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class20 - ated with this force is V r =-k r(10 with an...

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1 Central Force Motion REFERENCE: Arya, Sections 7.1 to 7.4 Review the derivation of the following important equations: 1. The effective single particle equation: μ ¨ r = F ( r r (1) 2. The force (system of) equations: F ( r ) = μ r - r ˙ θ 2 ) (2) 0 = μ ( r ¨ θ + 2 ˙ r ˙ θ ) (3) 3. The integrals (constants) of motion: Angular momentum: L = μr 2 ˙ θ = constant (4) Energy E = 1 2 μ ˙ r 2 + L 2 2 μr 2 + V ( r ) = constant (5) 4. The effective potential V eff ( r ) = V ( r ) + L 2 2 μr 2 (6) example: An asteroid of mass m = 1 moves in a gravitational field of the sun F ( r ) = - k r 2 ˆ r. (7) At large distances, it has speed v , with a perpendicular distance from the sun of a . Is the motion bounded? Find an equation for the distance of closest approach. solution: At large distance, the motion of the asteroid is unaffected by the force field. The integral of motion corresponding to the angular momentum about the sun L = av (8) and the total energy E = 1 2 v 2 (9) 1

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the ladder being the kinetic energy, which is a positive constant. Since the potential associ-
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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 2-k r = constant = 1 2 v 2 (12) where, since the apsidal distance is a turning point, ˙ r = 0, and L 2 2 μr 2-k 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 2-k r = 1 2 v 2 (14) ( av ) 2 2 r 2-k r = 1 2 v 2 (15) multiplying by 2 r 2 a 2 v 2-2 kr = r 2 v 2 (16) or v 2 r 2 + 2 kr-a 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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