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Analytical Mech Homework Solutions 76

Analytical Mech Homework Solutions 76 - 6.23 From Problem...

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6.23 From Problem 6.9, ( ) 2 4 3 GMm f r m r πρ = − Gr ( ) 3 2 4 3 GMm f r m r πρ = G ( ) ( ) 3 3 3 2 4 4 2 2 3 3 4 4 1 3 3 a GMma mG f a M f a a GMma mGa a M πρ πρ πρ πρ + = = + From equation 6.14.3, ( ) ( ) 1 2 3 f a a f a ψ π = + 1 1 3 2 3 2 3 3 4 4 1 4 2 3 3 3 4 4 1 1 3 3 a a M M a a M M πρ πρ ψ π π πρ πρ + + = + = + + 1 2 1 1 4 c c ψ π + = + , 3 4 3 a c M πρ = 6.24 We differentiate equation 6.11.1b to obtain ( ) dU r mr dr = − ±± For a circular orbit at r a , r ±± so = 0 = 0 r a dU dr = = For small displacements x from r a = , and r x
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