Chapter13 - 1. The magnitude of the force of one particle...

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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 : () 11 2 2 12 12 6.67 10 N m / kg 5.2kg 2.4kg 19m 2.3 10 N Gm m r F ×⋅ == = × .
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2. We use subscripts s, e, and m for the Sun, Earth and Moon, respectively. Plugging in the numerical values (say, from Appendix C) we find 2 2 2 30 8 22 4 1 1 / 1.99 10 kg 3.82 10 m 2.16. / 5.98 10 kg 1.50 10 m sm s m sm s em em e m em e sm FG m m rm r m m r §· ×× == = = ¨¸ ©¹
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3. The gravitational force between the two parts is () 2 22 == Gm M m G Fm M m rr which we differentiate with respect to m and set equal to zero: 2 =0= 2 =2 dF G M mM m dm r ¡ . This leads to the result m / M = 1/2.
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(a) The ratio of the moon’s gravitational pulls at the two different positions is 2 2 2 86 1 28 6 0 /( ) 3.82 10 m 6.37 10 m 1.06898. /( ) mM E E ME E E E M E E GM m R R FR R FG M m R R R R §· + × == = = ¨¸ +− × × ©¹ Therefore, the increase is 0.06898, or approximately, 6.9%. (b) The change of the gravitational pull may be approximated as 10 22 2 2 3 4 12 . () mm m m m E EE ME E ME E ME ME ME ME ME GM m GM m GM m GM m GM mR RR FF R R R R R −= + = −+ On the other hand, your weight, as measured on a scale on Earth is 2 E gE E GM m Fm g R . Since the moon pulls you “up,” the percentage decrease of weight is 3 3 22 6 75 24 8 7.36 10 kg 6.37 10 m 4 4 2.27 10 (2.3 10 )%. 5.98 10 kg 3.82 10 m m E M E M R FM R −− § · ×× = × × ¨ ¸ © ¹ 4. The gravitational force between you and the moon at its initial position (directly opposite of Earth from you) is 0 2 m ME E GM m F = + where m M is the mass of the moon, ME R is the distance between the moon and the Earth, and E R is the radius of the Earth. At its final position (directly above you), the gravitational force between you and the moon is 1 2 m ME E GM m F = .
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5. 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, Gm A m C r 2 = Gm A m B d 2 . We substitute in m B = 3 m A and m B = 3 m A , and (after canceling “ m A ”) solve for r . We find r = 5 d . Thus, particle C is placed on the x axis, to left of particle A (so it is at a negative value of x ), at x = –5.00 d .
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6. 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 () ( ) 88 8 net = 1.04 10 ,1.85 10 2.13 10 60.6 . F −− ×× ¡ ×∠° G (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 ° .
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7. At the point where the forces balance 22 12 // es 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 ( ) 1 1 =.
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This note was uploaded on 05/11/2010 for the course PHYS 2101,2102 taught by Professor Giammanco during the Spring '10 term at LSU.

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Chapter13 - 1. The magnitude of the force of one particle...

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