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nocks
#25
Nov21-09, 02:49 PM
P: 24
Quote Quote by George Jones View Post
...
Given

[tex]\left( \frac{dr}{d \lambda} \right)^2 &= E^2 - V^2(r) \right)[/tex]
and
[tex]V^2(r) = \left(1 - \frac{2M}{r} \right)\frac{L^2}{r^2}[/tex]

I have [tex] \frac{d^2r}{d\lambda^2} = -\frac{1}{2}\frac{d}{dr}V^2(r)[/tex]