solution_pdfsd

# solution_pdfsd - Patel(sap785 homework#6 shubeita(58645...

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Patel (sap785) – homework #6 – shubeita – (58645) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points Two satellites have circular orbits about the same planet. Their masses are m and 3 m , respectively, with orbits oF radii r and 2 r , respectively. 2 r r 3 m m What is the ratio oF the orbital speeds oF the two satellites? 1. v 3 m v m = 2 2. v 3 m v m = 1 2 correct 3. v 3 m v m = 1 2 4. v 3 m v m = 9 5. v 3 m v m = 3 6. v 3 m v m = 2 7. v 3 m v m = 1 9 8. v 3 m v m = 1 3 9. v 3 m v m = 3 10. v 3 m v m = 1 3 Explanation: The only Force acting on an orbiting satel- lite is the Force oF gravity F = GM m r 2 , where M is the planet’s mass and m is the mass oF the satellite itselF. Consequently, the satel- lite is in Free Fall with acceleration g = GM r 2 directed towards the planet’s center. ±or a circular orbit, the Free-Fall accel- eration equals the centripetal acceleration a c = v 2 r , so GM r 2 = v 2 r v = r GM r r 1 r , regardless oF the satellite’s mass m . Consequently, For two satellites in circular orbits around the same planet, v 3 m v m = r r m r 3 m = r r 2 r = 1 2 . 002 10.0 points An apparatus like the one Cavendish used to fnd G has a large lead ball that is 7 . 8 kg in mass and a small one that is 0 . 03 kg. Their centers are separated by 0 . 055 m. ±ind the Force oF attraction between them. The value oF the universal gravitational con- stant is 6 . 67259 × 10 11 N · m 2 / kg 2 . Correct answer: 5 . 16161 × 10 9 N. Explanation: Let : d = 0 . 055 m , G = 6 . 67259 × 10 11 N · m 2 / kg 2 , m 1 = 7 . 8 kg , and m 2 = 0 . 03 kg . By Newton’s universal law oF gravitation, F = G m 1 m 2 d 2 = (6 . 67259 × 10 11 N · m 2 / kg 2 ) (7 . 8 kg) (0 . 03 kg) 0 . 055 m 2 = 5 . 16161 × 10 9 N .

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Patel (sap785) – homework #6 – shubeita – (58645) 2 003 10.0 points Calculate the mass of the earth from the pe- riod of the moon (27 . 3 d) and its mean orbital radius of 3 . 84 × 10 8 m. The universal gravitational constant is 6 . 673 × 10 11 N · m 2 / kg 2 . Correct answer: 6 . 02115 × 10 24 kg. Explanation: Let : T m = 27 . 3 d , r m = 3 . 84 × 10 8 m , and G = 6 . 673 × 10 11 N · m 2 / kg 2 . Using Kepler’s third law T M 2 = 4 π 2 GM E r m 3 M s = 4 π 2 r m 3 GT m 2 = 4 π 2 (3 . 84 × 10 8 m) 3 6 . 673 × 10 11 N · m 2 / kg 2 × p 1 27 . 3 d P 2 p 1 d 24 h P 2 p 1 h 3600 s P 2 = 6 . 02115 × 10 24 kg . 004 10.0 points Given: G = 6 . 67259 × 10 11 N m 2 / kg 2 Calculate the work required to move a planet’s satellite of mass 1200 kg from a cir- cular orbit of radius 2 R to one of radius 3 R , where 7 . 1 × 10 6 m is the radius of the planet. The mass of the planet is 4
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## This note was uploaded on 04/16/2010 for the course PHY 317k taught by Professor Kopp during the Spring '07 term at University of Texas.

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solution_pdfsd - Patel(sap785 homework#6 shubeita(58645...

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