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{\bf Question}

Show that for a body moving horizontally near the surface of the
earth in the northern hemisphere the Coriolis Force acts towards
the right of the particle's motion.  What happens in the southern
hemisphere?

\vspace{.25in}

{\bf Answer}

Let $\ds {\bf V} = V_1 {\bf i} + V_2 {\bf j}$

\begin{eqnarray*} {\rm Coriolis\ Force} & = & -2m\omega(-\cos \lambda {\bf i}
+ \sin\lambda {\bf k})\times (V_1 {\bf i} + V_2 {\bf j}) \\ & =&
-2m[ \omega \sin \lambda V_1 {\bf k} \times {\bf i} - \omega \cos
\lambda V_2 {\bf i} \times {\bf j} + \\ & & \hspace{.5in} \omega
\sin \lambda V_2 {\bf k} \times {\bf j} \,] \\ & = & 2m\omega [
\sin \lambda({\bf i} V_2 - {\bf j} V_1) + V_2 \cos \lambda {\bf
k}]
\end{eqnarray*}

Horizontal component is $\ds {\bf H} = 2m\omega \sin \lambda ({\bf
i} V_2 - {\bf j} V_1)$

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\put(0,1){\vector(1,2){0.25}}

\put(0,1){\line(2,-1){1}}

\put(0,1){\vector(2,-1){0.5}}

\put(0.7,2){\makebox(0,0)[l]{$\bf V$}}

\put(1,0.5){\makebox(0,0)[l]{${\bf H} = 2m\omega \sin \lambda
({\bf i} V_2 - {\bf j} V_1)$}}

\end{picture}
\end{center}


Clearly to the right means that $\lambda >0$ and to the left is
when $\lambda <0$.  i.e. we are in the southern hemisphere.


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