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The two positive powers below require $\beta >0$; the negative power requires $\beta <0$, so that the force is attractive. To leading order, $r\simeq r_0[1+ϵ\cos (\sqrt{k+2}\varphi +\delta )]$. The azimuthal advance from one radial maximum to the next isFor $k=2$, ${\omega }_r/\Omega =2$: 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 $\Delta \varphi =4\pi$, 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.

Choose the orientation $J>0$. To leading order in the radial disturbance, the azimuthal angular frequency is ${\omega }_{\theta }=J/(m{r}_0^2)$, soFor either orientation the positive frequency ratio is ${\omega }_r/|{\omega }_{\theta }|$. At $k=-1$, the ratio is $1$: there is one radial oscillation per revolution. The bound Kepler orbit is an ellipse with the force centre at a focus, with one closest and one furthest approach per revolution. Its exact polar coordinates equation isAt $k=2$, the ratio is $2$: there are two radial oscillations per revolution. The isotropic harmonic oscillator trajectory is an ellipse centred on the force centre. An exact parametrization isHere $r^2=a^2{\cos }^2\Omega t+b^2{\sin }^2\Omega t$ repeats twice during one full circuit. The two different placements of the force centre distinguish these ellipses. These are the two all-bound-orbit cases of Bertrand's theorem.

In plane polar coordinates the radial component of Newton's second law is $m(\ddot{r}-r{\dot{\theta }}^2)=-U^{\prime }(r)$. A central force has zero torque, so angular momentum $J=m{r}^2\dot{\theta }$ is constant. ThereforeFor a nonzero-angular-momentum circular orbit at $r_0$, $nB{r}_0^{n+2}=J^2/m$. The curvature there isThus a stable circular orbit, and nearby bound orbits, require $\boxed{nB>0,n>-2}$, with $n=0$ excluded. For $n>0$, $B>0$ confines the particle at large radius; for $-2<n<0$, $B<0$ permits bound energies between the effective-potential minimum and zero. At $n=-2$ there is no isolated stable minimum.

The angular frequency is ${\Omega }^2=nB{r}_0^{n-2}/m$, and radial perturbations have frequency $\kappa =\sqrt{n+2}\Omega$. A rational ratio $\kappa /\Omega$ closes the linearized near-circular trajectory. It does not prove finite-amplitude closure. Bertrand's theorem supplies the stronger conclusion: every bound orbit is closed only for $n=-1$ and $n=2$ among these smooth attractive power-law potentials. Particular circular or resonant orbits must not be confused with that all-orbit statement.

For $n=2+ϵ$, the slowly rotating elliptical orientation has angular speedThis is retrograde for $ϵ>0$ in the near-circular approximation. Consecutive radial maxima are separated by approximately $\pi$ in azimuth; the ellipse's axis is defined modulo $\pi$, which accounts for the factor of two relative to a Kepler-style apsidal convention.

Stable circular power-law orbits require $nB>0$ and $n>-2$, excluding $n=0$. Bertrand's theorem selects $n=-1,2$ when every bound orbit must close. For $n=2+ϵ$, the near-circular ellipse orientation precesses at $-ϵ\Omega /8$ to first order.

Full derivation: solution.

For central potential $V(r)=\beta r^k$ and nonzero angular momentum $J$, a circular orbit satisfies $\beta k{r}_0^{k+2}=J^2/m$, so attraction requires $\beta k>0$. Twice differentiating the radial effective potential $V+J^2/(2m{r}^2)$ gives $V_{\mathrm{eff}}^{\prime \prime }(r_0)=(k+2)J^2/(m{r}_0^4)$. Its harmonic approximation therefore has radial angular frequency ${\omega }_r=\sqrt{k+2}|J|/(m{r}_0^2)$, whereas $|{\omega }_{\theta }|=|J|/(m{r}_0^2)$. This proves the ratio and shows stability precisely when $k>-2$. At $k=-2$ the tuned effective potential is flat; at $k<-2$ the radial disturbance grows exponentially. Rationality of this local frequency ratio is only a necessary condition for all nearby orbits to close; Bertrand's theorem makes a stronger finite-amplitude assertion.

For a central potential $V=\beta r^k$ and fixed nonzero angular momentum $L$, the effective potential $L^2/(2m{r}^2)+\beta r^k$ has a nondegenerate stable circular orbit when $\beta k>0$ and $k>-2$. Its radius obeys $r_0^{k+2}=L^2/(m\beta k)$. The radial small oscillation frequency satisfies ${\omega }_r^2=(k+2){\Omega }^2$, where $\Omega =|L|/(m{r}_0^2)$. Rational frequency ratios close the linearized near-circular sketch, but generally not the exact finite-amplitude orbit; Bertrand's theorem distinguishes the exceptional harmonic and Kepler potentials.