copyright 2014 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher Astrophysics and Cosmology (a) 33-17 From Section 33–2, a white dwarf with a mass equal to that of the Sun has a radius about the size of the Earth’s radius, 6380 km , and a neutron star with a mass equal to 1.5 solar masses has a radius of about 20 km For the black hole, we use the Schwarzschild radius formula R= 2GM c2 2(6.67 × 10−11 N ⋅ m /kg ) ⎡3(1.99 × 1030 kg) ⎤ ⎣ ⎦ = 8849 m ≈ 8.85 km = (3.00 × 108 m/s) (b) The ratio is 6380 : 20 : 8.85 = 721: 2.26 :1 ≈ 700 : :1 (a) All distant objects in the universe are moving away from each other, as indicated by the galactic redshift, indicating that the universe is expanding If the universe has always expanded, it must have started as a point The 25% abundance of He supports the standard Big Bang model The Big Bang theory predicted the presence of background radiation, which has since been observed The curvature of the universe determines whether the universe will continue expanding forever (open) or eventually collapse back in on itself (closed) Dark energy increases the total energy of the universe, increasing the probability that it is an open universe (b) (c) Each helium atom requires protons and neutrons and has a total mass of 4u Each hydrogen atom requires proton and has a mass of 1u If there are times more protons than neutrons, then for every neutrons there are 14 protons Two protons combine with the neutrons to produce a helium atom The other 12 protons produce 12 hydrogen atoms Therefore, there is 12 times as much hydrogen as helium, by number of atoms Each helium atom has a mass times that of the hydrogen atom, so the total mass of the hydrogen is only times the total mass of the helium About 380,000 years after the Big Bang, the initially very hot temperature of the universe would have cooled down to about 3000 K At that temperature, electrons could orbit bare nuclei and remain there, without being ejected by collisions Thus stable atoms were able to form This allowed photons to travel unimpeded through the universe, so the universe became transparent The radiation at this time period would have been blackbody radiation at a temperature of about 3000 K As the universe continued to expand and cool, the wavelengths of this radiation would have increased according to Wien’s law The peak wavelength now is much larger and corresponds to the blackbody radiation at a temperature of 2.7 K (a) Because the speed of the galaxy is small compared to the speed of light, we can use Eq 33–6 instead of Eq 33–3 The velocity of the galaxy is negative ⎛ −130 × 103 m/s ⎞ ⎛υ ⎞ Δλ = λrest ⎜ ⎟ = 656 nm ⎜ = −0.284 nm ⎜ 3.00 × 108 m/s ⎟⎟ ⎝c⎠ ⎝ ⎠ (b) (c) This is a blue shift because the wavelength has decreased and because the galaxy is approaching The time is equal to the distance between the galaxies divided by their relative speed This is assuming that they continue to move at the same present speed (They would actually accelerate due to the gravitational force between them.) t= d υ = 2.5 × 106 ly ⎛ 9.46 × 1015 m ⎞ ⎛ yr ⎜⎜ ⎟⎟ ⎜⎜ ly 130 × 10 m/s ⎝ ⎠ ⎝ 3.156 × 10 ⎞ ⎟ = 5.8 × 10 yr s ⎟⎠ © Copyright 2014 Pearson Education, Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher ...INSTRUCTOR SOLUTIONS MANUAL VOLUMES & VOLUME DOUGLAS C GIANCOLI S PHYSICS PRINCIPLES WITH APPLICATIONS 7TH EDITION BOB DAVIS TAYLOR UNIVERSITY J ERIK HENDRICKSON... chapters – 15 of Physics: Principles with Applications, 7th Edition, by Douglas C Giancoli At the end of the manual are grids that correlate the 6th edition questions and problems to the 7th edition... reproduced, in any form or by any means, without permission in writing from the publisher PREFACE This Instructor’s Solutions Manual provides answers and worked-out solutions to all end of chapter questions