CH-13 - Chapter 13 1. The magnitude of the force of one...

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Chapter 13 1. The magnitude of the force of one particle on the other is given by F = Gm 1 m 2 / r 2 , where m 1 and m 2 are the masses, r is their separation, and G is the universal gravitational constant. We solve for r : ( 29 ( 29 ( 29 11 2 2 1 2 12 6.67 10 N m / kg 5.2kg 2.4kg 19m 2.3 10 N Gm m r F - - = = = 2. We use subscripts s, e, and m for the Sun, Earth and Moon, respectively. 2 2 2 s m sm sm s em e m em e sm em Gm m F r m r Gm m F m r r = = Plugging in the numerical values (say, from Appendix C) we find 2 30 8 24 11 1.99 10 3.82 10 2.16. 5.98 10 1.50 10 × × = × × 3. The gravitational force between the two parts is ( 29 ( 29 2 2 2 = = Gm M m G F mM m r r - - which we differentiate with respect to m and set equal to zero: ( 29 2 = 0 = 2 = 2 dF G M m M m dm r - which leads to the result m / M = 1/2. 4. Using F = GmM/r 2 , we find that the topmost mass pulls upward on the one at the origin with 1.9 × 10 - 8 N, and the rightmost mass pulls rightward on the one at the origin with 1.0 × 10 - 8 N. Thus, the ( x, y ) components of the net force, which can be converted to polar components (here we use magnitude-angle notation), are ( 29 ( 29 8 8 8 net = 1.04 10 ,1.85 10 2.13 10 60.6 . F - - - а r 581
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CHAPTER 13 (a) The magnitude of the force is 2.13 × 10 - 8 N. (b) The direction of the force relative to the + x axis is 60.6 . . 5. At the point where the forces balance 2 2 1 2 / / e s GM m r GM m r = , where M e is the mass of Earth, M s is the mass of the Sun, m is the mass of the space probe, r 1 is the distance from the center of Earth to the probe, and r 2 is the distance from the center of the Sun to the probe. We substitute r 2 = d - r 1 , where d is the distance from the center of Earth to the center of the Sun, to find ( 29 2 2 1 1 = . e s M M r d r - Taking the positive square root of both sides, we solve for r 1 . A little algebra yields ( 29 9 24 8 1 30 24 150 10 m 5.98 10 kg = = = 2.60 10 m. + 1.99 10 kg + 5.98 10 kg e s e d M r M M G G G Values for M e , M s , and d can be found in Appendix C. 6. The gravitational forces on m 5 from the two 5.00g masses m 1 and m 4 cancel each other. Contributions to the net force on m 5 come from the remaining two masses: ( 29 ( 29 ( 29 ( 29 11 2 2 3 3 3 net 2 1 14 6.67 10 N m /kg 2.50 10 kg 3.00 10 kg 1.00 10 kg = 2 10 m =1.67 10 N. F - - - - - - - ״ The force is directed along the diagonal between m 2 and m 3 , towards m 2 . In unit-vector notation, we have 14 14 net net ˆ ˆ ˆ ˆ (cos45 i sin 45 j) (1.18 10 N)i (1.18 10 N) j F F - - + = E + r 7. We require the magnitude of force (given by Eq. 13-1) exerted by particle C on A be equal to that exerted by B on A . Thus, = . We substitute in
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CH-13 - Chapter 13 1. The magnitude of the force of one...

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