237_Dynamics 11ed Manual

237_Dynamics 11ed Manual - Engineering Mechanics - Dynamics...

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Engineering Mechanics - Dynamics Chapter 13 v 0 2 OA' OB OB 2 + OB OA' OB + () OA' GM p = v 0 36.5 10 3 × ft s = (speed at B ) v A' OB v 0 OA' = v A' 7.3 10 3 × ft s = hO B v 0 = h 385.5 10 9 × ft 2 s = Thus, T π OB OA' + h OB OA' = T 12.19 10 3 × s = T 2 1.69 hr = *Problem 13-120 The speed of a satellite launched into a circular orbit about the earth is given by Eq. 13-25. Determine the speed of a satellite launched parallel to the surface of the earth so that it travels in a circular orbit a distance d from the earth’s surface. Given: d 800 km = G 6.673 10 11 × N m 2 kg 2 = M e 5.976 10 24 kg × = r e 6378 km = Solution: v GM
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