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Chapter 13 Student Solutions Manual

Chapter 13 Student Solutions Manual - Chapter 13 Student...

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Chapter 13 – Student Solutions Manual 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 1 2 12 6.67 10 N m / kg 5.2kg 2.4kg 19m 2.3 10 N Gm m r F × = = = × 7. At the point where the forces balance , where M 2 1 / e s GM m r GM m r = 2 2 / 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 ( ) 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 ( ) 9 24 8 1 30 24 150 10 m 5.98 10 kg = = = 2.60 + 1.99 10 kg + 5.98 10 kg e s e d M r M M × × × × × 10 m. Values for M e , M s , and d can be found in Appendix C. 17. The acceleration due to gravity is given by a g = GM/r 2 , where M is the mass of Earth and r is the distance from Earth’s center. We substitute r = R + h , where R is the radius of Earth and h is the altitude, to obtain a g = GM / ( R + h ) 2 . We solve for h and obtain / g h GM a = R . According to Appendix C, R = 6.37 × 10 6 m and M = 5.98 × 10 24 kg, so ( )( ) ( ) 11 3 2 24 6 6 2 6.67 10 m / s kg 5.98 10 kg 6.37 10 m 2.6 10 m. 4.9m / s h × × = × = × 29. (a) The density of a uniform sphere is given by ρ = 3 M /4 π R 3 , where M is its mass and R is its radius. The ratio of the density of Mars to the density of Earth is 3 3 4 3 3 0.65 10 km = = 0.11 = 0.74. 3.45 10 km M M E E E M M R M R ρ ρ × ×
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