ch 12 sol

CHAPTER 12 Gravitation Answers to Understanding the Concepts Questions 1.

In this chapter we mentioned friction and spring forces. Other nonfundamental forces include: drag forces in fluids or gases, the tension in ropes or solid rods, the normal force that opposes gravity for objects at rest. These forces are all associated on the microscopic level with the electromagnetic forces that hold matter together.

2.

The initial speed of a cannonball is not nearly as much as what it would take to escape Earth, so the range of motion of the cannonball is very short compared with the size of Earth. For this reason when calculating the altitude reached by a cannon ball we assumed a flat Earth with uniform acceleration of gravity, which resulted in the angle dependency of the altitude. Should it be fired at some sufficient speed, the cannonball would indeed escape Earth, regardless of the launch angle –– assuming that the tremendous speed does not cause it to burn out in the atmosphere. In arriving at that result we of course have to take into consideration the inverse-squared nature of the acceleration of gravity.

3.

Tidal forces arise from the difference between the attraction to the Moon (and Sun) on one side of Earth and that on the other. Near the North Pole this difference is very tiny; it would actually vanish at the pole if the Sun and Moon moved in the plane of the equator and there would be no tidal effects at the pole.

4.

French Guiana is located in the equatorial region, where the linear speed due to the spin of Earth is the greatest anywhere on its surface. If a satellite is launched in the same general direction of the rotation of Earth, it can take the greatest advantage of this rotation to achieve a sufficient speed.

5.

If the satellite is launched near the equator. in the direction of Earth’s rotation, then it starts off with a velocity relative to the center of Earth that is the sum of the velocity of the launch pad due to Earth’s rotation and the velocity it acquires as a result of the acceleration produced by the burning fuel. This can be a substantial help. The science-fiction writer Arthur C. Clarke has pointed out that the ideal launching pad would be from the top of a very large mountain on the equator in Sri Lanka (Ceylon).

6.

As the shuttle orbits around Earth, the astronauts are subject to Earth’s gravitational pull, and that force acts as the centripetal force for the circular motion of the astronauts about the center of Earth. The same is true for the shuttle itself. Therefore the shuttle and the astronauts experience the same acceleration (of gravity), i.e., they fall together; and shuttle does not need to exert a supporting force on the astronauts, who can then float in the shuttle cabin.

7.

No. The constant depends on the mass of the object being revolved around. For a Jovian satellite it would depend on the mass of Jupiter, while for a planet in the solar system it depends on the mass of the Sun.

8.

The gravitational acceleration at the surface of a neutron star is incredibly large, beyond what anybody can possibly tolerate. In fact g = GM/R2, where M is over one solar mass, or about 107 the mass of Earth (M e); while R is only a few miles, or about 10–3 that of Earth (R E ). So g ≈ G(107 ME )/(10–3 RE ) 2 = 1013 (GME /RE 2) = 1013 gE . Nobody can make it there.

9.

r Consider a celestial object of mass m moving with velocity v . Draw a line passing through the center of r the Sun, parallel to v . Call the distance between v and that line r. Then L = mvr about the center f the

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Fishbane, Gasiorowicz, and Thornton

Sun. If L = 0 then r = 0, meaning that the object is either headed straight toward the center of the Sun or moving directly away from it. 10. Zero, if we assume that Earth is a uniform sphere. This is because the tungsten piece is being pulled evenly by Earth in all directions, resulting in a zero net gravitational force. 11. Angular momentum is conserved for any central force. and the text shows that this is equivalent to the equal areas in equal times law. The r-dependence is irrelevant for this purpose. 12. Burning occurs when an object, be it a satellite or a meteor, reaches excessive speeds in the atmosphere and encounters a substantial amount of friction, which quickly raises the temperature of the object. As a satellite returns to Earth it is being accelerated toward Earth, so its speed is even higher than that while it is in orbit. And it moves the fastest as it reaches the lower atmosphere where the air is denser (which makes it more vulnerable to burning). On its way up, however, the satellite is powered by a rocket and its speed gradually increases due to the propulsion of the rocket, passing through the denser, lower atmosphere at relatively low speeds. It does not reach orbital speed until it leaves the atmosphere and the rocket engines are shut down. So the speed of the satellite climbing up to its orbit is considerably less than its speed upon reentry into the atmosphere, and that is why it does not burn on its way up. 13. This means that, as the Moon completes one revolution around Earth (which takes about 27.3 days) it also completes one rotation about its own axis. So the length of a “day” on the Moon is about 27.3 days. This is due to the tidal force between the Moon and Earth. 14. The tidal force is due to the difference in gravitational attraction exerted by either the Moon or the Sun on the near- and far-sides of Earth. Since the gravitational force drops like r-2, the difference drops like D(dr-2/dr), which depends on r as r-3, which decreases more rapidly than r-2 as r increases. (Here D is the diameter of Earth.) The distance between the Sun and Earth is far greater than that between the Moon and Earth, which is the reason why the tidal force due to the Sun is less than that due to the Moon. 15. No astronaut to date has ever ventured far enough from Earth so as to be virtually free from its gravitational pull. A space shuttle, for example, cruises at only a couple of hundred miles above the surface of Earth, where the acceleration of gravity is not substantially different from that on the ground. So yes, they are indeed still subject to Earth’s gravitational force. The reason why they can float inside the spacecraft is because both the astronauts and the spacecraft are accelerating at the same rate (g) towards the center of Earth. (This does not mean they are crashing down; the acceleration of gravity serves as the centripetal acceleration that enables them to circle around the planet.) This is very much analogous to the case of a person being able to “float” inside an elevator whose cables have failed and is falling freely to the ground. 16. All three answers are correct. The Moon has no atmosphere, which means that no energy has to be expended to overcome the air friction. Also, the acceleration of gravity on the surface of the Moon is only about 1/6 that on Earth, so the escape speed from the Moon is considerably less than that on Earth. Answer (c ) is also correct. In fact it means that we can use the Moon as a staging ground, effectively a giant space station, to launch deep space missions. 17. In the text you learned that a spherically symmetric mass distribution exerts the same force on an external object as the same mass concentrated at the center point of the distribution. Let’s consider the simplest case of inhomogeneity, namely a small, point-like lump somewhere inside Earth. The mass distribution may be viewed as a superposition of a spherically symmetric mass distribution centered at Earth’s center plus a point mass somewhere inside Earth. Since forces add vectorially, the force may be viewed as one due to two masses, one large, centered at Earth’s center, and one small, at the lump location. The small mass will perturb the orbit in ways that are different for different orientations of © 2005 Pearson Education, Inc., Upper Saddle River, NJ. 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.

