# hw10 - zhou(cz3574 – HW10 – mackie –(10611 1 This...

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Unformatted text preview: zhou (cz3574) – HW10 – mackie – (10611) 1 This print-out should have 4 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 . 8 km / s. Determine the satellite’s 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: 2259 . 35 km. Explanation: Let : v = 6 . 8 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 . 8 km / s) 2 parenleftbigg 1 km 1000 m parenrightbigg 3 = 8629 . 35 km , and the height of the satellite above the Earth’s surface is h...
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## This note was uploaded on 03/25/2011 for the course PHYS 122 taught by Professor Hiyishi during the Spring '10 term at Temple College.

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hw10 - zhou(cz3574 – HW10 – mackie –(10611 1 This...

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