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Unformatted text preview: hernandez (ejh742) – oldhomework 18 – Turner – (58120) 1 This printout should have 12 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Note: The radius of the asteroid is very much smaller than radius of the orbit and can be considered negligible. It takes satellite 0 . 904 h to revolve around unknown asteroid in circular orbit of radius 110 km. How long it will take an object without initial speed to reach surface of the asteroid from this orbit? Correct answer: 0 . 32 h. Explanation: Trajectory of object when it falls from some height without initial velocity to the asteroid could be considered as extreme case of ellipse. The semimajor axis in this case is distance from initial point to the center of the asteroid and period is twice time of fall t . Therefore Kepler’s third law could be writ ten as (2 r ) 3 T 2 = r 3 (2 t ) 2 . Thus t = T 2 √ 2 = (0 . 904 h) 2 √ 2 = 0 . 319612 h . 002 10.0 points Given: G = 6 . 67259 × 10 11 N m 2 / kg 2 A small Moon of a planet has an orbital period of 2 . 08 days and an orbital radius of 5 . 04 × 10 5 km. From these data, determine the mass of the planet. Correct answer: 2 . 34532 × 10 27 kg. Explanation: From the Kepler’s third law for the case of negligible mass of the Moon, T 2 = parenleftbigg 4 π 2 GM parenrightbigg r 3 , where M is the mass of the planet, we obtain M = 4 π 2 r 3 GT 2 = 4 π 2 (5 . 04 × 10 5 km) 3 (6 . 67259 × 10 11 N m 2 / kg 2 )(2 . 08 days ) 2 = 2 . 34532 × 10 27 kg ....
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This note was uploaded on 06/14/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Physics, Work

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