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:
$$
\omega(\psi)=\Omega_0
=\sqrt{\frac{GM}{r_0^3}}.
$$

Put $r=r_0+\delta r$ and expand
$$
\Phi_{cg}=-\frac{GM}{\sqrt{r^2+z^2}}
-\frac12\Omega_0^2r^2.
$$
The linear radial term vanishes because $GM/r_0^2=\Omega_0^2r_0$. The radial and vertical second derivatives are respectively $-3\Omega_0^2$ and $\Omega_0^2$, so
$$
\boxed{\Phi_{cg}
\simeq\text{constant}
+\frac12\Omega_0^2[z^2-3(r-r_0)^2]}.
$$

If the field line makes angle $\theta$ from the vertical, then locally $\delta r=z\tan\theta$. Its quadratic potential change is
$$
\frac12\Omega_0^2z^2(1-3\tan^2\theta),
$$
which is negative exactly when
$$
\boxed{\theta>30^\circ}.
$$
This is the local <magnetocentrifugal acceleration> launching criterion.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 314 / 4 / e / Solution

{wiki=Magneto-centrifugal_launching}

Magnetocentrifugal acceleration occurs when matter tied to a rotating inclined magnetic field moves outward under the effective centrifugal force. For a Keplerian disc, cold launching is possible when the field tilts by more than $30^\circ$ from the vertical.


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 Incoming links: Magnetocentrifugal acceleration