 Solution

Ideal flux freezing anchors a field line in the highly conducting disc, so its field-line angular velocity naturally equals the angular velocity of its footpoint. A circular orbit in the central potential is Keplerian:

Put $r=r_0+\delta r$ and expandThe linear radial term vanishes because $GM/r_0^2={\Omega }_0^2{r}_0$. The radial and vertical second derivatives are respectively $-3{\Omega }_0^2$ and ${\Omega }_0^2$, so

If the field line makes angle $\theta$ from the vertical, then locally $\delta r=z\tan \theta$. Its quadratic potential change iswhich is negative exactly whenThis is the local magnetocentrifugal acceleration launching criterion.