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HW7_prob2_radial impulse - Problem Goldstein 3-12(3rd ed...

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Problem Goldstein 3-12 (3 rd ed. 3.18) From the result of Problem 2-9 (2.11 in 3 rd ed.) we can relate the change in · r incurred by the particle of reduced mass μ to the strength of the radial impulse S , ( μ · r ) 2 ( μ · r ) 1 = S , (1) where the subscripts 1 and 2 refer to values just before and after the impulse, respectively. Since the impulse occurs at perigee, ( μ · r ) 1 = 0 , and we fi nd · r 2 = S/μ . (2) The general expression for the energy of the orbit is E = μ · r 2 2 k r + l 2 2 μr 2 . (3) We can evaluate the initial energy at the perigee, r = r p : E 1 = k r p + l 2 2 μr 2 p . (4) Although r p is no longer the perigee of the new orbit after the impulse, we can still evaluate the fi nal energy at r = r p , because the two orbits share this point in common. We have the result, E 2 = S 2 2 μ k r p + l 2 2 μr 2 p = E 1 + S 2 2 μ . (5) Note that the two orbits have the same angular momentum, l . Because the radial impulse was applied at the perigee, it is parallel to the radius vector at this point and does not generate a torque.
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