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DarkEternal
Sep17-04, 11:31 PM
I have a central force motion with vector form r''=(k^2)*r, where k^2>0. It's trivial to solve for the vector r(t), but I'm having a little trouble solving for r(theta). I get a second order differential equation of the form r''=A/r^3+Br. Any tips on solving this?

Gokul43201
Sep18-04, 12:27 AM
r'' = \frac {A}{r^3} + Br = - \frac {d}{dr}(\frac{A}{2r^2} - \frac{Br^2}{2})

Multiply both sides with r' = dr/dt

LHS = r'r" = \frac{1}{2} \frac {d}{dt} r'^2
RHS = -\frac{dX}{dr}.\frac{dr}{dt} = -\frac{dX}{dt}
This gives :
r'^2 = -2(X) + const.

DarkEternal
Sep18-04, 02:36 AM
think i got it, thanks gokul