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quiz4 - Muraj Hamza Quiz 4 Due May 3 2006 10:00 pm Inst...

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Muraj, Hamza – Quiz 4 – Due: May 3 2006, 10:00 pm – Inst: Florin 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Two iron spheres, of mass m and 2 m , respectively, and equally spaced points “ r apart” are shown in the figure. A m B C D 2 m E r r r r r r At which location would the net gravita- tional force on an object due to these two spheres be a minimum? 1. D 2. C correct 3. B 4. E 5. A Explanation: Let F = G m 0 m r 2 , the mass of an object be m 0 , and the distance between adjacent locations and the centers of the spheres be r . F A = G m 0 m r 2 + G m 0 (2 m ) (5 r ) 2 = 27 25 F F B = G m 0 m r 2 - G m 0 (2 m ) (3 r ) 2 = 7 9 F F C = G m 0 (2 m ) (2 r ) 2 - G m 0 m (2 r ) 2 = 1 4 F F D = G m 0 (2 m ) r 2 - G m 0 m (3 r ) 2 = 17 9 F F E = G m 0 (2 m ) r 2 + G m 0 m (5 r ) 2 = 51 25 F Thus the gravitational force at C is mini- mum. 002 (part 1 of 1) 10 points Two satellites have circular orbits about the same planet. The masses of the two satellites are respec- tively m m = m and m 3 m = 3 m . Both satel- lites have circular orbits with respective radii r m = r and r 3 m = 2 r . 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 3 3. v 3 m v m = 1 9 4. v 3 m v m = 1 2 5. v 3 m v m = 3 6. v 3 m v m = 2 7. v 3 m v m = 3 8. v 3 m v m = 9 9. v 3 m v m = 1 3 10. v 3 m v m = 1 2 correct Explanation:
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Muraj, Hamza – Quiz 4 – Due: May 3 2006, 10:00 pm – Inst: Florin 2 The only force acting on an orbiting satel- lite is the force of gravity, F = G M m r 2 , where M is the planet’s mass and m is the mass of the satellite itself. Consequently, the satellite is in free fall with acceleration g = G M r 2 directed towards the planet’s center. Note: This acceleration is independent on the satellite’s mass m . For a circular orbit, the free-fall accelera- tion equals the centripetal acceleration, g = a c = v 2 r , and therefore G M r 2 = v 2 r = v = r G M r , regardless of the satellite’s mass m . Conse- quently, for two satellites in circular orbits around the same planet, v 3 m v m = r G M r 3 m r G M r m = r r m r 3 m = r r 2 r = 1 2 . 003 (part 1 of 1) 10 points A “synchronous” satellite, which always re- mains above the same point on a planet’s equator, is put in orbit about a planet similar to Jupiter. This planet rotates once every 5 . 3 h, has a mass of 1 . 9 × 10 27 kg and a radius of 6 . 99 × 10 7 m. Given that G = 6 . 67 × 10 - 11 N m 2 / kg 2 , calculate how far above Jupiter’s surface the satellite must be. Correct answer: 3 . 54316 × 10 7 m. Explanation: Basic Concepts: Solution: According to Kepler’s third law: T 2 = 4 π 2 G M r 3 where r is the radius of the satellite’s orbit. Thus, solving for r r = G M T 2 4 π 2 1 3 = { 6 . 67 × 10 - 11 N m 2 / kg 2 } 1 3 × (1 . 9 × 10 27 kg) (19080 s) 2 4 (3 . 1415926536 ) 2 1 3 = 1 . 05332 × 10 8 m .
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