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Unformatted text preview: Phys 263/Amath 261 Assignment 6 SOLUTIONS 1. For the orbit r = ke (1) (a) We use the path equation to determine the force law. First note, d d 1 r = d d e k = e k (2) and d 2 d 2 1 r = 2 e k = 2 r (3) then from the path equation F ( r ) = L 2 r 2 2 r + 1 r (4) or F ( r ) = L 2 r 3 ( 2 + 1 ) (5) and inverse cube. (b) From the first integral of motion = L r 2 = L k 2 e 2 (6) which we can use to get e 2 d = L k 2 dt (7) integrating: e 2 2 = Lt k 2 + C (8) where C is the constant of integration. Multiply by 2 and let C = 2 C , e 2 = 2 Lt k 2 + C (9) 1 taking the log of both sides: ( t ) = 1 2 ln 2 Lt k 2 + C (10) This is our answer for the angle. The constants C and L must be determined from initial conditions. We can similarly solve for r ( t ) by noting r 2 k 2 = e 2 = 2 Lt k 2 + C (11) then r ( t ) = s 2 Lt + Ck 2 (12) (c) We can find the energy from the integral of motion for energy E = 1 2 r 2 + L 2 2 r 2 + V ( r ) (13) if we can find...
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 Spring '10
 RogerMelko

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