Analytical Mech Homework Solutions 76

Analytical Mech - 6.23 From Problem 6.9 f r = GMm 4 mGr 3 r2 2GMm 4 mG 3 r3 4 4 a 3 2GMma 3 mG 2 f(a 3 3M = = 4 f a GMma 2 mGa 4 a 3 a 1 3 3M f(r =

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6.23 From Problem 6.9, () 2 4 3 GMm f rm r πρ =− G r 3 24 3 GMm f r G 3 3 3 2 44 22 33 4 4 1 3 3 a GMma mG fa M a GMma mGa a M −− + ==  +   From equation 6.14.3, 1 2 3 a ψπ  =+   1 1 3 2 3 2 4 4 14 2 3 3 3 11 a a M M aa MM π + −+ = ++ 1 2 1 c c + = + , 3 4 3 a c M = 6.24 We differentiate equation 6.11.1b to obtain ( ) dU r mr dr ±± For a circular orbit at ra , r so = 0 = 0 dU dr = = For small displacements x from = , and rxa rx = From Appendix D … ()( )( ) 2 2 x fxa fa x f a ′′ += + + + Taking f r to be dU dr , 2 2 dU fr dr = Near = 2 2 dU dU d U x dr dr dr + 2 2 mx x dr = This represents a “restoring force,” i.e., stable motion, so long as
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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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