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Chapter 12: Gravitation

the orbit, and from the study of the orbit the location and magnitude of the small mass may be inferred. More general inhomogeneities may be constructed as a superposition of lumps, and the orbits will then be ones in the field of a massive point mass at Earth’s center accompanied by a number of smaller masses (some of which may have to be chosen negative!) in the vicinity of the large one. It should be noted that a spherical mass distribution –– a lump that is extended in a spherically symmetric way –– is equivalent to a point of the same total mass, but unless we know the density of the spherical distribution, we will not know its size. Thus the above construction still leaves us ignorant of some aspects of Earth’s interior. That is why seismology is essential for the detailed investigation of Earth’s mass distribution. 18. As far as the apparent weight is concerned, it is still greater at the poles, since this is where one does not have to consider the effect of the spin of Earth, which effectively causes an upward centrifugal force at the equator, reducing the apparent weight. The true weight of the person, which is the actual gravitational attraction of Earth, would be the same in magnitude wherever on Earth, if Earth were a uniform, perfect sphere. 19. If we assume uniform density for Earth, then surface measurements of g may yield information about the shape of the surface. The answer to Question 17 also applies to this question. 20. Since the impact delivers no angular original circular orbit impulse to the satellite, its angular where collision occurs momentum remains unchanged: L = mvθ r new elliptical orbit = constant, where vθ is the component of its velocity that is perpendicular to the radial direction. There is a linear impulse in the radial direction, however; center of Earth and that causes it to acquire a radial velocity (inward). As a result it moves inward, reducing r, just after the collision. This causes vθ to increase as L cannot change. The satellite’s orbit changes from circular to an elliptical one, with variable distance to the center of Earth and variable speed. But its angular momentum remains the same. 21. The correct answer is (b). The equation of motion for the two-mass system is M2 g – µk M1gsinθ = (M1 + M2)a, with µ k the coefficient of kinetic friction between the inclined plane and M1, and θ the angle of inclination. Divide each term by M 2 to obtain g – µk (M1/ M2)g sinθ = (1 + M1/ M2)a,, which gives a in terms of M1/ M2. 22. The Sun also causes (smaller) tides, so that if the Moon were not present the slowing down would take longer, but it would still occur. 23. If the surface of Jupiter were as rigid as that of Earth, then the length of a day, which is the time it takes to complete one rotation about its own axis, would the same throughout its surface. So the surface of Jupiter is not as rigid as that of Earth, although the small percent difference between the lengths of a day in the two regions of Jupiter suggests that the rigidity of its surface is not all that low, either. 24. Yes. The spin angular momentum of Earth is not conserved since Earth alone is not an isolated system. But the combined angular momentum of the Earth-Moon system is still conserved (if we neglect the effect of the Sun). 25. For a circular orbit the force is always perpendicular to the displacement, so that no work is done at any instant. For (closed) elliptical orbits, the net work done is also zero, since otherwise the object in orbit would lose or gain energy, giving rise to a change in the orbit. © 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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26. It takes an external torque to change the angular momentum of an object. In this case the asteroid does exert an external force on Earth –– through its center of mass, which is at the geometrical center of Earth since we assumed Earth to be an isotropic, spherically symmetric object. This force obviously provides no torque about Earth’s axis of rotation, which passes through its center; so the spin angular momentum of Earth cannot be changed by the passing asteroid. Therefore the rate of Earth’s rotation remains the same. 27. Suppose your spaceship is at rest in a region of negative potential energy. (This could happen if the spaceship left Earth with a velocity somewhat less than the terminal velocity, and reached the turnaround point). Can you eject some material and put your spaceship in a region of larger potential energy? From momentum conservation we can see that by ejecting some material in the direction of Earth, the spaceship could acquire a velocity in a direction away from Earth. This can only happen if there is some energy stored in the spaceship. It may be chemical or nuclear, but that energy can be converted into kinetic energy. The spaceship will then move to a region of higher potential energy, losing some of its newly acquired kinetic energy in the process. No law of nature is violated by this scenario. 28. If the cannonball’s speed exceeds that value but is still below the escape speed of Earth then the cannonball will not follow the curvature of the surface of Earth; rather, it would travel further than before while dropping the same vertical distance. The cannonball can thus reach a higher altitude before turning back. The resulting trajectory of the cannonball would no longer be circular but rather elliptical, with the center of Earth at one of its focal points. If the cannonball were to reach the escape speed of Earth then it would be able to travel to infinity, and the trajectory would be parabolic. If the speed is still greater then the trajectory becomes part of a hyperbola, and the cannonball once again would be able to reach infinity –– neglecting the effects of the Sun as well as all other stellar objects, of course. 29. According to Newton’s shell theorem, the acceleration of gravity inside a uniform spherical shell is zero. Everything would be weightless inside a hollow Earth. 30. Tides arise from the difference in gravitational attraction of the Moon (and, to a lesser degree, the Sun) on different parts of the ocean. Compared with a unit mass at the center of Earth, a unit mass on the surface of Earth that’s closer to the Moon (and the Sun) experiences greater attraction toward the Moon (and the Sun) and therefore the ocean surface in that region bulges out, creating a high tide. This effect is more dramatic when the Moon is aligned with the Sun on the same side. If they are on opposite sides, then the Moon will cause a bulge on the nearer side of Earth while the Sun causes another one opposite to it, and the region midway in between will have low tides.


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Solutions to Problems

1.

If we use the data for the Earth's orbit around the Sun, we have C = T2 /R3 = (3.156 × 107 s)2 /(1.496 × 101 1 m)3 = 2.975 × 10– 1 9 s2 /m3 .

2.

From Kepler’s third law, T2 = 4π2 R 3 /GM, we have TIo2 /RIo3 = TEuropa2 /REuropa3 ; (152,874 s)2 /(422,000 km)3 = T2 Europa/(671,400 km)3 , which gives TEuropa = 306,787 s.

3.

( a ) We write the dimensions of F(r) = h/r3 as [F] = [h] [r– 3] or [MLT– 2] = [h][L– 3], which gives [h] = [ML4 T– 2]. (b) For the circular motion, the central force provides the centripetal acceleration: F r = mar; h/R 3 = mv2 /R, which gives v = h/m /R. The angular momentum is L = mvR = m( h/m/R )R = mh . (c) The period of the circular motion is T = 2πR/v = 2πR/( h/m/R ) = 2π m/h R 2 , which we write as T/R2 = 2π m/h = a constant.

4.

r r r r r We can write the angular momentum as L = r × p = r × mv = constant. r r r r From the definition of the cross product, we know that r is perpendicular to L , and v = drr / d t is r r r r perpendicular to L . Consequently, r and changes in r lie in the plane perpendicular to L so that r must remain in this plane.

5. ( a ) The gravitational force between them is F = GMm/r2 = (6.67 × 10– 1 1 N· m 2 /kg2 )(95 kg)(68 kg)/(0.48 m)2 = 1.9 × 10– 6 N. (b) The attraction toward each other must be the same. Remember Newton’s third law. 6.

The gravitational force between the proton and the electron is F = GMm/r2 = (6.67 × 10– 1 1 N · m 2 /kg2 )(1.67 × 10– 2 7 kg)(9.1 × 10– 3 1 kg)/(0.6 × 10– 1 0 m)2 = 2.8 × 10– 4 7 N .

7.

The gravitational force between the spheres is F = GMm/r2 = (6.67 × 10– 1 1 N · m 2 /kg2 )(1 kg)(1 kg)/(0.40 m)2 = 4.2 × 10– 1 0 N . We estimate the weight of a fly: (0.1 × 10– 3 kg)(9.8 m/s2 ) ≈ 1 × 10– 3 N, so the attraction is ≈ 10– 7 W fly .

8.

The acceleration due to gravity on the surface of a planet is g = GM/R2 . ( a ) gMoon = (6.67 × 10––11 N · m 2 /kg2 )(7.35 × 102 2 kg)/(1.74 × 106 m)2 = 1.62 m/s2 . (b) gMars = (6.67 × 10– 1 1 N · m 2 /kg2 )(6.42 × 102 3 kg)/(3.40 × 106 m)2 = 3.70 m/s2 . (c) gJupiter = (6.67 × 10– 1 1 N · m 2 /kg2 )(1.90 × 102 7 kg)/(7.14 × 107 m)2 = 24.9 m/s2 . (d) gSun = (6.67 × 10– 1 1 N · m 2 /kg2 )(1.99 × 103 0 kg)/(6.96 × 108 m)2 = 274 m/s2 .

© 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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9. We find the radius from Kepler’s third law: T2 = 4π2 R 3 /GM ; [(90 min)(60 s/min)]2 = 4π2 R 3 /[(6.67 × 10– 1 1 N · m 2 /kg2 )(5.98 × 1024 kg)], which gives R = 6.6 × 106 m.

