HW8 Solution

HW8 Solution - lew (dl9564) – hk8 – Opyrchal –...

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Unformatted text preview: lew (dl9564) – hk8 – Opyrchal – (41104) 1 This print-out should have 10 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 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 parenleftbigg 1 km 1000 m parenrightbigg 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....
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This note was uploaded on 01/22/2009 for the course PHYS Phys 106 taught by Professor Opyrachal during the Fall '08 term at NJIT.

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HW8 Solution - lew (dl9564) – hk8 – Opyrchal –...

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