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Unformatted text preview: ated with this force is V ( r ) =k r (10) with an eﬀective potential V eﬀ ( r ) =k r + L 2 2 μr 2 (11) we know that the motion is bounded for all E < 0. Hence the asteroid’s motion is unbounded. To get the apsidal distance, use the integral of motion corresponding to energy: E = 1 2 μ ˙ r 2 + L 2 2 μr 2k r = constant = 1 2 v 2 (12) where, since the apsidal distance is a turning point, ˙ r = 0, and L 2 2 μr 2k r = 1 2 v 2 (13) also, considering the mass of the sun as very large, the reduced mass μ ≈ m asteroid = 1, and L 2 2 r 2k r = 1 2 v 2 (14) ( av ) 2 2 r 2k r = 1 2 v 2 (15) multiplying by 2 r 2 a 2 v 22 kr = r 2 v 2 (16) or v 2 r 2 + 2 kra 2 v 2 = 0 (17) which is a quadratic equation we can solve for r to get the distance of closest approach. This is simpliﬁed if we are given the constants a and v . 2...
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This note was uploaded on 09/20/2010 for the course AMATH 261 taught by Professor Rogermelko during the Spring '10 term at Waterloo.
 Spring '10
 RogerMelko

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