Consider the case of a single circular loop, establishing a spherical coordinate system $(r, \alpha)$ with the loop axis as the polar axis. Then:  $$ V = {\frac{1}{4\pi\varepsilon_{0}}}\int_{0}^{2\pi}{\frac{Q d\theta}{2\pi}}{\frac{1}{\sqrt{r^{2}+R^{2}-2R r \sin\alpha \cos\Theta}}}={\frac{Q}{4\pi\varepsilon_{0}}}\left({\frac{1}{r}}+{\frac{R^{2}}{r^{3}}}{\frac{1-3\cos^{2}\alpha}{2}}\right)\#\left({\frac{1}{r}}+{\frac{R^{2}}{r^{2}}}\right). $$  Next, consider the superposition of multiple circular loops. Based on geometric relations:  $$ V_{i}={\frac{Q}{4\pi\varepsilon_{0}}}\left({\frac{1}{r}}+{\frac{R^{2}}{r^{3}}}{\frac{1-3\sin^{2}\theta\cos^{2}\left(\varphi-{\frac{\mathrm{i}\pi}{\mathrm{n}}}\right)}{2}}\right) $$  Summing results in:  $$ V={\frac{n Q}{4\pi\varepsilon_{0}}}\left({\frac{1}{r}}-{\frac{R^{2}}{r^{3}}}{\frac{1-3\cos^{2}\theta}{4}}\right) $$  On the equatorial plane, $\textstyle{\theta={\frac{\pi}{2}}}$  $$ V={\frac{n Q}{4\pi\varepsilon_{0}}}\left({\frac{1}{r}}-{\frac{R^{2}}{4r^{3}}}\right) $$  Assuming the angular momentum is $L=m\omega_{\theta}r_{0}^{2}$, the effective potential energy is  $$ \mathrm{V}_{\mathrm{eff}}=-{\frac{n Q q}{4\pi\varepsilon_{0}}}\left({\frac{1}{r}}-{\frac{R^{2}}{4r^{3}}}\right)+{\frac{L^{2}}{2m r^{2}}} $$  The first derivative set to zero:  $$ -{\frac{n Q q}{4\pi\varepsilon_{0}}}{\left({\frac{-1}{r^{2}}}+{\frac{3R^{2}}{4r^{4}}}\right)}-{\frac{L^{2}}{m r^{3}}}=0 $$  Solving gives:  $$ \omega_{\theta}^{2}=\frac{n Q q}{4\pi\varepsilon_{0}m r_{0}}\left(\frac{1}{r_{0}^{2}}-\frac{3R^{2}}{4r_{0}^{4}}\right) $$  The second derivative:  $$ -{\frac{n Q q}{4\pi\varepsilon_{0}}}\left({\frac{2}{r^{3}}}-{\frac{3R^{2}}{r^{5}}}\right)+{\frac{3L^{2}}{m r^{4}}}=m\omega_{r}^{2} $$  Solving gives:  $$ \omega_{r}^{2}=\frac{n Q q}{4\pi\varepsilon_{0}m r_{0}}\left(\frac{1}{r_{0}^{2}}+\frac{3R^{2}}{4r_{0}^{4}}\right) $$  $$ T_{r}-T_{\oplus}=\frac{2\pi}{\omega^{2}}(\omega_{\theta}-\omega_{r})=-\frac{3\pi R^{2}}{2r_{0}^{2}}\sqrt{\frac{4\pi\varepsilon_{0}m r_{0}^{3}}{n Q q}} $$