HW7_prob4_orbit1-3potential

HW7_prob4_orbit1-3potential - Problem: Orbits with a 1-3...

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Problem: Orbits with a 1-3 central potential The equation of motion for the one-dimensional radial problem is μ ·· r = f ef ( r )= dV dr + l 2 μr 3 , (1) where l is the constant angular momentum, μr 2 · θ ,and μ is the reduced mass of the two particle system. The 1-3 potential is de f ned as V ( r )= k r a 3 r 3 , (2) where k and a are positive contants. With a =0 , we recover the unperturbed Kepler problem. The attractive r 3 term arises from general relativistic correc- tions to Newtonian dynamics. It causes the Keplerian orbits to precess. The radial equation of motion can be written explicitly as μ ·· r = f ef ( r )= k r 2 a r 4 + l 2 μr 3 = dV 0 dr , (3) where V 0 is the e f ective potential for this problem, V 0 = V ( r )+ l 2 2 μr 2 . (4) (a) Circular Orbits: The circular orbit condition is that the e f ective force must vanish when r = r c , f ef ( r c )= [ dV 0 ( r c ) /dr c ]=0 ,where r c is the radius of the circular orbit, f ef ( r c )= k r 2 c a r 4 c + l 2 μr 3 c =0 . (5) This equation can be simpli f ed and solved as follows. First, introduce a scaled radial coordinate x c = r c /r 0 , (6) where r 0 is the radius of the circular orbit when a =0 , r 0 = l 2 /μk. (7) Then Eq.(5) reduces to x 3 c x 2 c cx 4 c =0 , (8) where c is the positive, dimensionless constant c = a/ ( kr 2 0 ) . (9) Then, after multiplying Eq.(8) by x 4 c , we obtain a quadratic equation for the variable x c , x 2 c x c + c =0 , (10) 1
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with solutions given by the quadratic formula 2 x c =1 ± 1 4 c. (11) Real roots are possible only for c 1 / 4 . The larger root always corresponds to a stable circular orbit (see the subsequent analysis and plot of
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This note was uploaded on 12/19/2010 for the course PHYSICS ph 409 taught by Professor Wilemski during the Fall '10 term at Missouri S&T.

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HW7_prob4_orbit1-3potential - Problem: Orbits with a 1-3...

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