Past exam of the mathematics course of the University of Oxford / 2023 / Prelims / Mathematics IV / 5 / c / Solution

Use the dimensionless variablesThe stated $R>a/2$ gives $0<A<1$. Since $f(r)=4Rx(1-x)$, the inequality from b factorizes asThe two additional possible turning points areThey are real exactly when $\beta +A\le 1$. This exposes the intended critical valuebut the ordering of the initial radius $A$ relative to these roots must also be checked.

First suppose $0<A<1/3$. Then $A<(1-A)/2$. If $\beta <1-A$, the allowed branch starting at $A$ is trapped either between $x_{-}$ and $A$ or between $A$ and $x_{-}$, according to its initial direction. At the coincident-root value ${\beta }^2=4A(1-2A)$ it is exactly a circular orbit. For $\beta =0$, the inner radial bound is the axis rather than a nonzero turning point.

At $\beta =1-A$, the additional roots coalesce at $x_{\ast }=(1-A)/2>A$. The particle starts outward and approaches $x_{\ast }$, orwithout reaching it in finite time. Indeed $D$ is proportional to $(x-A)(x-x_{\ast }{)}^2$ and the radial energy equation gives $dt/dx$ proportional to $1/|x-x_{\ast }|$ near $x_{\ast }$. Its integral diverges. This is a bounded separatrix approaching an unstable circular orbit. For $\beta >1-A$, there is no outer root and the particle escapes. Thus for $0<A<1/3$, boundedness holds exactly when $\beta +A\le 1$, with the equality case included. At $A=1/3$, the same inequality characterizes boundedness, but equality is the exact circular orbit at the starting radius.

The source's weaker assumption $A<1$ is insufficient to make this single threshold a general criterion. For $1/3<A<1/2$, the initial radius lies beyond the coalescing-root position. The transition is instead the unstable circular orbit atBelow this speed the initial radial acceleration is inward and the particle remains in the inner allowed interval. At equality it stays exactly circular; above it the initial acceleration is outward and the accessible branch extends to infinity, even if $\beta +A<1$. For $A>1/2$, the surface slopes downward outward and every such launch escapes. At $A=1/2$, only $\beta =0$ remains at rest on the summit circle.

Consequently the complete classification for the intended azimuthal launch, including zero launch speed and exact equilibria, isFor a concrete counterexample to the unqualified proposed criterion, take $A=3/4$ and $\beta =1/10$. Then $\beta +A=17/20<1$, but $f^{\prime }(a)<0$ and the outward motion has no accessible outer barrier. This satisfies the printed radius assumption and escapes. A sufficient additional radius restriction for the intended threshold is $a\le 2R/\sqrt{3}$, with the equality and circular orbit qualifications above.