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Unformatted text preview: Phys 263/Amath 261 Assignment 6 SOLUTIONS 1. For the orbit r = ke αθ (1) (a) We use the path equation to determine the force law. First note, d dθ 1 r = d dθ e αθ k = αe αθ k (2) and d 2 dθ 2 1 r = α 2 e αθ k = α 2 r (3) then from the path equation F ( r ) = L 2 μr 2 α 2 r + 1 r (4) or F ( r ) = L 2 μr 3 ( α 2 + 1 ) (5) and inverse cube. (b) From the first integral of motion ˙ θ = L μr 2 = L μk 2 e 2 αθ (6) which we can use to get e 2 αθ dθ = L μk 2 dt (7) integrating: e 2 αθ 2 α = Lt μk 2 + C (8) where C is the constant of integration. Multiply by 2 α and let C = 2 αC , e 2 αθ = 2 αLt μk 2 + C (9) 1 taking the log of both sides: θ ( t ) = 1 2 α ln 2 αLt μk 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 αθ = 2 αLt μk 2 + C (11) then r ( t ) = s 2 αLt μ + Ck 2 (12) (c) We can find the energy from the integral of motion for energy E = 1 2 μ ˙ r 2 + L 2 2 μr 2 + V ( r ) (13) if we can find ˙...
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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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