Analytical Mech Homework Solutions 67

Analytical Mech Homework Solutions 67 - 4 M d = r3 3 GMm 4...

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3 4 3 d M r π ρ = () 2 4 3 GMm F r mGr r πρ =− 6.10 11 k ue rr θ == D k du k e dr D 22 2 2 k du k ek D u From equation 6.5.10 … 2 21 1 uk uu fu du ml u += 2 3 () ( ) 12 2 1 f um l k + u 3 1 ml k fr r + The force varies as the inverse cube of r. From equation 6.5.4, 2 rl = ± 2 2 k dl e dt r = D 2 2 k l ed d r = D t 2 2 1 2 k lt eC kr =+ D 2 ln 2 klt C    D varies logarithmically with t. 6.11 3 3 k f rk r u From equation 6.5.10 …
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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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