Discussion
A particle of mass \(m\), and angular momentum \(\ell\) is moving in a circular orbit of radius \(r_{0}\) under the influence of an attractive force \(\vec{F}(r) = -\frac{k}{r^{2}}\hat{r}\). Keeping its angular momentum unchanged, the particle is displaced radially by a small distance \(\delta r \ll r_{0}\), due to which its radial distance varies periodically. The corresponding time period is:
Answer: A
The period of the radial oscillation is \(\frac{2\pi\ell^{3}}{mk^{2}}\).
With \(\ell\) fixed, the radial motion is one-dimensional motion in the effective potential:
\[ U_{\text{eff}}(r) = \frac{\ell^{2}}{2mr^{2}} - \frac{k}{r} \]
The circular orbit sits at its minimum, \(U'_{\text{eff}}(r_{0}) = 0\), which gives \(r_{0} = \frac{\ell^{2}}{mk}\). The curvature there sets the oscillation frequency:
\[ \begin{aligned} U''_{\text{eff}}(r_{0}) &= \frac{3\ell^{2}}{mr_{0}^{4}} - \frac{2k}{r_{0}^{3}} \\ &= \frac{3k - 2k}{r_{0}^{3}} = \frac{k}{r_{0}^{3}} \\ \omega^{2} &= \frac{U''_{\text{eff}}}{m} = \frac{k}{mr_{0}^{3}} \\ &= \frac{m^{2}k^{4}}{\ell^{6}} \end{aligned} \]
So \(\omega = \frac{mk^{2}}{\ell^{3}}\) and \(T = \frac{2\pi}{\omega} = \frac{2\pi\ell^{3}}{mk^{2}}\). This equals the period of the circular orbit itself, which is why inverse-square orbits close into ellipses.
\(2\pi\sqrt{m/k}\) is not even a time: \(k\) has units of N m2, so \(m/k\) is not in s2. The forms with 3 or 5 in the denominator do not follow from the curvature \(k/r_{0}^{3}\); they come from keeping the wrong terms of \(U''_{\text{eff}}\).
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