A particle of mass $m$ moves under an attractive central force $Kr^4$ with angular momentum $L.$ For what energy will the motion be circular and what is the radius of the circle?
In order to find the radius I have been taught to use $F=ma$, multiply by $\dot r$ and then integrate. Then set $\frac{dV}{dr}=0.$ I have done what I can of this below.
Set $r^2\dot\theta=h.$
$$Kr^4=m(\ddot r-r\ddot\theta^2)$$
$$\Rightarrow Kr^4=m\bigg(\ddot r-\frac{h^2}{r^3}\bigg)$$
$$\Rightarrow Kr^4\dot r=mr\dot r -\frac{h^2}{r^3}\dot r$$
$$\frac{Kr^5}{5}=\frac{1}{2}\dot r^2+\frac{h^2}{2r^2}$$
$$\frac{1}{2}m\dot r^2+\frac{h^2}{2r^2}-\frac{kr^5}{5}=0$$
Then $\frac{dV}{dr}=0$ or $\frac{h^2}{r^3}=-Kr^4.$
And so, $$r=\sqrt[7]{\frac{-h^2}{K}}$$
I'm not convinced that this is correct. Could someone verify this and also explain how to go about finding the energy required for circular orbit?