= 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,
$$
\Phi_{\rm eff}(R,z)=-\frac{GM}{\sqrt{R^2+z^2}}-\frac12\Omega_0^2R^2.
$$
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 is
$$
\Phi_{\rm eff}(s)=-\frac{GM}{\sqrt{R_0^2+2R_0s\sin i+s^2}}-\frac{GM}{2R_0^3}(R_0+s\sin i)^2.
$$
Its first <derivative> at $s=0$ is zero by the <Keplerian rotation> balance. At the footpoint the meridional <Hessian> is $\operatorname{diag}(-3\Omega_0^2,\Omega_0^2)$ with zero mixed derivative. Hence
$$
\boxed{\Phi_{\rm eff}(s)=-\frac{3GM}{2R_0}+\frac12\Omega_0^2(1-4\sin^2i)s^2+O(\Omega_0^2s^3/R_0).}
$$
When $0\leq i<30^\circ$, 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:
$$
\boxed{-\frac{d\Phi_{\rm eff}}{ds}=\Omega_0^2(4\sin^2i-1)s+O(\Omega_0^2s^2/R_0)>0\quad(s>0\text{ small},\ i>30^\circ).}
$$
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 $-7\Omega_0^2s^3/(16R_0)$, as in <marginal straight-line launch at thirty degrees>. Global wind escape additionally depends on conditions farther along the line.
Back to article page