10. Using the data from Problem 8, the radius of the circle is R = 1.74 × 106 m + 0.09× 106 m = 1.83 × 106 m. We find the period from Kepler’s third law: T2 = 4π2 R 3 /GM = 4π2 (1.83 × 106 m)3 /[(6.67 × 10– 1 1 N · m 2 /kg2 )(7.35 × 1022 kg)], which gives T = 7.02 × 103 s = 1.95 h .

11. We find the period from Kepler’s third law: T2 = 4π2 R 3 /GM = 4π2 (6.67 × 106 m)3 /[(6.67 × 10– 1 1 N · m 2 /kg2 )(5.98 × 102 4 kg)], which gives T = 5.42 × 103 s = 1.51 h.

12. The acceleration due to gravity on the surface of a planet is g = GM/R2 = GρV/R2 = Gρ()πR3 )/R2 = )πGρR = )π(6.67 × 10– 1 1 N · m 2 /kg2 )(5400 kg/m3 )(15 × 103 m) = 0.023 m/s2 .

13. We express the kinetic energy as K = !mv2 = !m(2πR/T)2 = 2mπ2 R 2 /T2 . When we combine this with Kepler's third law, T2 = CR3 , we get K = 2mπ2 /CR. For the two orbits, we have R 2 /R1 = K1 /K2 = !, so R 2 = !R1 = ![3(6.37 × 106 m)] = 9.56 × 106 m.

14. We assume that the weight lifter can exert the same force: F = mg = m′g′, so we have m′ = (g/g′)m. ( a ) mMoon = [(9.8 m/s2 )/(1.62 m/s2 )](138 kg) = 835 kg . (b) mSun = [(9.8 m/s2 )/(274 m/s2 )](138 kg) = 4.9 kg .

15. With the reference level for potential energy at the ground, we use energy conservation to relate the maximum height to the initial speed: K i + Ui = Kf + Uf; ! m v0 2 + 0 = 0 + mgh, which gives v0 2 = 2gh. Because we assume that the initial speed is constant, with gMars from Problem 8, we have gMars h Mars = gE h E , or h Mars = (gE /gMars) h E = [(9.8 m/s2 )/(3.70 m/s2 )](1.85 m) = 4.9 m .


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16. Because the gravitational force is always attractive, the two forces will be in opposite directions. If we let x be the distance from the smaller mass M2 where the magnitudes of the forces are equal, we have GM1 m/(R – x)2 = GM2 m/x2 , which becomes M1 x2 = M2 (R – x)2 . ( a ) For the Earth-Sun system, we have MSx2 = ME(R – x)2 ; (1.99 × 103 0 kg)x2 = (5.98 × 102 4 kg)[(1.496 × 1011 m) – x]2 , which gives x = 2.60× 108 m from the Earth’s center. (b) For the M-2M system, we have Mx2 = 2M(R – x)2 ; x2 = 2(R – x)2 , which gives x = 0.41 R from M, where R is the distance between M and 2M. 17. The semimajor axis is a = !(rmax + rmin), and the eccentricity is e = !(rmax – rmin)/a = (rmax – rmin)/(rmax + rmin). Thus (r max – rmin)/(rmax + rmin) = e = 0.25. 18. The semimajor axis is a = !(rmax + rmin), and the eccentricity is e = !(rmax – rmin)/a . Thus rmax = a(1 + e) = (149.6 × 106 km)(1 + 0.017) = 1.521 × 108 km. 19. Because the escape speed is the speed at the surface necessary to get far away with no final speed, we use energy conservation, with the reference level for potential energy at infinity: K i + Ui = Kf + Uf; ! m vesc2 – GMsm/Rs = 0 + 0, which gives vesc = 2GM/R =

–11

2(6.67 × 10

2

30

8

N·m 2/kg )(1.99 × 10 kg)/(6.96 × 10 m) =

5

6.18 × 10 m/s.

20. From Problem 19, we have vesc = 2GM/R . ( a ) For the Moon, we have vesc,Moon = 2(6.67 × 10–11 N· m2/kg2)(7.35 × 1022 kg)/(1.74 × 106 m) = 2.37 × 103 m/s. (b) For Mars, we have vesc,Mars = 2(6.67 × 10–11 N· m2/kg2)(6.42 × 1023 kg)/(3.40 × 106 m) = 5.02 × 103 m/s. (c) For Jupiter, we have vesc,Jupiter = 2(6.67 × 10–11 N·m 2/kg 2)(1.90 × 1027 kg)/(7.14 × 107 m) = 5.96 × 104 m/s.

21. From Problem 19, we have v esc = (2GM/R)1/2= (8GπρR2/3 )1/2 =[8π (6.67×10-11 Nm2/kg2)(5400kg/m3)(1.5×104 m)2/3]1/2 = 26 m/s.

22. From Problem 19, we have vesc = 2GM/R . For the two systems we have v esc, neutron = vesc, Earth[(Mn/ME )(R E /Rn)]1/2 = (11.2 × 103 m/s) [(1.8×105)(750)]1/2 = 1.3 × 108 m/s. (Here for simplicity we pretended that we could neglect relativistic effects. See Chapter 39.)


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23. For the material to just barely stay on the surface, the gravitational attraction must provide the centripetal acceleration for the circular motion: GMm/R2 = mv2 /R = mRω2 , so we have ω 2 = GM/R3 = Gρ()πR3 )/R3 = )Gρπ = )(6.67 × 10– 1 1 N · m 2 /kg2 )(4800 kg/m3 )π, which gives ω = 1.6 × 10– 3 rad/s.

24. We use conservation of energy, with the reference level for potential energy at infinity: K i + Ui = Kf + Uf; ! m v0 2 – GMm/R = 0 – GMm/(R + h), which gives R + h = RGM/(GM – !v0 2 R) = R/[1 – v0 2 R/(2GM)] = R/(1 – v0 2 /vesc2 ) . We use the escape speed from the Earth: h = R/(1 – v0 2 /vesc2 ) – R = R[1/(1 – v0 2 /vesc2 ) – 1] = (6.37 × 106 m){1/[1 – 6.8 km/s)2 /(11.2 km/s)2 ] – 1} = 3.7 × 106 m.

25. We use conservation of energy, with the reference level for potential energy at infinity: K i + Ui = Kf + Uf; ! m v0 2 – GMm/D = !mv2 – GMm/R, or v2 = v0 2 + 2GM(1/R – 1/D); v 2 = (2000 m/s)2 + 2(6.67 × 10– 1 1 N · m 2 /kg2 )(5.98 × 1024 kg)[1/(6.37 × 106 m) – 1/(8.00 × 107 m)], which gives v = 1.09 × 104 m/s.

26. We use conservation of energy, with the reference level for potential energy at infinity: K i + Ui = Kf + Uf; ! m v0 2 – GMm/R = 0 + 0, which gives

! m v0 2 = GMm/R = (6.67 × 10– 1 1 N·m2 /kg2 )(7.35 × 102 2 kg)(3800 kg)/(1.74 × 106 m) = 1.07 × 101 0 J.

27. From Kepler’s third law, T2 = 4π2 R 3 /GM, we can relate the periods at different radii: (T/TMoon) 2 = (R/RMoon) 3 . ( a ) (T/TMoon) 2 = (1/3)2 = [R1 /(3.84 × 108 m)]3 , which gives R 1 = 1.85 × 108 m. (b) (T/TMoon) 2 = 32 = [R2 /(3.84 × 108 m)]3 , which gives R 2 = 7.99 × 108 m. (c) (T/TMoon) 2 = (1/1000)2 = [R3 /(3.84 × 108 m)]3 , which gives R 3 = 3.84 × 106 m, which is less than the radius of the Earth and thus impossible.

