Analytical Mech Homework Solutions 75

# Analytical Mech Homework Solutions 75 - k 2 1 m RE a and...

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2 21 E k v mR a  =−   D and solving for a 2 2 E E R a R mv k = D noting that … (3) E E EE kG M gR mR R == 3 2 4.49 10 1.426 2 E RR ak v gR = D m The eccentricity ε can be found from the angular momentum per unit mass, l , equation 6.5.19, and the data on ellipses defined in figure 6.5.1 … () 1 1 2 2 2 2 1 sin E ka k lr vR mm ε α θθ  = =   DD ± where v , D θ D are the launch velocity, angle Solving for ε (using (3) above) 22 1 2 sin 0.795 vv gR gR εθ = D 0.892 = Inserting these values for a, ε into (2) and using (3) gives (a) 1 1 2 31 2 1 1 1 4.61 10 1 E E Ra vg R k m s ∆= −− = + (b) 3 12 . 0 9 1 E ha R k m =+−= ⋅ 0 {altitude above the Earth … at perigee}
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## This note was uploaded on 01/06/2012 for the course PHYSICS 4360C taught by Professor Johnanderson during the Fall '11 term at UNF.

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