Answer Happy

6) For hyperbolic trajectories (e > 1), the time t is related to the true anomaly as 1 eve-Isin 1+c86 Ve+1+√e-1tan f Vell √e Itan H)]. (e² - 1)³/2 which can be put in the hyperbolic Kepler's equation form as Mesinh F-F where the hyperbolic mean anomaly is M₁ = (²-1)/2 t, and the hyperbolic eccentric anomaly F is F =sinh-1 ve²-1 sin 01 1+ e cos = lu Ve+1+√e-1tan √e+1-√e-1tang By using this result, first show that cosh F= cos + e 1+ e cos 6¹ by using the hyperbolic trigonometric identity cosh" - sinh" z = 1. Then, show that the hyperbolic eccentric anomaly F can be put another form as F = 2tanh-¹ tan e+1 by using the hyperbolic trigonometric identity tanh elliptic version sinh F = 1+cosh!! This form is analog of the E /1 c 8 tan1+e tana ¹2 By putting M₁ and F in place in the hyperbolic Kepler's equation, show that one can obtain (9.42) as t = { (√)] esinh 2 tanh tan 2 tanh tan- 4² (²1)3/2 cl