232_Dynamics 11ed Manual

232_Dynamics 11ed - Engineering Mechanics Dynamics Chapter 13 Given 6 h = 4 10 m G = 66.73 10 12 m 3 2 kg s Me = 5.976 10 24 kg R e = 6378 km

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Engineering Mechanics - Dynamics Chapter 13 Given: h 410 6 m = G 66.73 10 12 × m 3 kg s 2 = M e 5.976 10 24 kg × = R e 6378 km = Solution: Circular orbit: v C GM e R e h + = v C 6.199 km s = Parabolic orbit: v e 2 e R e h + = v e 8.766 km s = Δ vv e v C = v 2.57 km s = Problem 13-113 Prove Kepler’s third law of motion. Hint: Use Eqs. 13–19, 13–28, 13–29, and 13–31. Solution: From Eq. 13-19, 1 r C cos θ () GM s h 2 + = For 0deg = and 180deg = 1 r ρ C s h 2 + = 1 r a C s h 2 + = Eliminating C , From Eqs. 13-28 and 13-29, 2 a b 2 2 s h 2 = From Eq. 13-31, T π h 2 a b = Thus, b 2 T 2 h 2 4 2 a 2
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This note was uploaded on 08/16/2011 for the course EGN 3321 taught by Professor Christianfeldt during the Spring '08 term at University of Central Florida.

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