28. From the analysis in Problem 27, we have (T/TMoon) 2 = (R/RMoon) 3 ; (T/27.32 d)2 ={[(6.37 + 0.37) × 106 m]/(3.84 × 108 m)}3 , which gives T = 0.0636 d = 1.53 h . Because the gravitational force that provides the centripetal acceleration is proportional to the mass, the relationship between period and radius does not depend on the mass. The period would still be 1.53 h.


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29. For Newton’s second law, we have GMm/r2 = mv2 /r, or v 2 = GM/r = (6.67 × 10– 1 1 N · m 2 /kg2 )(5.98 × 102 4 kg)/[(6.37 + 0.20) × 106 m], which gives v = 7.68 × 103 m/s = 7.79 km/s tangent to the orbit. For the kinetic energy, we get K = !mv2 = !(500 kg)(7.79 × 103 m/s)2 = 1.52 × 101 0 J. For the angular momentum, we get L = mvr = (500 kg)(7.79 × 103 m/s)(6.57 × 106 m) = 2.56 × 101 3 kg · m2 /s perpendicular to the orbit.

30. ( a ) For the satellite to stay over the same point, it must rotate with the Earth, so we have T = 24 h = 8.64 × 104 s, and ω = 2π/T = 2π/(8.69 × 104 s) = 7.27 × 10– 5/s. (b) The gravitational force must provide the centripetal acceleration: GMm/r2 = mrω2 , or r = (GM/ω2 ) 1/3 = [(6.67 × 10– 1 1 N · m 2 /kg2 )(5.98 × 102 4 kg)/(7.27 × 10– 5/s)2 ]1/3 = 4.22 × 107 m. (c) For Kepler’s third law we have satellite: C = T 2 /R3 = (8.64 × 104 s)2 /(4.22 × 107 m)3 = 9.9 × 10– 1 4 s2 /m3 . Moon: C = T 2 /R3 = [(27.3 day)(8.64 × 104 s/day)]2 /(3.84 × 108 m)3 = 9.8 × 10–14 s2 /m3 .

31. From Kepler’s third law, we have T 2 = 4π2 R 3 /GM = 4π2 [(3393 + 95) × 103 m]3 /[(6.67 × 10– 1 1 N · m 2 /kg2 )(6.42 × 1023 kg)], which gives T = 6.25 × 103 s = 1.74 h.

32. From Kepler’s third law, we have T2 = 4π2 R 3 /GM; [(27.3 d)(24 h/d)(3600 s/h)]2 = 4π2 R 3 /[(6.67 × 10– 1 1 N · m 2 /kg2 )(7.35 × 102 2 kg)], which gives R = 3.83 × 108 m.

33. From Kepler’s third law, we have T2 = 4π2 R 3 /GM; [(365 d)(24 h/d)(3600 s/h)]2 = 4π2 R 3 /[(6.67 × 10– 1 1 N · m 2 /kg2 )(1.99 × 103 0 kg)], which gives R = 1.49 × 101 1 m.

34. The period of the satellite (1 day = 86400 s) satisfies T2 = 4π2r3/GM, so the orbital radius is r = (GMT2/4π2) 1/3 and the altitude of the orbital is h = r – RE = (GMT2/4π2) 1/3 – RE = [(6.67 × 10– 1 1 N · m2 /kg2 )(5.96 × 102 4 kg)(86400 s)2/4π2]1/3 – 6.37 × 106 m = 3.58 × 107 m = 35800 km. Suppose that it passes over a certain point A over the equator at a given moment of the day. 12 hours later, the satellite once again passes over the equator, at a point diametrically opposite to the initial location of point A. Meanwhile point A has just completed half a revolution about the axis of Earth and also reaches the same spot on the equator, diametrically opposite to its initial location. So the satellite passes over the same point A on the equator, twice a day. For any other latitude, the satellite passes over one spot (B) each day and a different spot (C) later in the same day, but the same two spots (B and C) are passed over and over every 24 hours.


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35. To escape from the solar system the escape speed needed, vesc, satisfies !Mvesc2 – GmME/RE – GmMS/rES = 0, or vesc = [2G (ME /RE + MS /rES )]1/2 . Plug in ME = 5.98 × 1024 kg, MS = 1.99 × 1030 kg, RE = 6.38 × 106 m, and rES = 1.50 × 1011 m, to obtain vesc = 43.5 km/s. Earth moves once around the Sun in a near-circular orbital of radius r = 1.5 × 101 1 m in t = one year = 3.156 × 107 s, so its (average) orbital speed is V E = 2π rES /t = 2π (1.5 × 101 1 m)/(3.156 × 107 s) = 2.99 × 104 m/s = 29.9 km/s, relative to the Sun. To escape from the Sun with a minimum possible speed vmin, launch the r r rocket r in the same direction as v E , so the speed of the rocket relative to the Sun is v = vE + vmin. (If v E and v min are not aligned then v would be less.) Set v = vesc to obtain v min = vesc – vE = 43.5 km/s – 29.9 km/s = 13.4 km/s ≈ 13 km/s . (To be more accurate, one needs to consider the linear speed of Earth at the equator due to its spin. This is 0.464 km/s. So if the launch takes place at the equator we may further decrease vmin by that amount. But since rES is known to only 2 significant figures this does not affect the answer of 13 km/s.) 36. ( a ) For the conservation of angular momentum, we have L = mv0 R = mvr, which gives v = (R/r)v0 . For the conservation of energy, we have K i + Ui = Kf + Uf; ! m v0 2 – GMm/R = !mv2 – GMm/r; v 0 2 – 2GM/R = v2 – 2GM/r; Using the escape speed, given by vesc2 = 2GM/R, and the result from angular momentum conservation, we get v 0 2 – vesc2 = (R/r)2 v 0 2 – vesc2 (R/r), which we can write as a quadratic equation for r/R: (v 0 2 – vesc2 )(r/R )2 + vesc2 (r/R ) – v0 2 = 0; [(10.5 km/s)2 – (11.2 km/s)2 ](r/R )2 + (11.2 km/s)2 (r/R ) – (10.5 km/s)2 = 0, which gives r/R = 7.33 or 1.0. The maximum distance from the center of the Earth is r = (7.33)(6.4 × 103 km) = 47 × 103 km. (b) When the satellite is fired vertically, the angular momentum is zero. For energy conservation, we have: K i + Ui = Kf + Uf; ! m v0 2 – GMm/R = 0 – GMm/r ; v 0 2 – vesc2 = – vesc2 R/r, (10.5 km/s)2 – (11.2 km/s)2 = – (11.2 km/s)2 (R/r), which gives r/R = 8.33. The maximum distance from the center of the Earth is r = (8.33)(6.4 × 103 km) = 53 × 103 km. (c) When the satellite is fired at an angle of 30°, the angular momentum is L = mv0 (cos 41°) R = mvr, which gives v = 0.755(R/r)v0 . (d) For the conservation of energy, we have (b) K i + Ui = Kf + Uf; (c) ! m v0 2 – GMm/R = !mv2 – GMm/r; N v 0 2 – vesc2 = [0.755v0 (R/r)]2 – vesc2 (R/r), which we can write as a quadratic equation for r/R : S (v 0 2 – vesc2 )(r/R)2 + vesc2 (r/R) – 0.570v0 2 = 0; [(10.5 km/s)2 – (11.2 km/s)2 ](r/R)2 + (11.2 km/s)2 (r/R) – 0.570(10.5 km/s)2 = 0, (a) which gives r/R = 7.79. The maximum distance from the center of the Earth is r = (7.79)(6.4 × 103 km) = 50 × 103 km. © 2005 Pearson Education, Inc., Upper Saddle River, NJ. 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.

