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 12 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 / es 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 ( ) 22 1 1 =. MM r dr Taking the positive square root of both sides, we solve for r 1 . A little algebra yields ( ) 92 4 8 1 30 24 150 10 m 5.98 10 kg = 2 . 6 0 + 1.99 10 kg + 5.98 10 kg e se dM r ×× × 1 0 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 hG M a = R . According to Appendix C, R = 6.37 × 10 6 m and M = 5.98 × 10 24 kg, so ( ) 11 3 2 24 66 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 34 33 0.65 10 km = = 0.11 = 0.74. 3.45 10 km E EE M MR ⎛⎞ × ⎜⎟ × ⎝⎠
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(b) The value of a g at the surface of a planet is given by a g = GM/R 2 , so the value for Mars is () 2 24 22 23 0.65 10 km = = 0.11 9.8 m/s = 3.8 m/s . 3.45 10 km ME gg E EM MR aM a ⎛⎞ × ⎜⎟ × ⎝⎠ (c) If v is the escape speed, then, for a particle of mass m 2 12 =. 2 mM GM mv G v R R ⇒= For Mars, the escape speed is ( ) 11 3 2 24 3 6 2(6.67 10 m /s kg) 0.11 5.98 10 kg = 5.0 10 m/s. 3.45 10 m v ×⋅ × × 37. (a) We use the principle of conservation of energy. Initially the particle is at the surface of the asteroid and has potential energy U i = GMm/R , where M is the mass of the asteroid, R is its radius, and m is the mass of the particle being fired upward. The initial kinetic energy is 2 1 2 mv . The particle just escapes if its kinetic energy is zero when it is infinitely far from the asteroid. The final potential and kinetic energies are both zero.
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This homework help was uploaded on 04/10/2008 for the course PHYSICS 150 taught by Professor Naik during the Winter '08 term at University of Michigan.

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

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