I am told that a particle moves under the action of an attractive central force $F=\frac{-k}{r^2}\hat r,$ with angular momentum $L.$
I am asked to find the energy for which the motion is circular and the radius of the circle. So far I have said the following:
$$L=\vec r \times m\vec v = \begin{vmatrix} \hat r & \hat\theta & \hat z \\ r & 0 & 0 \\ m \dot r & m r \dot \theta & 0 \end{vmatrix} = mr^2\dot\theta\hat z .$$
Using $F=ma,$ we have that $$\frac{-k}{r^2}=m(\ddot r-r \dot \theta^2)$$
Then using angular momentum substitute in for $\dot\theta$ giving $$\frac{-k}{r^2}=m(\ddot r-\frac{L^2}{m^2r^3})$$
Rearranging this then gives $$m\ddot r-\frac{L^2}{mr^3}+\frac{k}{r^2}=0$$
Then multiply by $\dot r$ to give $$m\ddot r \dot r-\frac{L^2\dot r}{mr^3}+\frac{k\dot r}{r^2}=0$$
Then we have $$\frac{1}{2}m\dot r^2+\frac{L^2}{2mr^2}-\frac{k}{r}=\text{ constant }=E$$
To find the radius I then set $\frac{dP}{dr}=0$ where P is the effective potential $(\frac{L^2}{2mr^2}-\frac{k}{r}).$
This gives me $$\frac{L^2}{mr^3}=\frac{k}{r^2}$$
And so the radius is $$r=\frac{L^2}{mk}$$
How do I find for what energy the motion is circular though?