HW07-solutions - baum(lmb2768 HW07 distler(56295 This...

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baum (lmb2768) – HW07 – distler – (56295) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 3) 10.0 points Consider two planets of mass m and 2 m , respectively, orbiting the same star in circular orbits. The more massive planet is 4 times as far from the star as the less massive one. What is the ratio F 1 : F 2 of the gravita- tional forces exerted on the star by the two planets? Correct answer: 8. Explanation: Let : m 1 = m , m 2 = 2 m , and r 2 = 4 r 1 . F = G m 1 m 2 r 2 m 1 m 2 r 2 , so F 1 F 2 = M S m r 1 2 M S (2 m ) (4 r 1 ) 2 = m r 1 2 · (4 r 1 ) 2 2 m = 8 . 002 (part 2 of 3) 10.0 points What is the ratio v 1 : v 2 of the speeds of the two planets? Correct answer: 2. Explanation: Gravitational force supplies the centripetal force, so F = G M S m r 2 = m v 2 r v = G M r 1 r v 1 v 2 = r 2 r 1 = 4 = 2 . 003 (part 3 of 3) 10.0 points What is the ratio T 1 : T 2 of the orbital periods of the two planets? Correct answer: 0 . 125. Explanation: T = 2 π r v r V , so T 1 T 2 = r 1 v 1 r 2 v 2 = r 1 r 2 v 2 v 1 = 1 4 · 1 2 = 0 . 125 . 004 10.0 points An object is projected upward from the sur- face of the earth with an initial speed of 3 km / s. The acceleration of gravity is 9 . 81 m / s 2 . Find the maximum height it reaches. Correct answer: 494 . 312 km. Explanation: Let : R E = 6 . 37 × 10 6 m , v i = 3 km / s = 3000 m / s , and g = 9 . 81 m / s 2 . g = G M E R 2 E and the potential energy at a distance r from the surface of the earth is U ( r ) = - G M E m r Using conservation of energy to relate the initial potential energy of the system to its energy at its maximum height ( K f = 0), K f - K i + U f - U i = 0 - K ( R E ) + U ( R E + h ) - U ( R E ) = 0 - 1 2 mv 2 i + GM E m R E + h - GM E m R E = 0
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baum (lmb2768) – HW07 – distler – (56295) 2 1 2 v 2 i = G M E h R E ( R E + h ) · R E R E v 2 i = 2 G M E R 2 E · h R E R E + h v 2 i = 2 g h R E R E + h ( R E + h ) v i 2 = 2 g h R E h = - v 2 i R E v 2 i - 2 g R E Since v 2 i - 2 g R E = (3000 m / s) 2 - 2 (9 . 81 m / s 2 ) × (6 . 37 × 10 6 m) = - 1 . 15979 × 10 8 , h = - (3000 m / s) 2 (6 . 37 × 10 6 m) - 1 . 15979 × 10 8 · 1 km 1000 m = 494 . 312 km . 005 10.0 points An earth satellite remains in orbit at a dis- tance of 17925 km from the center of the earth. What speed would it have to main- tain? The universal gravitational constant is 6 . 67 × 10 - 11 N · m 2 / kg 2 and the mass of the earth is 5 . 98 × 10 24 kg. Correct answer: 4717 . 2 m / s. Explanation: Let; r = 17925 km = 1 . 7925 × 10 7 m , G = 6 . 67 × 10 - 11 N · m 2 / kg 2 , and M = 5 . 98 × 10 24 kg . The gravitational force supplies the cen- tripetal force, so G M m r 2 = m v 2 r v = G M r = 6 . 67 × 10 - 11 N · m 2 / kg 2 1 . 7925 × 10 7 m × 5 . 98 × 10 24 kg = 4717 . 2 m / s . 006 10.0 points A curve of radius 56 . 3 m is banked so that a car of mass 2 . 3 Mg traveling with uniform speed 63 km / hr can round the curve without relying on friction to keep it from slipping on the surface. 2 . 3 Mg μ 0 θ At what angle is the curve banked? The acceleration due to gravity is 9 . 8 m / s 2 .
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