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

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Unformatted text preview: 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 m r 2 , the mass of an object be m , and the distance between adjacent locations and the centers of the spheres be r . F A = G m m r 2 + G m (2 m ) (5 r ) 2 = 27 25 F F B = G m m r 2- G m (2 m ) (3 r ) 2 = 7 9 F F C = G m (2 m ) (2 r ) 2- G m m (2 r ) 2 = 1 4 F F D = G m (2 m ) r 2- G m m (3 r ) 2 = 17 9 F F E = G m (2 m ) r 2 + G m 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: 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 = GM m r 2 , where M is the planets mass and m is the mass of the satellite itself. Consequently, the satellite is in free fall with acceleration g = GM r 2 directed towards the planets center. Note: This acceleration is independent on the satellites mass m . For a circular orbit, the free-fall accelera- tion equals the centripetal acceleration, g = a c = v 2 r , and therefore GM r 2 = v 2 r = v = r GM r , regardless of the satellites mass m . Conse- quently, for two satellites in circular orbits around the same planet, v 3 m v m = r GM r 3 m r GM 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 planets 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 Nm 2 / kg 2 , calculate how far above Jupiters surface the satellite must be. Correct answer: 3 . 54316 10 7 m....
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This note was uploaded on 04/12/2010 for the course PHY 58195 taught by Professor Turner during the Spring '09 term at University of Texas at Austin.

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

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