Analytical Mech Homework Solutions 82

# Analytical Mech Homework Solutions 82 - 1 2 1 2 sin = 2 RV...

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( ) ( ) 1 2 2 2 2 2 2 2 1 sin sin 2 sin 1 RV RV θ ε φ φ ε = + Again from Example 6.5.3 … ( ) 1 2 2 2 2 1 si V RV R ε φ = + n 2 ( ) 2 2 2 2 1 sin 2 sin RV RV ε φ φ = + ( ) ( ) 1 2 2 2 2 2 2 1 sin sin sin RV RV θ φ ε = φ 2 1 sin sin cos RV θ φ φ ε = 2 2 2 cos sin 1 sin sin cos RV RV θ φ θ φ φ = 2 2 cot tan sin 2 RV θ φ φ = 1 2 2 cot tan csc2 RV θ φ φ = For , 0.5 V = 4 R = , : 30 φ = D ( ) 1 2 2 cot tan30 csc60 4 .5 θ = D D ( ) 1 1 1 2 cot cot 3 3 3 2 = = 30 θ = − 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.
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