233_Dynamics 11ed Manual

233_Dynamics 11ed Manual - Engineering Mechanics - Dynamics...

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Engineering Mechanics - Dynamics Chapter 13 tangent to the earth’s surface at perigee and the period of its orbit. Given: h p 800 km = G 66.73 10 12 × m 3 kg s 2 = h a 2400 km = M e 5.976 10 24 kg × = s 1 6378 km = Solution: r p h p s 1 + = r p 7178km = r a h a s 1 + = r a 8778km = r a r p 2 GM e r p v 0 2 1 = v 0 1 r a r p r p 2 + 2 r p r a r p + () r a e = v 0 7.82 km s = hr p v 0 = h 56.12 10 9 × m 2 s = T π h r p r a + r p r a = T 1.97 hr = Problem 13-115 The rocket is traveling in free flight along an elliptical trajectory The planet has a
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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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