<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}
\newcommand{\ds}{\displaystyle}
\newcommand{\pl}{\partial}
\newcommand{\ud}{\underline}
\begin{document}
\parindent=0pt


{\bf Question}

Show that the small angular deviation $\epsilon$ of a plumb line
from the direction to the centre of the earth at a point on the
earth's surface at a latitude $\lambda$ is $$\epsilon = \frac{Rw^2
\sin \lambda \cos \lambda}{g_0 - Rw^2 \cos^2\lambda}$$ where $R$
is the radius of the earth.  What is the value of the maximum
deviation and at what latitude does this occur?

\vspace{.25in}

{\bf Answer}

$\ds {\bf g} = {\bf g}_0 - \mbox{\boldmath $\omega$} \times
(\mbox{\boldmath $\omega$} \times \ud r)$

$\ds {\bf g} = {\bf g}_0 - [(\mbox{\boldmath $\omega$} \cdot {\bf
r})\mbox{\boldmath $\omega$} - \omega^2 {\bf r}]\ \ \  (*)$

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\times(\mbox{\boldmath $\omega$} \times {\bf r})$}}

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\bigskip

Taking the cross product with ${\bf g}_0$

\begin{eqnarray*} {\bf g} \times {\bf g}_0 &amp; = &amp; {\bf g}_0 \times \bf
g_0 + \omega^2 {\bf r} \times {\bf g}_0 - (\mbox{\boldmath
$\omega$} \cdot {\bf r}) \mbox{\boldmath $\omega$} \times {\bf
g}_0 \\ &amp; = &amp;- (\ud \omega \cdot {\bf r}) \mbox{\boldmath
$\omega$} \times {\bf g}_0 \hspace{.3in} {\rm as\ } {\bf r}, {\bf
g}_0\  {\rm are\ parallel} \\
\\ g g_0 \sin \epsilon &amp; = &amp; \omega R \cos \left( \frac{\pi}{2} -
\lambda\right) \omega g_0 \sin \left( \frac{\pi}{2} -
\lambda\right) \hspace{.3in} {\rm by\ defn\ of\ } \epsilon \\ \sin
\epsilon &amp; = &amp; \frac{\omega^2 R}{g} \sin \lambda \cos \lambda
\end{eqnarray*}

Take the dot product of (*) with itself to find g:

\begin{eqnarray*} g^2 &amp; = &amp; g_0^2 - 2 {\bf g}_0 \cdot [\omega^2 \bf
r - (\mbox{\boldmath $\omega$} \cdot {\bf r}) \mbox{\boldmath
$\omega$}]+ {\rm terms\ including\ }\omega^4 \\ &amp; \approx &amp; g_0^2
- 2g_2[\omega^2R - \omega R \sin \lambda \omega \sin \lambda] \\ &amp;
\approx &amp; g_0^2 - 2g_0 \omega^2 R \cos^2 \lambda \\ g &amp; \approx &amp;
g_0 \left(1 - 2 \frac{\omega^2 R}{g_0} \cos^2 \lambda
\right)^{\frac{1}{2}} \\ &amp; \approx &amp; g_0 + \omega^2 R \cos^2
\lambda \end{eqnarray*}

Thus $\ds \sin \epsilon \approx \frac{R\omega^2 \sin \lambda \cos
\lambda}{g_0 - R \omega^2 \cos^2 \lambda}$

$\epsilon$ is clearly small so $\sin \epsilon \approx \epsilon,$
hence the result.



\end{document}
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