### Rigid Body

The equilibrium position is along the radius, when the center of mass moves on the circular orbit. The angular velocity $\omega_0$ on the orbit is obtained from $m\omega_0^2r=GMm/r^2$, that is, $\omega_0=\sqrt{GM/r^3}$. Since the graviational force and the centrifugal force depend on the radius, different forces are applied to the two masses. The total force on a mass at the radius $R$, rotating with the rod, is $F=-GMm/R^2+m\omega^2R$, so that the difference is $\Delta F=(dF/dR)l=(2GMm/r^3+\omega_0^2)l=(3GMm/r^3)l$. If the rod rotates by a small angle $\theta$ the resulting torque is $N=-\Delta F l\theta$, so that one has
$I\ddot{\theta}=-(3GMml^2/r^3)\theta$
where $I=2m(l/2)^2=ml^2/2$. Thus,
$\omega^2=6GM/r^3=6\omega_0^2$