Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 4 / 10C / iii

Table of contents
- Solution iii

Solution

For

V (r) = - \frac{km}{r - r _{0}},

Binet's equation becomes

u^{''} + u = \frac{K}{( 1 - r _{0} u ) ^{2}}, K = \frac{k m ^{2}}{ℓ ^{2}}, 0 < u < \frac{1}{r _{0}} .

A circular orbit has constant

u = u_{c}

and therefore satisfies

K = u_{c} (1 - r_{0} u_{c})^{2} .

The right side has maximum

4/ (27 r_{0})

at

u_{c} = 1/ (3 r_{0})

. Thus there are two circular orbits when

0 < K < 4/ (27 r_{0})

, one marginal orbit at equality, and none above it.

u = u_{c} + η

and linearizing gives

η^{''} + \frac{1 - 3 r _{0} u _{c}}{1 - r _{0} u _{c}} η = 0.

Hence the outer orbit $u_{c} < 1/ (3 r_{0})$ , equivalently $r_{c} > 3 r_{0}$ , is stable; the inner orbit is unstable. These are the circular-orbit branches of the shifted inverse-radius potential.

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