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37. For the conservation of angular momentum, we have L = mv (cos 45°) R = mvfr, which gives v = (r/R) vf/cos 45°. For the conservation of energy, we have K i + Ui = Kf + Uf; ! m v2 – GMm/R = !mvf2 – GMm/r; Using the escape speed, given by vesc2 = 2GM/R, and the result from angular momentum conservation, with r = 2R, we get [(r/R) vf/cos 45°]2 – vesc2 = vf2 – vesc2 (R/r); [2vf/cos 45°]2 – vesc2 = vf2 – !vesc2 , which reduces to v f2 = vesc2 /14 = (11.2 km/s)2 /14, which gives v f = 2.99 km/s. 38. For the conservation of energy, we have K a + Ua = Kp + Up ; ! m va2 – GMm/ra = !mvp 2 – GMm/rp , which can be rearranged to ! v p 2 – !va2 = GM (1/rp – 1/ra) ; !(81 × 103 m/s)2 – !(54 × 103 m/s)2 = (6.67 × 10– 1 1 N · m2 /kg2 )M[1/(4 × 101 0 m) – 1/(6 × 101 0 m)], which gives M = 3.2 × 103 0 kg. 39. ( a ) r v0

d

r vmax

(b) At the nearest approach, the speed will be maximum. For the conservation of angular momentum, we have L = mv0 d = mvmaxrmin , which gives rmin = v0 d/vmax . For the conservation of energy, we have K 0 + U0 = K + U; ! m v0 2 + 0 = !mvmax2 – GMm/rmin . Using the result from angular momentum conservation, we get vmax2 – 2GMvmax/v0 d – v0 2 = 0; v max2 – 2(6.67 × 10– 1 1 N · m 2 /kg2 )(1.99 × 1030 kg)v max/[(12 × 103 m/s)(3 × 101 1 m)] – (12 × 103 m/s)2 = 0. Solving this quadratic equation for vmax , we get v max = – 1.92 × 103 m/s, and + 7.56 × 104 m/s. The positive answer is the physical result: vmax = 75.6 km/s perpendicular to the radius. (c) From the conservation of angular momentum, we get rmin = v0 d/vmax = (12 km/s)(3 × 108 km)/(75.6 km/s) = 4.76 × 107 km. © 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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40. With v 1 the initial speed, before the firing, the angular momentum is L 1 = mv1 R 1 . In the circular orbit, for Newton’s second law, we have GMm/R1 2 = mv1 2 /R1 , or v1 2 = GM/R1 . If the firing does not change the total energy, the speed immediately after firing is still v1 , since there has been no change in the position and thus no change in the potential energy. The firing changed the direction of the satellite to decrease the angular momentum: L 2 = mv1 R 1 cos θ = !mv1 R 1 . At apogee and perigee, the velocity is perpendicular to the radius, so the angular momentum is L 2 = mvr = !mv1 R 1 , which gives v = !(R1 /r)v1 . Using the result for v1 2 , we can write this as v2 = (GM/4R1 )(R 1 /r)2 . For the conservation of energy, we have K 1 + U1 = K + U; ! m v1 2 – GMm/R1 = !mv2 – GMm/r; !GM/R1 – GM/R1 = !(GM/4R1 )(R 1 /r)2 – GMm/r, which reduces to (R 1 /r)2 – 8(R1 /r) + 4 = 0. When we solve this quadratic equation, we get R 1 /r = 7.46 and 0.336, which gives r = 0.134R1 , 1.86R1 .

41. ( a ) In the circular orbit, for Newton’s second law, we have GMm/r0 2 = mv0 2 /r0 , or v 0 2 = GM/r0 = (6.67 × 10– 1 1 N · m 2 /kg2 )(6 × 1024 kg)/[(6.37 + 2.00) × 106 m], which gives v 0 = 6.91 × 103 m/s. (b) The angular momentum is L = mv0 r0 = (300 kg)(6.91 × 103 m/s)(8.37 × 106 m) = 1.74 × 101 3 kg · m2 /s. (c) If the direction of motion immediately after the firing is unchanged and the speed is v1 = !v0 , the angular momentum is L = mv1 r0 = !mv0 r0 = !(1.74 × 101 3 kg · m2 /s2 ) = 8.68 × 101 2 kg · m2 /s. (d) The orbit will be elliptical. For the conservation of angular momentum for the new orbit, at the lowest point of the orbit we have mvr = L = mv1 r0 = !mv0 r0 , which gives v = !(r0 /r)v0 . For the conservation of energy, we have K 1 + U1 = K + U; ! m v1 2 – GMm/r0 = !mv2 – GMm/r; ! ( ! v 0 ) 2 – GM/r0 = !(!v0 ) 2 (r 0 /r)2 – GM/r. Using the expression from part (a), we have (1/8)(GM/r0 ) – GM/r0 = (1/8)(GM/r0 )(r 0 /r)2 – GM/r, which reduces to (r 0 /r)2 – 8(r0 /r) + 7 = 0. When we solve this quadratic equation, we get r0 /r = 1 and 7. Thus the perigee is r0 /7 = (8.37 × 106 m)/7 = 1.20 × 106 m. Because this is less than the radius of the Earth, the satellite crashes.


Page 12-12


42. Because the comet’s motion is periodic, with varying distance from the Sun, the orbit is elliptical. ( a ) We use Kepler’s third law to find the semimajor axis: T2 /a3 = 4π2 /GM; [(76 yr)(3.16 × 107 s/yr)]2 /a3 = 4π2 /[(6.67 × 10– 1 1 N · m 2 /kg2 )(1.99 × 1030 kg)], which gives a = 2.67 × 101 2 m. (b) From the properties of an ellipse, we use rmin = a(1 – e); 8.9 × 1010 m = (2.67 × 101 2 m)(1 – e), which gives e = 0.967. (c) To find the aphelion, we use rmax = a(1 + e) = (2.67 × 101 2 m)(1 + 0.967) = 5.25 × 101 2 m.

43. ( a ) In the circular orbit, for Newton’s second law, we have GMm/r0 2 = mv0 2 /r0 , or v 0 2 = GM/r0 = (6.67 × 10– 1 1 N · m 2 /kg2 )(5.98 × 1024 kg)/[(6.37 + 0.30) × 106 m], which gives v 0 = 7.73 × 103 m/s. (b) The angular momentum is L = mv0 r0 = (2000 kg)(7.73 × 103 m/s)(6.67 × 106 m) = 1.03 × 101 4 kg · m2 /s. (c) The total energy is E = !mv0 2 – GMm/r0 = – !GMm/r0 = – !(6.67 × 10– 1 1 N · m 2 /kg2 )(5.98 × 1024 kg)(2000 kg)/(6.67 × 106 m) = – 5.98 × 101 0 J. (d) Because the force is radial, the torque is zero. (e) Because the torque is zero, the angular momentum will be conserved: mvr = mv0 r0 . This shows that the speed will decrease as r increases, but vr will remain constant. From part (a) we see that for circular orbits, v2 r is constant. If r changes, both of these cannot be true. The orbit cannot be circular.

44. Because we are outside a spherically symmetric mass, we have mg = GMm/r2 , or g = GM/r2 . The height of the mountain is h = (14 × 103 ft)(0.305 m/ft) = 4.27 × 103 m. Because this is so small compared to the radius of the Earth, we can approximate the changes by differentials: ∆g/∆r ≈ dg/dr = – 2GM/r3 = – 2g/r. We see that g decreases with height; the magnitude of the change in g is ∆g ≈ 2gh/r = 2g(4.27 × 103 m)/(6.37 × 106 m) = 1.34 × 10– 3 g.

45. For the approximate form, we have g(h)/g(0) ≈ 1 – 2h/RE = 1 – 2(10 × 103 m)/(6.37 × 106 m) = 0.996860. For the exact form, we have g(h)/g(0) = R E2 /(RE + h) 2 = (6.37 × 106 m)2 /[(6.37 + 0.01) × 106 m]2 = 0.996868 .


