HW20 - schneiter (tcs639) HW20 Ross (89152) 1 This...

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Unformatted text preview: schneiter (tcs639) HW20 Ross (89152) 1 This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points A satellite moves in a circular orbit around the Earth at a speed of 6 . 9 km / s. Determine the satellites altitude above the surface of the Earth. Assume the Earth is a homogeneous sphere of radius 6370 km and mass 5 . 98 10 24 kg. The value of the universal gravitational constant is 6 . 67259 10- 11 N m 2 / kg 2 . Correct answer: 2011 . 03 km. Explanation: Let : v = 6 . 9 km / s , R e = 6370 km , M e = 5 . 98 10 24 kg , and G = 6 . 67259 10- 11 N m 2 / kg 2 . The gravitational force provides the cen- tripetal acceleration, so GmM e r 2 = mv 2 r r = GM e v 2 = (6 . 67259 10- 11 N m 2 / kg 2 ) 5 . 98 10 24 kg (6 . 9 km / s) 2 parenleftbigg 1 km 1000 m parenrightbigg 3 = 8381 . 03 km , and the height of the satellite above the Earths surface is h = r- R e = 8381 . 03 km- 6370 km = 2011 . 03 km . 002 10.0 points The planet Saturn has a mass 95 . 2 times that of the earth and a radius 9 . 47 times that of the earth. Find the escape speed for objects on the surface of Saturn. Correct answer: 35 . 5109 km / s. Explanation: Let : M S = 95 . 2 M E and R S = 9 . 47 R E . The escape speed from Saturn is v e.S = radicalBigg 2 GM S R S (1) and the escape speed from Earth is v e.E = radicalBigg 2 GM E R E . (2) Dividing, v e.S v e.E = radicalBigg R E R S M S M E v e.S = radicalBigg R E R S M S M E v e.E = radicalbigg 1 9 . 47 95 . 2 1 (11 . 2 km / s) = 35...
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This note was uploaded on 04/23/2009 for the course PHYS 152 taught by Professor Button during the Spring '08 term at IUPUI.

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HW20 - schneiter (tcs639) HW20 Ross (89152) 1 This...

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