HW6_prob2_probGPS11_G3

# HW6_prob2_probGPS11_G3 - r 2 = k 1 R 1 r , (7) where r = 0...

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Problem Goldstein 3-3 (3 rd ed. 3.11) We can think of this two particle system as a single particle of reduced mass μ in a circular orbit of period τ ,rad ius R , and angular momentum l .W ec an relate these quantities using Kepler’s Law of equal areas in equal time dA dt = l 2 μ , (1) which, for a circular orbit of area π R 2 ,integratesto π R 2 = l τ 2 μ . (2) Because we have no information about l , we would like to eliminate it from consideration. We can do this using another property of a circular orbit, namely, that at r = R , dV 0 /dr =0 ,wh e r e V 0 is the e f ective potential for the one- dimensional radial motion, V 0 = k r + l 2 2 μr 2 . (3) Setting the f rst derivative of V 0 equal to zero results in R = l 2 μk , (4) andwhencomb inedw ithEq .(2)g ivesus R 3 = k μ ³ τ 2 π ´ 2 . (5) At the instant that the circular motion is halted, the angular momentum becomes zero ( l =0 ) and the radial EOM simpli f es to μ ·· r =

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Unformatted text preview: r 2 = k 1 R 1 r , (7) where r = 0 at r = R . Now solve this equation for r , and rearrange it into the form s 2 k R 3 dt = dr R 3 / 2 q 1 r 1 R = dx q 1 x 1 , (8) 1 where x = r/R , (9) and we choose the minus sign because t will increase as r decreases. If we make the variable substitution, x = sin 2 , dx = 2 sin cos d , (10) the integration of Eq.(7) becomes straightforward. We integrate the left side from t = 0 to t = t c and the right side from = / 2 ( x = 1 ) to = 0 ( x = 0 ) to obtain s 2 k R 3 t c = 2 Z / 2 sin 2 d = Z / 2 (1 cos 2 ) d = 2 . (11) Thus, the time until the particles collide is t c = 2 r R 3 2 k = 4 2 , (12) where the right hand side is found after we substitute Eq.(5) for R. 2...
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## HW6_prob2_probGPS11_G3 - r 2 = k 1 R 1 r , (7) where r = 0...

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