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Unformatted text preview: hanna (brh687) – hk 13 – Opyrchal – (11105) 1 This printout should have 11 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points An artificial satellite circling the Earth com pletes each orbit in 94 minutes. What is the value of g at the location of this satellite? The mass of the earth is 5 . 98 × 10 24 kg and the universal gravitational constant is 6 . 67259 × 10 11 N · m 2 / kg 2 . Correct answer: 8 . 50223 m / s 2 . Explanation: Let : M earth = 5 . 98 × 10 24 kg , T = 94 min , and G = 6 . 67259 × 10 11 N · m 2 / kg 2 . ω = 2 π T = 2 π 94 min parenleftbigg 1 min 60 s parenrightbigg = 0 . 00111404 rad / s . The centripetal force is F c = mr ω 2 = GM earth m r 2 , so r = 3 radicalbigg GM earth ω 2 = 3 radicalBigg 6 . 67259 × 10 11 N · m 2 / kg 2 (0 . 00111404 rad / s) 2 × 3 radicalbig 5 . 98 × 10 24 kg = 6 . 85064 × 10 6 m is the radius of the satellite’s orbit. At this distance from the Earth, F = GM earth m r 2 = mg , so g = GM earth r 2 = 6 . 67259 × 10 11 N · m 2 / kg 2 (6 . 85064 × 10 6 m) 2 × (5 . 98 × 10 24 kg) = 8 . 50223 m / s 2 . 002 10.0 points Planet X has a mass 7 . 17 times that of the Earth and a radius 2 . 77 times the radius of the Earth. What is the acceleration due to gravity on the surface of Planet X, measured in units of g (the gravitational acceleration on the surface of the Earth)? Correct answer: 0 . 934458 g . Explanation: Let : M X = 7 . 17 M E and R X = 2 . 77 R E . The acceleration due to gravity is a = GM R 2 ∝ M R 2 , so g X g E = R 2 E R X 2 M X M E g X = parenleftbigg R E 2 . 77 R E parenrightbigg 2 parenleftbigg 7 . 17 M E M E parenrightbigg g Earth = . 934458 g ....
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This note was uploaded on 04/07/2012 for the course PHYSICS 121 taught by Professor James during the Spring '12 term at University of Texas.
 Spring '12
 James
 Physics

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