# Effective potential energy minimum

Effective potential energy is defined by
$$U^*(\rho)=\frac{L^2}{2m\rho^2}+U(\rho)$$
in many problems I found that particle will have stable circular orbit if $$U^*(\rho)$$ has minimum.

1. Why is that a case? Why circle? Why not ellipse for example?

2. Is this condition equivalent with
$$\frac{f'(\rho)}{f(\rho)}+\frac{3}{\rho}>0$$?

vanhees71
Gold Member
Around a minimum ##\rho_0## you (very often) have
$$U(\rho)=U_0(\rho_0)+\frac{m \omega^2}{2}$$
with ##m \omega^2=U''(\rho)>0##. This implies that if the particle is close to the minimum ##\rho_0##, it behaves as in a harmonic-oscillator potential. A particular solution is of course ##\rho(t)=\rho_0=\text{const}##. Then you have a circular orbit. Otherwise the particle oscillates around such an orbit.

And when this is not the case? I am really trying to understand this but I have difficulties.

vanhees71