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{\bf Question}

Find the gravitational field along the axis of a uniform density
circular hoop as a function of distance from the centre of the
hoop.

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{\bf Answer}

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The gravitational field at A is parallel to the axis (by
symmetry). Therefore the only component we have to consider is the
{\bf i} component.

\begin{eqnarray*}  & = & -Gm \int_C
\frac{\rho}{r^2}\cos \phi \, ds \\ & = & -Gm \rho \int_{\theta =
0}^{\theta = 2 \pi} \frac{\cos \phi}{r^2}R \, d\theta
\hspace{.1in}{\rm as\ } ds = r \, d\theta \\ & = & -Gm\rho
\frac{\cos \phi}{r^2} R \times 2\pi  \\ & &   \\ & = & -
\frac{2\pi Gm\rho R x}{(R^2 + x^2)^{\frac{2}{3}}}\\&&{\rm Since\ }
r^2 = R^2 + x^2 {\rm and\ } \cos \phi = \frac{x}{\sqrt {r^2 +
x^2}}
\end{eqnarray*}


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