Chapter 12: GravityAnswers to Even-Numbered Conceptual Questions2. A person passing you on the street exerts a gravitational force on you, but it is so weak (about 10–7 N or less)that it is imperceptible.4. As the tips of the fingers approach one another, we can think of them as two small spheres (or we canreplace the finger tips with two small marbles if we like). As we know, the net gravitational attractionoutside a sphere of mass is the same as that of an equivalent point mass at its center. Therefore, the twofingers simply experience the finite force of two point masses separated by a finite distance.6. No. A satellite must be moving relative to the center of the Earth to maintain its orbit, but the North Pole isat rest relative to the center of the Earth. Therefore, a satellite cannot remain fixed above the North Pole.8. Yes. The rotational motion of the Earth is to the east, and therefore if you launch in that direction you areadding the speed of the Earth’s rotation to the speed of your rocket.10. As the astronauts approach a mascon, its increased gravitational attraction would increase the speed of thespacecraft. Similarly, as they pass the mascon, its gravitational attraction would now be in the backwarddirection, which would decrease their speed.12. It makes more sense to think of the Moon as orbiting the Sun, with the Earth providing a smaller force thatmakes the Moon “wobble” back and forth in its solar orbit.Solutions to Problems and Conceptual Exercises1. Picture the Problem: Three systems contain different masses that are separated by different distances.Strategy: Use the Universal Law of Gravity (equation 12-1) to determine the ranking of the force magnitudes.Solution: 1. Apply equation 12-1: FA=Gm1m2rA2=Gm2r22. Repeat for system B: FB=Gm 2m( )2r( )2=Gm22r23. Repeat for system C: FC=G2m( )3m( )2r( )2=G3m22r24. Repeat for system D: FD=G4m( )5m( )3r( )2=G20m29r25. By comparing the forces we arrive at the ranking B < A < C < D.Insight: Changing the magnitude of the separation distance has the greatest effect upon the force.2. Picture the Problem: The two apples attract each other gravitationally.Strategy: Use the Universal Law of Gravity (equation 12-1) to find the force between the apples.Solution: 1. (a) Apply equation 12-1: F =Gm1m2r2= 6.67×10−11 N ⋅m2/kg2( )0.16 kg( )0.16 kg( )0.25 m( )2= 2.7×10−11 NCopyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. Noportion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.12 – 1Chapter 12: Gravity James S. Walker, Physics, 4th Edition2. (b) Repeat for the new distance: F =Gm1m2r2= 6.67×10−11 N ⋅m2/kg2( )0.16 kg( )0.16 kg( )0.50 m( )2= 6.8×10−12 NInsight: Doubling the distance between the apples cut the gravitational force by a factor of 4.Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. Noportion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.12 – 2Chapter 12: Gravity James S. Walker, Physics, 4th Edition3. Picture the Problem: The two bowling balls attract each other gravitationally.Strategy: Use the Universal Law of Gravity (equation 12-1) to find the force between the bowling balls, then solve the same equation for distance to answer part (b).Solution: 1. (a) Apply equation 12-1: F =Gm1m2r2= 6.67×10−11 N ⋅m2/kg2( )6.1 kg( )7.2 kg( )0.75 m( )2= 5.2×10−9 N2. (b) Solve equation 12-1 for r: r =Gm1m2F=6.67×10−11 N ⋅m2/kg2( )6.1 kg( )7.2 kg( )2.0×10−9 N=1.2 mInsight: Increasing the distance between the balls from 0.75 m to 1.2 m decreased the force from 5.2 nN to 2.0 nN.4. Picture the Problem: The Earth and the satellite attract each other gravitationally.Strategy: Use equation 5-5 to find the weight of the satellite on the Earth’s surface, and then equation 12-1 to find the gravitational force on the satellite while it is in orbit.Solution: 1. (a) Apply equation 5-5: Ws=mg = 480 kg( )9.81 m/s2( )=4700 N = 4.7 kN2. (b) Apply equation 12-1 directly: F =GmMEr2= 6.67×10−11 N ⋅m2/kg2( )480 kg( )5.97×1024 kg( )35×106 m( )2= 0.16 kNInsight: The distance between the satellite and the center of the Earth has increased by a factor of 5.5 so that the gravitational force on the satellite has decreased by a factor of 5.52 = 30.5. Picture the Problem: You and the asteroid attract each other gravitationally.Strategy: Estimate that your mass ≈70 kg. Apply equation 12-1 to find the gravitational force between you and Ceres.Solution: Apply equation 12-1: F =Gm1m2r2= 6.67×10−11 N ⋅m2/kg2( )8.7×1020 kg( )70 kg( )14×106 m( )2= 0.021 NInsight: If you stood on the surface of Ceres (radius 500 km) the force would be 16 N (3.6 lb).6. Picture the Problem: The apple and the orange attract each other gravitationally.Strategy: Use the Universal Law of Gravity (equation 12-1) to find the force between the two fruits.Solution: 1. (a) Apply equation 12-1: F =Gm1m2r2= 6.67×10−11 N ⋅m2/kg2( )0.11 kg( )0.24 kg( )0.85 m( )2= 2.4×10−12 N2. (b) The force the apple exerts on the orange is equal and opposite to the force the orange exerts on the apple, so its magnitude must be 2.4 ×10−12 N. Insight: Halving the distance between the two fruits would quadruple the force between them, but it would still be tiny.Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. Noportion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.12 – 3Chapter 12: Gravity James S. Walker, Physics, 4th Edition7. Picture the Problem: The spaceship is attracted gravitationally to both the Earth and the Moon.Strategy: Use the Universal Law of Gravity (equation 12-1) to relate the attractive forces from the Earth and the Moon. Set the force due to the Earth equal to twice the force due to the Moon when the spaceship is at a distance r from the center of the Earth. Let R =3.84×108 m, the distance between the centers of the Earth and Moon. Then solve the expression for the distance r.Solution: 1. (a) Set FE=2FM usingequation 12-1 and solve for r: GmsmEr2=2GmsmMR−r( )2mER −r( )2=2mMr2R −r = 2mMmErr =R1+ 2mMmE=3.84×108 m1+ 2×7.35×1022