chap12

# chap12 - Note: to obtain the numerical results given in...

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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 earth-moon radius to the sun-moon radius. Using the earth-sun radius as an average for the sun-moon 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 satellites 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 earth-sun 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 - = +- =-- 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 = -- 12.8: ( 29 ( 29 4 2 10 03 . 6 500 , 23 000 , 333- = 12.9: Denote the earth-sun separation as 1 r and the earth-moon separation as 2 r ....
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## chap12 - Note: to obtain the numerical results given in...

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