Quote by Reshma
I don't have a clue. The force can be shown as [itex]\vec F = \vec \nabla V[/itex]. The work done will be: [itex]W = \int \vec F \cdot dr[/itex]. Will this equal the energy?

[tex]T=(1/2)mv^2[/tex]
Now, what is this is polar coordinates? Well, [tex]v^2=\dot{r^2}+r^2 \dot{\theta^2}[/tex]
And, of course, E = T + V(r)...
Dan
BTW, you can get rid of the [tex]\dot{\theta}[/tex] by using angular momentum conservation.