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Closest approach from initial velocity and impact parameter
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[QUOTE="Kelli Van Brunt, post: 6240779, member: 653509"] ##\dot r## seems it would equal zero at the point of closest approach, since that will also be a turning point for the particle, so the radial distance is not changing at that instant. I assume this is true for all basic two-body orbits? So, using ##E = \frac{m \dot r^2}{2} + \frac{L^2}{2mr^2} - \frac{GMm}{r}## and substituting ##E = \frac{mv_0^2}{2}## and ##\dot r = 0##, after multiplying through by ##r^2## and using the quadratic formula, we get $$r = \frac{-GM - \sqrt{(GM)^2+v_0^4b^2}}{v_0^2}$$ I assume we are taking the negative root of the solution here, since we want the minimum value of r. The eccentricity of this orbit is given by $$e = \sqrt{1 + \frac{2EL^2}{(GMm)^2}}$$ Substituting the expressions for E and L, we get $$e = \frac{\sqrt{(GM)^2+v_0^4b^2}}{GM}$$ Combining the expression for e with the expression for r gives, finally, $$r = \frac{GM(e+1)}{v_0^2}$$ This is closer to the actual answer in that I managed to get eccentricity in there, and at least the units are correct, but it is not equivalent to the given ##\frac{(v_0b)^2}{GM(1+e)}##. I can't tell where I'm making a mistake here, unless my initial assumption of ##\dot r = 0## at the closest approach is incorrect? [/QUOTE]
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Closest approach from initial velocity and impact parameter
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