Analytical Mech Homework Solutions 82

Analytical Mech - 1 2 1 2 sin = 2 RV 2 sin 2 2 RV 2 sin 2 1 Again from Example 6.5.3 1 2 2 2 = 1 V 2 RV sin R 2 = 1 RV 2 sin 2 RV 2 sin 2 2 1 2 22

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() 1 2 2 22 2 2 2 1 sin sin 2 sin 1 RV RV θε φ ε  =− +   Again from Example 6.5.3 … 1 2 2 2 2 1s i VR V R εφ  =+   n 2 2 2 i n 2 s i n RV RV φφ ( ) 1 2 2 1 sin sin sin RV RV θφ 2 1 sin sin cos RV θ = 2 cos sin 1 sin sin cos RV RV = 2 2 cot tan sin 2 RV 1 2 2 cot tan csc2 RV For , 0.5 V = 4 R = , : 30 = D 1 2 2 cot tan30 csc60 4.5 DD 11 12 cot cot 3 33 2 −− = 30 D 6.32 The picture at left shows the orbital transfer and the position of the two satellites at the moment the transfer is initiated. Satellite B is “ahead” of satellite A by the angle θ 0 r 2 r 1 θ 0 2 rr a + = is the semi-major axis of the elliptical transfer orbit. From Kepler’s 3
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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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