Gravitation-solutions

# Gravitation-solutions - le (txl65) Gravitation mohanty...

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le (txl65) – Gravitation – mohanty – (81440) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points In a distant solar system, a giant planet has a small moon moving in a circular orbit oF radius 4 . 31 × 10 5 km with an orbital period oF 63 . 8 hrs. ±ind the mass oF the planet. You will need G = 6 . 67 × 10 11 m 3 / kg / s 2 . Correct answer: 8 . 98296 × 10 26 kg. Explanation: The centripetal Force oF the moon is F c = mv 2 r = 2 r , and the gravitational Force acting on the moon by the planet is F g = GM m r 2 . Since the two Forces should be equal, we have M = r 3 ω 2 G = (4 . 31 × 10 8 m) 3 × (2 . 73563 × 10 5 s 1 ) 2 6 . 67 × 10 11 m 3 / kg / s 2 = 8 . 98296 × 10 26 kg . 002 10.0 points Two iron spheres oF mass m and 2 m , respectively, and equally spaced points “ r apart” are shown in the fgure. A m B C D 2 m E r r r r r r At which oF these locations would the net gravitational Force on an object due to these two spheres be the greatest? 1. B 2. D 3. C 4. E correct 5. A Explanation: Let the mass oF an object at the point in question be m , with a distance r between ad- jacent locations and the centers oF the spheres. F = G m m r 2 , so 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, oF the points A-±, the gravitational Force at E is largest. Note that the Force will simply get larger as you approach closer to the 2 m mass, so this does not represent a maximum oF the Force. 003 10.0 points Planet X has Four times the diameter and seven times the mass oF the earth. What is the ratio g X : g e oF gravitational acceleration at the surFace oF planet X to the gravitational acceleration at the surFace oF the Earth? Correct answer: 0 . 4375. Explanation: Let : M x = 7 M e and R x = 4 R e .

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le (txl65) – Gravitation – mohanty – (81440) 2 mg e = GM e m R 2 e g e = GM e R 2 e . Similarly, g x = GM x R 2 x . The ratio is g x g e = M x R 2 e M e R 2 x = 7 M e R 2 e M e (4 R e ) 2 = 7 16 = 0 . 4375 . 004 (part 1 of 2) 10.0 points A 1 . 5 kg mass weighs 14 . 1 N on the surface of a planet similar to Earth. The radius of this planet is roughly 6 . 7 × 10 6 m. Calculate the mass of of this planet. The value of the universal gravitational constant is 6 . 67259 × 10 11 N · m 2 / kg 2 . Correct answer: 6 . 32387 × 10 24 kg. Explanation: Let : G = 6 . 67259 × 10 11 N · m 2 / kg 2 , m = 1 . 5 kg , r = 6 . 7 × 10 6 m , and W = 14 . 1 N . By Newton’s law of universal gravitation,
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## This note was uploaded on 04/05/2011 for the course SD ds taught by Professor Sdsd during the Spring '11 term at Tarrant County.

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Gravitation-solutions - le (txl65) Gravitation mohanty...

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