= Solution
The two positive powers below require $\beta>0$; the negative power requires $\beta<0$, so that the <force> is attractive. To leading order, $r\simeq r_0[1+\epsilon\cos(\sqrt{k+2}\,\phi+\delta)]$. The azimuthal advance from one radial maximum to the next is
$$
\Delta\phi=\frac{2\pi}{\sqrt{k+2}}.
$$
For $k=2$, $\omega_r/\Omega=2$: there are two radial oscillations per revolution. The <isotropic harmonic oscillator> has orbits that are exact <ellipses> centred on the force centre, and the perturbed circle is such an <ellipse>.
For $k=7$, $\omega_r/\Omega=3$: the leading perturbation has three radial lobes per revolution, with consecutive radial maxima separated by $2\pi/3$. Its small-amplitude sketch is a rounded three-lobed orbit.
For $k=-7/4$, $\omega_r/\Omega=1/2$: the radius completes one oscillation while the body makes two revolutions. The leading curve closes after $\Delta\phi=4\pi$, with a double-wound, generally self-crossing shape; the nearest and furthest points occur in the same spatial direction one revolution apart.
The rational ratios for $k=7$ and $k=-7/4$ describe closure of the linearized near-circular curves. They do not imply that every finite-amplitude orbit closes exactly: higher-order changes in the radial-to-orbital frequency ratio generally produce <apsidal precession>. <Bertrand's theorem> identifies the harmonic potential $k=2$ and the Kepler potential $k=-1$ as the exceptional attractive central potentials with all nearby bounded orbits closed.
\Image[/past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2015-power-law-near-circular-orbits.png]
{title=Leading near-circular orbit shapes for the three powers in Question 8}
{description=Original small-amplitude curves $r/r_0=1+0.08\cos(\sqrt{k+2}\,\phi)$. The $k=-7/4$ panel follows two revolutions. For $k=7$ and $k=-7/4$, exact closure of these linearized illustrations is not a claim of exact finite-amplitude closure.}
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