# Potential for particle circular motion

1. Mar 30, 2010

### MartinK

Hi everyone!

I have a small question that bothers me. Consider a test particle in the Earths gravitational field on circullar obrit. Speciffic effective potential is
$$\widetilde{V}_{\rm eff(1)} = \frac{1}{2} \frac{l^2}{m^2} \frac{1}{r^2} - \frac{GM}{r}, \nonumber$$
where
$$l = m r^2 \Omega = {\rm const.}, \nonumber$$
because $$\theta$$ is cyclic coordinate. But we have partice in circular orbit, so $$\Omega =$$const. too, and we can write
$$\widetilde{V}_{\rm eff(2)} = \frac{1}{2} \Omega^2 r^2 - \frac{GM}{r}.\nonumber$$
Now we have correct form:
$$\frac{ \partial^2 \widetilde{V}_{\rm eff(1)}}{\partial r^2} = \frac{3 l^2}{m^2 r^4} - \frac{GM}{r} = 3 \Omega^2 - \frac{2GM}{r^3}.\nonumber$$
and incoorect form
$$\frac{ \partial^2 \widetilde{V}_{\rm eff(2)}}{\partial r^2} = \Omega^2 - \frac{2GM}{r^3}.\nonumber$$

Why is there a difference between $$\partial^2 \widetilde{V}_{\rm eff(1)} / \partial r^2,$$ and $$\partial^2 \widetilde{V}_{\rm eff(2)} / \partial r^2,$$?

Thanks for any replies, have a nice day with physics :-)
Martin

2. Apr 7, 2010

### MartinK

I found the answer with the help of my friends.
Second potential is wrong, because $\Omega$ is constant only with time - it depends on $r$.