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Unformatted text preview: Note: to obtain the numerical results given in this chapter, the following numerical values of certain physical quantities have been used; kg. 10 97 . 5 and s m 80 . 9 , kg m N 10 673 . 6 24 E 2 2 2 11 × = = ⋅ × = m g G Use of other tabulated values for these quantities may result in an answer that differs in the third significant figure. 12.1: The ratio will be the product of the ratio of the mass of the sun to the mass of the earth and the square of the ratio of the earthmoon radius to the sunmoon radius. Using the earthsun radius as an average for the sunmoon radius, the ratio of the forces is . 18 . 2 kg 10 5.97 kg 10 1.99 m 10 1.50 m 10 84 . 3 24 30 2 11 8 = × × × × 12.2: Use of Eq. (12.1) gives N. 10 67 . 1 m) 10 6.38 m 10 (7.8 kg) 2150 )( kg 10 97 . 5 ( ) kg m N 10 673 . 6 ( 4 2 6 5 24 2 2 11 2 2 1 g × = × + × × ⋅ × = = r m m G F The ratio of this force to the satellite’s weight at the surface of the earth is %. 79 79 . ) s m 80 . 9 )( kg 2150 ( ) N 10 67 . 1 ( 2 4 = = × (This numerical result requires keeping one extra significant figure in the intermediate calculation.) The ratio, which is independent of the satellite mass, can be obtained directly as , 2 E 2 E 2 E = = r R g r Gm mg r m Gm yielding the same result. 12.3: . ) ( ) )( ( 12 2 12 2 1 2 12 2 1 F r m m G nr nm nm G = = 12.4: The separation of the centers of the spheres is 2 R , so the magnitude of the gravitational attraction is . 4 ) 2 ( 2 2 2 2 R GM R GM = 12.5: a) Denoting the earthsun separation as R and the distance from the earth as x , the distance for which the forces balance is obtained from , ) ( 2 E 2 S x m GM x R m GM = which is solved for m. 10 59 . 2 1 8 E S × = + = M M R x b) The ship could not be at equilibrium for long, in that the point where the forces balance is moving in a circle, and to move in that circle requires some force. The spaceship could continue toward the sun with a good navigator on board. 12.6: a) Taking force components to be positive to the right, use of Eq. (12.1) twice gives ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 Ν × = + ⋅ Ν × = 11 2 2 2 2 11 10 32 . 2 , m 0.600 kg . 10 m 0.400 kg 00 . 5 kg 100 . kg m 10 673 . 6 g F with the minus sign indicating a net force to the left. b) No, the force found in part (a) is the net force due to the other two spheres. 12.7: ( 29 ( 29 ( 29 ( 29 . 10 4 . 2 m 10 78 . 3 kg 10 35 . 7 kg 70 kg m . 10 673 . 6 3 2 8 22 2 2 11 Ν × = × × Ν × 12.8: ( 29 ( 29 4 2 10 03 . 6 500 , 23 000 , 333 × = 12.9: Denote the earthsun separation as 1 r and the earthmoon separation as 2 r ....
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This homework help was uploaded on 02/26/2008 for the course PHYS 11 and 21 taught by Professor Hickman during the Spring '08 term at Lehigh University .
 Spring '08
 Hickman
 The Land

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