 Solution

Matter corotating with the footpoint has fixed angular velocity ${\Omega }_0=(GM/R_0^3{)}^{1/2}$. Its effective potential in that rotating frame is the sum of Newtonian gravitational potential and centrifugal potential,Along an outward-inclined straight magnetic field line, write $R=R_0+s\sin i$ and $z=s\cos i$, where $s$ is distance from the footpoint. The exact potential along it isIts first derivative at $s=0$ is zero by the Keplerian rotation balance. At the footpoint the meridional Hessian is $\mathrm{diag}(-3{\Omega }_0^2,{\Omega }_0^2)$ with zero mixed derivative. HenceWhen $0\le i<30^{\circ }$, the potential initially rises and a cold parcel must overcome a barrier. When $i>30^{\circ }$, its quadratic curvature is negative, so the footpoint is a local maximum along the field and a small outward displacement is accelerated further outward:This is the thirty-degree magnetocentrifugal launching criterion. It describes the lack of a local cold-launch barrier, not a nonzero force on a parcel exactly at rest at $s=0$. At exactly $30^{\circ }$ the quadratic term vanishes; for the specified straight line the next term is $-7{\Omega }_0^2{s}^3/(16R_0)$, as in marginal straight-line launch at thirty degrees. Global wind escape additionally depends on conditions farther along the line.