Page 12-13


h = (37000 ft)(0.305 m/ft) = 1.13 × 104 m, where g (h)/g(0) ≈ 1 – 2h/ RE = 1 – 2(1.13 × 104 m)/(6.37 × 106 m) = 0.99645, so ∆g = g(h) – g(0) = (1 – 0.99645)(9.8 m/s2) = – 0.035 m/s2, i.e., the acceleration of gravity is reduced by 0.035 m/s2 , or 0.35%. (b) Zero gravity occurs at 9/10 of the way to the Moon (see Example 12-4), i.e, at r = (9/10)dM E = (9/10)(3.84 × 108 m) = 3.46 × 108 m from Earth, where the acceleration of gravity due to Earth is g (r) = g(0) (RE /r)2 = g(0) [6.37 × 106 m/(3.46 × 108 m)]2 = 3.40 × 10– 4 g(0), so g is reduced to 0.034% of its value at the surface of Earth, i.e., by 1 – 0.034% = 99.97% .

46. (a)

47. The speed v of the object upon arriving at the surface of the sphere is found from conservation of energy: Ei (at infinity)= E f (on the surface of the sphere), or 0 = !mv2 – GMm/R; so v = (2GM/R)1/2. Once it enters the hollow sphere, the gravitational acceleration of the object becomes zero (due to Newton’s shell theorem), and the speed of the object does not change until it emerges from the other end of the sphere. Since the object has no angular momentum with respect to the sphere it must be aiming directly at the center of the sphere and therefore it travels a distance of 2R inside the sphere. The time it takes to travel through the sphere is then t = 2R/v = 2R/(2GM/R)1/2 = (2R3/GM)1/2 . 48. From Example 12-9, the attractive force on a mass m inside the sphere is F = – GMinsidem/r2 = – )πGmρr = – GmMr /R3 . To lift the mass, we apply a force opposite to this. We integrate to find the work: R

W= R/2

=

R

F · dr = 3 (1 8

R/2

GmM r dr = GmM R3 R3

1 2

R 2–

1 2

R 2

2

= 38 GmM R

kg)(6.67 × 10 – 11 N·m 2/kg 2)(1 kg)(5.98 × 10 24 kg) = 2.36 × 10 7 J. 6.37 × 10 6 m

49. Let the center-to-center distance between Jupiter and the Sun be dSJ. The gravitational force exerted by the Sun on an object of mass m at the center of Jupiter is Fg = GMSm/dSJ2. The tidal force on Jupiter is due to the difference in Fg between the near- and far-sides of Jupiter relative to the Sun: F tide ≈ GMS m/(dSJ – RJ) 2 – GMS m/(dSJ + RJ) 2 ≈ 2GMS m RJ /dSJ3. Thus the ratio of the tidal forces on Jupiter and on Earth is (2GMS m RJ /dSJ3)/ (2GMS m RE /dSE3) = (RJ / RE )(d SE/dSJ) 3 = (11.2)(1/5.2)3 = 0.080 = 8.0%, i.e., the tidal force on Jupiter is about 8% that on Earth. 50. The ratio of the gravitational accelerations of Earth due to the Moon and the Sun is gEM / gES = (GMM ME /dE M2)/(GMS ME /dES2) = (MM /MS)( dES /dE M) 2 = [7.36 × 102 2 kg/(1.99 × 103 0 kg)][1.50 × 101 1 m/(3.82 × 108 m)]2 = 5.7 × 10– 3 ; and the ratio of the gravitational accelerations of Earth due to Jupiter and the Moon is gEJ / gE M = (GMJME /dEJ2)/(G MM ME /dE M2) = (MJ /ME )( dEM /dEJ ) 2 ≈ [1.90 × 102 7 kg/(7.36 × 102 2 kg)][3.82 × 108 m/(6.28 × 101 1 m)]2 ≈ 0.01. Note that, the mean distance between Jupiter and the Sun is 7.78 × 101 1 m, while that between Earth and the Sun is 1.50 × 108 m. The distance between Earth and Jupiter varies, depending upon the position of each of the planets in their respective orbital. We chose to take the difference between the two orbital radii for d EJ –– that would be the closest possible distance between Earth and Jupiter, at which gEJ would be the greatest. Even so, it is only about 1% of gEM .


Page 12-14


51. From Example 12–9, we have F = – )πGmρr. The spring force, F = – kx, has a potential energy: !kx2 . Because the gravitational force in the tunnel has the same form, it also has a potential energy, given by U = !()πGmρ)r2 , with U = 0 at the center. Because the density is ρ = M/()πR3 ), this becomes U = !(GMm/R3 )r 2 . For conservation of energy as the mass falls from the surface to the center, we have K i + Ui = Kf + Uf; 0 + !(GMm/R3 )R 2 = !mv2 + 0; (6.67 × 10– 1 1 N · m 2 /kg2 )(5.98 × 1024 kg)/(6.37 × 106 m) = v2 , which gives v = 7.9 × 103 m/s. Note that the result is independent of the mass m. 52. ( a ) From Example 12–9, we have F = – )πGmρr. Because the density is ρ = M/()πR3 ), this becomes F = – (GMm/R3 )r. Tunnel Because this force is radial, we can use the definition of potential energy from Section 7–2, with the origin as our x m reference point and U(0) = 0: r

U = U(0) –

r

F · dr′ = –

– (GMm/R 3) r′ dr′, which gives

Earth

d

Re

0 0 r U = GMmr2 /2R3 , U = 0 at r = 0. (b) In terms of x, the result from part (a) is U = GMm(x2 + d2 )/2R3 . To change our reference level, we can add a constant C and have U(x = 0) = 0: U = GMm(x2 + d2 )/2R3 + C; 0 = (GMmd2 /2R3 ) + C, which gives C = – GMmd2 /2R3 , and U = GMmx2 /2R3 , with U = 0 at x = 0. We could also integrate the component of the force along the tunnel to get the same result.

53. We choose the origin at the Sun’s center, so we have X = ∑mixi/∑mi = [MS(0) + ME (d)]/(MS + ME ) = [ 0 + (5.98 × 102 4 kg)(1.50 × 108 km)]/(1.99 × 103 0 kg + 5.98 × 102 4 kg) = 4.51 × 102 km from the Sun’s center, which is inside the Sun. 54. The ratio of gravitational forces is F Saturn/FSun = [GMSaturnMJupiter/(RSaturn – R Jupiter) 2 ]/(GMSunMJupiter/RJupiter2 ) = (MSaturn/MJupiter)[R Jupiter/(RSaturn – RJupiter)]2 = (0.00029)[(7.78 × 108 km)/(14.2 × 108 km – 7.78 × 108 km)]2 = 0.000426 . 55. The acceleration on the pendulum bob due to the truck (of mass m) a distance r away from the bob is g’r≈ Gm/r2 ≈ (6.67 × 10– 1 1 N · m2 /kg2 )(20000 kg)/(20 m)2 = 3 × 10– 9 m/s2. r Since g ’ is horizontal and perpendicular to g , the pendulum bob will make a small angle θ with the vertical, where θ ≈ tan θ = g’/g ≈ (3 × 10– 9 m/s2)/(9.8 m/s2) = 3 × 10– 1 0 rad. Thus g’/g ≈ θ ≈ 3 /101 1, which is the same order of magnitude as 1/101 1, so yes, one does have to worry about the trucks near the lab.


