5-) For elliptical orbits (0
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5-) For elliptical orbits (0
5-) For elliptical orbits (0<e < 1), the time t is related to the true anomaly 0 as 1 1-e 1+e tan ev1-esin 6 1+ e cos 0 (1-2)/2 which can be put in the elliptic Kepler's equation form us M-E-esin E, where the elliptic mean anomaly Me is M. - (1-e²), and the elliptic eccentric anom- aly E is E2tan tan- te By putting M, and E in the elliptic Kepler's equation, show that one can obtain (9.28) or (9.30) as + = = {2-²¹ (VEF) - 200² (√)} 1 }; t= tan esin tan 1+e 1+e Hint: You may use T = and a = 2
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