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Analytical Mech Homework Solutions 74

# Analytical Mech Homework Solutions 74 - Integrate LHS by...

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Integrate LHS by parts 2 0 0 0 1 1 1 mr r mr dt F rdt τ τ τ τ τ τ = ± ± The first term is zero – since the quantity has the same value at 0 and τ . Thus 2 T F = − r where denote time average of the quantity within brackets. but dV k r F r r V dr r V = = = = − ⋅∇ hence 2 T V = − but 2 2 V V E T V V = + = − + = hence 2 V = E but 2 k E constant a = − = and 0 1 2 k E Edt E a τ τ = = = − so 2 k E a = − Thus: k V a = − as before and therefore 1 2 2 k T V a = − = 6.21 The energy of the initial orbit is 2 1 2 2 k k mv E r a = = − (1) 2 2 1 k v m r a = Since ( 1 a r a ) ε = + at apogee, the speed , at apogee is 1 v ( ) ( ) ( ) 2 1 1 2 1 1 1 k k v m a a ma ε ε ε = = + + To place satellite in circular orbit, we need to boost its speed to
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