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56. The value of C is about the same for Earth (E) and Mars (M). So gE h E ≈ gM h M , and the highest mountain peak on Mars would be h M ≈ gE h E / gM = (9.8 m/s2)(10 km)/(3.7 m/s2) = 26 km, in good agreement with 25 km. The C value for the Moon is (gE /6)(2 km)(hE /10 km) = gE h E /30, i.e., it is only 1/30 that of Earth. This suggests that the rocks of which the Moon is made is considerably different from those of Earth, in the sense that they take significantly less energy to melt. 57. The gravitational force provides the centripetal acceleration: GMM/(2R)2 = MRω2 , which gives ω 2 = GM/4R3 . The period of the motion is T = 2π /ω = 4π (R 3/GM)1/2 . 58. The pilot would not be able to tell the difference between an increased value of gravitational force and the effect of a noninertial centrifugal force.

r F

R

R

cm

r F

r F r mg

59. The time for the laser beam to cross the room is t = L/c. In this time the beam will fall a distance y = !gt2 = !g(L/c)2 = !(9.8 m/s2 )[(8 m)/(3 × 108 m/s)]2 = 3.5 × 10– 1 5 m. Because the size of an atom ˛ 10– 1 0 m, this is not a feasible experiment to test Einstein’s theory. 60. The time for the laser beam to cross the room is t = w/c. In this time the elevator will move a distance y = !gt2 = !g(w/c)2 , and the light ray will appear to have deflected this distance. We find the angle of deflection from tan θ = y/w = gw/2c2 . 61. We want the magnitude of the centripetal acceleration to have the magnitude of terrestrial gravity: a = rω 2 = g; 9.8 m/s2 = (60 m)ω2 , which gives ω = 0.40 rad/s. 62. ( a ) For the acceleration due to gravity, we have g = GM/R2 = Gρ)πR3 /R2 = 4GρπR/3. With the same density, 1.6 times the radius of Earth, g2 = 1.6g1 = 1.6(9.8 m/s2 ) = 16 m/s2 . (b) When we use Kepler’s third law, T2 = 4π2 R 3 /GM, for the two Moons of the same planet, we get (T Io/TGanymede) 2 = (RIo/RGanymede) 3 ; [TIo/(7.16 d)]2 = [(262,000 mi)/(660,000 mi)]3 , which gives TIo = 1.79 d. (c) The period of a planet is T = 2πR/v. When we use Kepler’s third law, T2 = 4π2 R 3 /GM, we get (T A/TB) 2 = (RAv B/RBv A) 2 = (RA/RB) 3 , which gives R A/RB = (vB/vA) 2 = (3/2)2 = 2.25. Thus RA = 2.25RB . 63. ( a ) For a "weightless" circular orbit, the gravitational force provides the centripetal acceleration: GM/R2 = )πGmρR = mv0 2 /R, or R 2 = 3v0 2 /4πGρ = 3(2.0 m/s)2 /[4π(6.67 × 10– 1 1 N · m 2 /kg2 )(5.2 × 103 kg/m3 )], which gives R = 1.66 × 103 m. (b) The escape speed is v esc = (2GM/R)1/2 = v0 √2 = (2.0 m/s)√2 = 2.8 m/s. (c) The surface speed at the equator is v = 2πR/T = 2π(1.66 × 103 m)/[(12 h)(3600 s/h)] = 0.24 m/s. By walking in the direction of the rotation, he would need a speed of 1.76 m/s to orbit the asteroid. © 2005 Pearson Education, Inc., Upper Saddle River, NJ. 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.

Page 12-16


64. ( a ) g = GM/R2 = (6.67 × 10– 1 1 N · m 2 /kg2 )(5.4 × 103 0 kg)/(12 × 103 m)2 = 2.5 × 101 2 m/s2 . (b) The gravitational force is F = GMm/r2 . Because a change in r of 1 mm is much smaller than the radius of 10 km, we approximate the changes by differentials: ∆F/∆r =#dF/dr = – 2GMm/r3 = – 2mg/r. Thus ∆F = – 2mg ∆r/r = – 2(1 × 10– 3 kg)(2.5 × 101 2 m/s2 )(1 × 10– 3 m)/(12 × 103 m) = – 0.4 kN.

65. ( a ) For a circular orbit, we must have F = k/rn = ma = mv2 /r, which will be satisfied for a speed of v = k/mrn – 1 . Circular orbits are supported. (b) The period of the circular motion is T = 2πr/v = 2πr/ k/mrn– 1 , which we write as T2 /rn+1 = 4π2 (m/k) = a constant .

66. ( a ) The total mass of the cluster is Mc = ρV = ρ()πR3 ) . Because the galaxy of mass M is at the edge, the cluster can be treated as a mass Mc at the center. The energy of M is E = K + U = !Mv2 – GMMc/R = !Mv2 – )πρGMR2 . (b) For the conservation of energy, we have K i + Ui = Kf + Uf. At the critical density, at infinity the final kinetic and potential energies will be zero, so we have ! M(HR)2 – )πρcGMR2 = 0, which gives ρc = 3H2 /8πG = 3[(15 km)/(106 ly)(103 m/km)(1 ly/9.46 × 101 5 m)]2 /[8π(6.67 × 10– 1 1 N · m 2 /kg2 )] = 4.5 × 10– 2 7 kg/m3 .

67. For a circular orbit, the attractive gravitational force provides the centripetal acceleration: GMm/r2 = mv2 /r, which gives v2 = GM/r. The energies are K = !mv2 = GMm/2r, U = – GMm/r, and E = K + U = GMm/2r – GMm/r = – GMm/2r. If the radius changes slowly, we can approximate the change in energy by differentiating: ∆E ≈ dE = – (– GMm/2r2 ) dr ≈ (GMm/2r2 ) ∆r. For the work-energy theorem, for one revolution, we have W f = ∆E; – f(2πr) = (GMm/2r2 ) ∆r, which gives ∆r = – 4πfr3 /GMm. The changes in the energies are ∆U ≈ dU = – (– GMm/r2 ) dr ≈ (GMm ∆r)/r2 = – 4πfr. ∆E = – 2πfr. ∆K ≈ dK = (– GMm/2r2 ) dr ≈ – (GMm ∆r)/r2 = 2πf r.


