HW8 Solution

HW8 Solution - lew(dl9564 hk8 Opyrchal(41104 This print-out...

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lew (dl9564) – hk8 – Opyrchal – (41104) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points A satellite moves in a circular orbit around the Earth at a speed oF 5 . 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: 5491 . 5 km. Explanation: Let : v = 5 . 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 (5 . 8 km / s) 2 p 1 km 1000 m P 3 = 11861 . 5 km , and the height oF the satellite above the Earth’s surFace is h = r - R e = 11861 . 5 km - 6370 km = 5491 . 5 km . 002 (part 1 oF 2) 10.0 points A 2 . 5 kg mass weighs 23 . 75 N on the surFace oF a planet similar to Earth. The radius oF this planet is roughly 7 . 7 × 10 6 m. Calculate the mass oF oF this planet. The value oF the universal gravitational constant is 6 . 67259 × 10 11 N · m 2 / kg 2 . Correct answer: 8 . 44132 × 10 24 kg. Explanation: Let : G = 6 . 67259 × 10 11 N · m 2 / kg 2 , m = 2 . 5 kg , r = 7 . 7 × 10 6 m , and W = 23 . 75 N .
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HW8 Solution - lew(dl9564 hk8 Opyrchal(41104 This print-out...

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