<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}
\usepackage{epsfig}
\newcommand{\ds}{\displaystyle}
\begin{document}
\parindent=0pt


{\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.

\vspace{.25in}

{\bf Answer}

\begin{center}
\epsfig{file=215-2-1.eps, width=60mm}
\end{center}

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


\end{document}
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