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68. ( a ) For a circular orbit, the attractive electric force provides the centripetal acceleration: e2 /r2 = mv2 /r, which gives v2 = e2 /mr. Using the gravitational force as an analogy, with GMm replaced by e2 , this inverse-square force is also conservative, with a potential energy of U = – e2 /r. The total energy is E = K + U = !mv2 – e2 /r = !e2 /r – e2 /r = – !e2 /r; – 1.6 × 10–18 J = – !(2.3 × 10–28 N · m 2 )/r, which gives r = 7.2 × 10– 1 1 m. (b) Using the analogy to Kepler’s third law, we have T2 = 4π2 r3 /(e2 /m) = 4π2 r3 m/e2 = 4π2 (7.2 × 10– 1 1 m)3 (9.1 × 10– 3 1 kg)/(2.3 × 10–28 N · m 2 ), which gives T = 2.4 × 10– 1 6 s. (c) For the total energy, we have E2 /E1 = r1 /r2 = 1/3, or E2 = – (1.6 × 10– 1 8 J)/3 = – 5.3× 10– 1 9 J. For the period, we have T2 /T1 = (r2 /r1 ) 3/2 = 33/2, or T2 = 33/2(2.4 × 10– 1 6 s) = 1.2 × 10– 1 5 s. 69. ( a ) While the astronaut throws the wrench, linear momentum is conserved. Using the initial reference frame of the astronaut, we have (m + mW )(0) = m(– vi) + mW v W ; 0 = (115 kg)(– 0.05 m/s) + (3 kg)vW , which gives v W = 1.92 m/s relative to himself, which is 1.97 m/s relative to the ship. (b) We find his average acceleration toward the ship from F av = GMm/rav2 = maav; a av = (6.67 × 10– 1 1 N · m 2 /kg2 )(105 kg)/(12 m + 0.5 m)2 = 4.3 × 10– 8 m/s2 . To find the time, we use x = !aavt 2 ; 1 m = !(4.3 × 10– 8 m/s2 )t 2 , which gives t = 6.8 × 103 s = 1.9 h. 70. At the equilibrium point, we have r GME m/rE 2 = GMM m/rM 2 . r m FE FM When the mass is displaced by x, the two forces become x φ θ F E = GME m/r1 2 = GME m/(rE 2 + x2 ) and r rM E F M = GMM m/r2 2 = GMM m/(rM 2 + x2 ) , Moon Earth with the directions indicated on the diagram. When we add the components along the line joining Earth and the Moon, we get F || = – FE cos θ + FM cos φ = – GME mrE /(rE 2 + x2 ) 3/2 + GMM mrM /(rM 2 + x2 ) 3/2. Using the approximation (r2 + x2 ) n ≈ r2n (1 + nx2 /r2 ) with n = – *, we get F || = – (GME m/rE 2 )[1 – *(x2 /rE 2 )] + (GMM m/rM 2 )[1 – *(x2 /rM 2 )] = – GME m/rE 2 + GMM m/rM 2 – term in x2 ≈ 0, x2 << r2 . When we add the components perpendicular to the line joining Earth and the Moon, we get F ⊥ = – FE sin θ – FM sin φ = – GME mx/(rE 2 + x2 ) 3/2 + GMM mx/(rM 2 + x2 ) 3/2. Using the same approximation, we get F ⊥ = (– MeGmx/rE 3 )[1 – *(x2 /rE 2 )] – (GMM x/rM 3 )[1 – *(x2 /rM 2 )] = – Gm(ME /rE 3 + (MM /rM 3 )x, x2 << r2 . The net force has magnitude F net = Gm(ME /rE 3 + MM /rM 3 )x, with a direction toward the original equilibrium point. © 2005 Pearson Education, Inc., Upper Saddle River, NJ. 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.

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71. The gravitational acceleration is the sum of the Jupiter contribution from Io (mass m, radius r = 0.2846 RE ) and Jupiter (mass M, radius R). Denote the centerIo to-center separation between the two as d. R r (a) At a point A on Io closest to Jupiter, the g B A gravitational acceleration due to Io is g1 = g g g Gm/r2, pointing towards the center of Io, while that due to Jupiter is g2 = GM/(d – r)2, d pointing towards the center of Jupiter (away from that of Io). The net value is gA = g1 – g2 = Gm/r2 – GM/(d – r)2 = (GME /RE 2)[(m/ME )/(r/RE )2 – (M/ME )/(d/RE – r/RE ) 2] = (9.8 m/s2) [(4.7 × 10–5)(318)/(0.2846)2 – (318)/(5.95 × 26 – 0.2846)2] = 1.68 m/s2 . (b) At a point B on Io furthest away from Jupiter, the gravitational acceleration due to Io is still g1 = Gm/r2, pointing towards the center of Io, while that due to Jupiter is g2‘= GM/(d + r)2, pointing towards the center of Jupiter (in the same direction as g1). The net value is gB = g1 + g2‘ = Gm/r2 + GM/(d + r)2 = (GME /RE 2)[(m/ME )/(r/RE ) 2 + (M/ME )/(d/RE + r/RE ) 2] = (9.8 m/s2) [(4.7 × 10–5)(318)/(0.2846)2 + 318/(5.95 × 26 + 0.2846)2] = 1.94 m/s2 . 1

2'

1

2

72. Because of the large distances, we assume that all objects can be treated as point masses. The change in gravitational force from Jupiter depends on the change in separation of the Earth and Jupiter. In a time t the Earth moves a distance t=0 ∆s e = 2πRE t/TE tangent to its orbit. In a time t Jupiter moves a distance ∆sJ ∆s J = 2πRJt/TJ tangent to its orbit. ∆st In the position of closest approach, if Jupiter were to move RJ ∆sE through the same angle as the Earth, there would be no change in separation. The angle turned through by the Earth is RE θ = ∆sE /RE , so the the corresponding distance for Jupiter is ∆s J′ = RJθ = (RJ/RE )∆s E . Thus the change in tangential separation is ∆s t = ∆sJ′ – ∆sJ = (RJ/RE )2πR E t/TE – 2πRJt/TJ = 2πRJt (1/TE – 1/TJ) = 2π(7.8 × 101 1 m)(1 d)[(1/1 yr) – (1/11.9 yr)](1 yr/365 d) = 1.23 × 101 0 m. The separation of the two planets, which was ∆R = RJ – RE , is now (∆R)2 + (∆st) 2 . The change in gravitational force on a baby of mass m is ∆F = GMJm/[(∆R)2 + (∆st) 2 ] – GMJm/(∆R)2 = [GMJm/(∆R)2 ]{1/[1 + (∆st/∆R)2 ] – 1}. Because (∆st/∆R)2 << 1, we can use the approximation 1/(1 + x) =#1 – x to get a magnitude of ∆F ≈ – [GMJm/(∆R)2 ](∆s t/∆R)2 = – (6.67 × 10– 1 1 N · m 2 /kg2 )(1.90 × 1027 kg)m(1.23 × 101 0 m)2 /[(7.8 – 1.5) × 101 1 m]4 = – 1.2m × 10– 1 0 N . A 4-ton truck has a mass M of (8 × 103 lb)(4.44 N/lb)/(9.8 m/s2 ) = 3.6 × 103 kg. The gravitational force from the truck is ∆F = GMm/R2 = (6.67 × 10– 1 1 N · m 2 /kg2 )(3.6 x103 kg)m/(75 m)2 = 4.3m × 10– 1 1 N . This is 0.36 of the change due to the planets, when they were in their most effective positions. © 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

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73. ( a ) From the symmetry of the arrangement, we see that the net force on a star is toward the center of the triangle. Thus the net force can M provide the necessary centripetal acceleration, so the arrangement is possible. D 30° R r (b) If D is the separation of a pair of stars, for Newton's second law we have r F F net = 2F cos 30° = 2(GM2 /D2 ) cos 30° = MR(2π/T)2 . Fnet The distance R from the center to a star is !D/cos 30°, so we have r M F M T = π(D3 /GM)1/2/ cos 30° 1 1 3 – 1 1 2 2 3 0 1/2 = π[(1.5 × 10 m) /(6.67 × 10 N · m /kg )(1.99 × 10 kg)] /cos 30° = 1.83 × 107 s (212 days). (c) To test the stability, we consider small changes in D. When D decreases, the force increases, but the required centripetal force decreases; thus the star system will collapse. When D increases, the force decreases, but the required centripetal force increases; thus the star system will separate. The system is unstable.

74. With the Earth pictured as an onion, we find the energy required to remove each layer and take it to infinity. A layer of radius r will have a mass dm = ρ 4πr2 dr, where the density is ρ = ME /()πRE 3 ) . Because the layer being removed is outside the remaining mass m, the potential energy of the layer is dU = – Gm dm/r, with the reference level at infinity. Thus the energy required to remove this layer is dE = 0 – (– Gm dm)/r = Gm dm/r = G[ρ()πr3 )ρ (4πr2 dr)]/r = @[Gρ2 (4π)2 r4 dr]. We find the total energy required by integrating over all of the layers: RE R E Gρ 2 (4π ) 2 r 4 G ρ 2 (4π ) 2 ⎡ r 5 ⎤ E= dr = ⎢5⎥ 0 3 3 ⎣ ⎦0

∫

4π 3 ⎞ 2 ⎛ 3 ⎞ 3GME 2 = G⎛⎜ ρ R ⎟ = . ⎝ 3 E ⎠ ⎜⎝ 5RE ⎟⎠ 5RE


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