1
$\begingroup$

I was reading Goldstein chapter 3 central force motion. I tried to do the exercise 21 on p. 130 of 3rd ed. but cant think of any way how to do it. any clue would be very helpful.

Show that the motion of a particle in the potential field $$V(r)=-\frac{k}{r}+\frac{h}{r^2}$$ is the same as that of the motion under the Kepler potential alone when expressed in terms of a coordinate system rotating or precessing around the enter of force.

Thinking about the Euler Lagrange equation gives me $$ m\ddot{r}-\frac{l^2}{mr^3}=\frac{k}{r^2}-\frac{2h}{r^3} $$ Now it seems natural to do $$ m\ddot{r}-\frac{l^2}{mr^3}+\frac{2h}{r^3}=\frac{k}{r^2} $$ and setting $h=\frac{l^2}{2m}$ we get $m\ddot{r}=\frac{k}{r^2}$ i dont see how rotating coordinate coming? help please

$\endgroup$
  • 1
    $\begingroup$ The signs of the terms on the right hand side of your application of the Euler-Legrange equation are incorrect. Having corrected those signs you should aim to get an expression of the form $m\ddot{r}=\frac{(L+M)^2-M^2}{mr^3}+\frac{k}{r^2}$ where $M$ is an additional angular momentum term which is much smaller than $L$ and then make the appropriate approximation to get an expression which looks like a perturbed Keplerian equation of motion. $\endgroup$ – Farcher Oct 9 '18 at 14:14
  • $\begingroup$ didnt understand sir.can you work out a little more $\endgroup$ – jowadul kader Oct 10 '18 at 5:30