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Is there a difference in Binet's equation when the force acting on a body is attractive and repulsive?

I mean, if the force has a magnitude $F(r)$, is it Binet's equation always $$\frac{d^2u}{d\theta ^2}+u=-\frac{F(u^{-1})}{mh^2u^2}, u=r^{-1},$$ no matter if the force is attractive or impulsive?

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    $\begingroup$ You can simply change the mass sign to negative to describe repulsion. $\endgroup$
    – TMS
    Commented Jul 22, 2013 at 9:39
  • $\begingroup$ so, the sign in formula changes when it's repulsive? When is it $-$ and when $+$ at the right side? $\endgroup$
    – gov
    Commented Jul 22, 2013 at 20:13
  • $\begingroup$ The one you have above is derived for gravity, which is attractive force. $\endgroup$
    – TMS
    Commented Jul 23, 2013 at 15:44

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This formula actually holds for any central force, $$\vec F=F(r)\hat r,$$ not only for gravity. For instance, for an attractive (repulsive) inverse square force, $$\vec F=\mp\frac{K}{r^2}\hat r,\quad K>0,$$ we have $$\frac{d^2u}{d\theta ^2}+u=\mp\frac{K}{mh^2}.$$ As you can see they are different. The solutions for the repulsive case are unbounded while the attractive case can be either unbounded or bounded.

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