Analytical Mech Homework Solutions 74

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

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Integrate LHS by parts 2 0 00 11 1 mr r mr dt F rdt ττ τ  ⋅− = ⋅  ∫∫ ±± The first term is zero – since the quantity has the same value at 0 and . Thus 2 TF =− r where denote time average of the quantity within brackets. but dV k rF r r V dr r V −⋅ = = = = ⋅∇ hence 2 TV but 22 VV ETV V =+= −+= hence 2 V = E but 2 k E constant a = and 0 1 2 k EE d t E a == = so 2 k E a = − Thus: k V a as before and therefore 1 k a = 6.21 The energy of the initial orbit is 2 1 kk mv E ra −== (1) 2 21 k v mr a    Since ( 1 a ) ε =+ at apogee, the speed , at apogee is 1 v () ( ) 2 1 1 v ma a m a = ++ To place satellite in circular orbit, we need to boost its speed to
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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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