Chapter 12

# Chapter 12 - Note to obtain the numerical results given in...

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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 satellite’s weight at the surface of the earth is %. 79 79 . 0 ) 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 =

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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 Ν × - = + - Ν × = - - 11 2 2 2 2 11 10 32 . 2 , m 0.600 kg 0 . 10 m 0.400 kg 00 . 5 kg 100 . 0 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 earth-sun separation as 1 r and the earth-moon separation as 2 r . a) ( 29 , 10 30 . 6 ) ( 20 2 2 E 2 2 1 S M Ν × = + + r m r r m Gm toward the sun. b)The earth-moon distance is sufficiently small compared to the earth- sun distance ( r 2 << r 2 ) that the vector from the earth to the moon can be taken to be perpendicular to the vector from the sun to the moon. The components of the gravitational force are then , 10 99 . 1 , 10 34 . 4 20 2 2 E M 20 2 1 S M Ν × = Ν × = r m Gm r m Gm and so the force has magnitude Ν × 20 10 77 . 4 and is directed ° 6 . 24 from the direction toward the sun.

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