 Solution

ID: past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2015/8/c/solution

Write

r = r_{0} + η

and keep

L

ω_{r}^{2} = \frac{V _{eff}^{''} ( r _{0} )}{m} = \frac{3 L ^{2}}{m ^{2} r _{0}^{4}} + \frac{β k ( k - 1 )}{m} r_{0}^{k - 2} = \frac{β k ( k + 2 )}{m} r_{0}^{k - 2} .

The circular orbital angular frequency has magnitude

Ω = ∣ L ∣/ (m r_{0}^{2})

, so

Ω^{2} = (β k / m) r_{0}^{k - 2}

. Hence

ω_{r} = Ω k + 2, \frac{ω _{r}}{Ω} = k + 2, f_{r} = \frac{Ω}{2 π} k + 2 .

A restoring radial oscillation exists for

k > - 2

, subject to the attractive-force condition in part (b). For

k < - 2

the radial stationary point is unstable;

k = - 2

is the degenerate boundary. This agrees with circular-orbit stability for a power-law central force, since the actual force is proportional to

- r^{k - 1}

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