Analytical Mech Homework Solutions 81

# Analytical Mech Homework Solutions 81 - E 1 + m c 2 = 1 k 1...

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2 2 1 1 1 E mc k E a  +   =  ++   D D () 2 2 3 fa E a k E a + += D D 1 2 3 a ψπ =+ 1 2 2 1 k a E + D 6.31 From equation 6.5.18a 2 (1 ) 1c o s a r ε θ = + (Here θ is the polar angle of conic section trajectories as illustrated by the coordinates in Figure 6.5.1) … and the data on ellipses in Figure 6.5.1 0 ) ra = so 1 o s com rr + = + D From equation 6.5.18b o s r α = + and at 0 = D 0 1 r = + And from equation 6.5.19 2 ml k = so 2 0 ) ml r k = + 1 mm kG M mG M == From Example 6.5.3, and lr 2 ee GM a v = 22 2 2 sin com com v φ = ( ) 2 2 sin 1 11 c o s com com com rv r av εθ + = 1 1s i n o s RV = + 1 cos sin 1
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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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