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Potential due to uniform ring

Consider a uniform ring of mass $ M$ , radius $ a$ , and negligible cross-sectional area, centered on the origin, and lying in the $ x$ -$ y$ plane. Let us consider the gravitational potential $ {\mit\Phi}(r)$ generated by such a ring in the $ x$ -$ y$ plane (which corresponds to $ \theta = 90^\circ$ ). It follows, from Section 3.4, that for $ r>a$ ,

$\displaystyle {\mit\Phi}(r) = - \frac{G\,M}{a}\sum_{n=0,\infty} [P_n(0)]^2\left(\frac{a}{r}\right)^{n+1}.$ (3.75)

However, $ P_0(0)=1$ , $ P_1(0) = 0$ , $ P_2(0)=-1/2$ , $ P_3(0)=0$ , $ P_4(0)=3/8$ , $ P_5(0)=0$ , $ P_6(0)=-5/16$ , $ P(7)=0$ , and $ P_8(0)=35/128$ (Abramowitz and Stegun 1965b). Hence,

$\displaystyle {\mit\Phi}(r) = - \frac{G\,M}{r}\left[1 + \frac{1}{4}\left(\frac{...
...rac{a}{r}\right)^6+ \frac{1225}{16384}\left(\frac{a}{r}\right)^8+\cdots\right].$ (3.76)

Likewise, for $ r<a$ ,

$\displaystyle {\mit\Phi}(r) = - \frac{G\,M}{a}\sum_{n=0,\infty} [P_n(0)]^2\left(\frac{r}{a}\right)^{n},$ (3.77)

giving

$\displaystyle {\mit\Phi}(r) = - \frac{G\,M}{a}\left[1 + \frac{1}{4}\left(\frac{...
...rac{r}{a}\right)^6+ \frac{1225}{16384}\left(\frac{r}{a}\right)^8+\cdots\right].$ (3.78)



Richard Fitzpatrick 2016-03-31