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Practice Exam 7

# Practice Exam 7 - practicework 07 – BAUTISTA ALDO – Due...

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Unformatted text preview: practicework 07 – BAUTISTA, ALDO – Due: May 9 2006, 10:00 am 1 Version number encoded for clicker entry: V1:1, V2:4, V3:2, V4:4, V5:3. Question 1 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. A 2. D 3. B 4. C correct 5. E 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. Question 2 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 = 1 √ 3 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 = √ 2 6. v 3 m v m = 3 7. v 3 m v m = 9 8. v 3 m v m = 2 9. v 3 m v m = √ 3 10. v 3 m v m = 1 √ 2 correct practicework 07 – BAUTISTA, ALDO – Due: May 9 2006, 10:00 am 2 Explanation: 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 . Question 3 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 9 . 7 h, has a mass of 2 . 1 × 10 27 kg and a radius of 69900000 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: 9 . 30463 × 10 7 m (tolerance ± 1 %). 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....
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Practice Exam 7 - practicework 07 – BAUTISTA ALDO – Due...

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