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Unformatted text preview: alexander (jra2623) – oldhomework 17 – Turner – (92510) 1 This printout should have 11 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A geosynchronous moon appears to remain over one spot on a planet similar to Earth with a day the same length of time but with a different density. The moon has an orbital radius of 4 . 03 × 10 7 m . Calculate its speed in orbit. Correct answer: 2930 . 7 m / s. Explanation: Since the moon is geosynchronous, its pe riod T = 24 hours. Its speed, therefore, is v = 2 π r T = 2 π (4 . 03 × 10 7 m) (86400 s) = 2930 . 7 m / s . 002 (part 2 of 2) 10.0 points The mass of the moon is 1 . 4 × 10 20 kg. What is the mass of the planet? Correct answer: 5 . 18744 × 10 24 kg. Explanation: The mass of the moon is not relevant in determining the mass of the planet. Kepler 3 rd law is T 2 = 4 π 2 G parenleftbigg r 3 M parenrightbigg . Since T = 2 π r v , we have M = v 2 r G = (2930 . 7 m / s) 2 (4 . 03 × 10 7 m) (6 . 67259 × 10 − 11 N m 2 / kg 2 ) = 5 . 18744 × 10 24 kg . 003 10.0 points How much work would it take to carry an object of mass M from the surface of the earth to an effectively infinite distance away? The earth has mass M E and radius R E , and U G ( r ) = − G M E M r = − M g R 2 E r . 1. M g R E correct 2. GM E R 2 E 3. Because the force vanished like 1 r 2 and r × 1 r 2 = 1 r → 0 as r → ∞ , the work required is zero....
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This note was uploaded on 10/20/2011 for the course PHY 302K taught by Professor Kaplunovsky during the Summer '08 term at University of Texas at Austin.
 Summer '08
 Kaplunovsky
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