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Potential Due to a Uniform Ring

Consider a uniform ring of mass $M$ and radius $r$, centered on the origin, and lying in the $x$-$y$ plane. Let us consider the gravitational potential $\Phi(r)$ generated by such a ring in the $x$-$y$ plane (which corresponds to $\theta = 90^\circ$). It follows, from Section 12.3, that for $r>a$,
\Phi(r) = - \frac{G\,M}{a}\sum_{n=0,\infty} [P_n(0)]^2\left(\frac{a}{r}\right)^{n+1}.
\end{displaymath} (1014)

However, $P_0(0)=1$, $P_1(0) = 0$, $P_2(0)=-1/2$, $P_3(0)=0$, and $P_4(0)=3/8$. Hence,
\Phi(r) = - \frac{G\,M}{r}\left[1 + \frac{1}{4}\left(\frac{a...
...ght)^2 + \frac{9}{64}\left(\frac{a}{r}\right)^4+\cdots\right].
\end{displaymath} (1015)

Likewise, for $r<a$,
\Phi(r) = - \frac{G\,M}{a}\sum_{n=0,\infty} [P_n(0)]^2\left(\frac{r}{a}\right)^{n},
\end{displaymath} (1016)

\Phi(r) = - \frac{G\,M}{a}\left[1 + \frac{1}{4}\left(\frac{r...
...ght)^2 + \frac{9}{64}\left(\frac{r}{a}\right)^4+\cdots\right].
\end{displaymath} (1017)

Richard Fitzpatrick 2011-03-31