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{\bf Question}

Find the Coriolis Force acting on a particle on the Earth's
surface at a latitude, $\lambda$, that
\begin{description}
\item[(a)] moves horizontally  due north with speed $V$;
\item[(b)] move vertically upwards with speed $V$.
\end{description}

\vspace{.25in}

{\bf Answer}

\begin{center}
$ \begin{array}{l} \epsfig{file=215-10-1.eps, width=40mm}
\end{array}
\ \ \
\begin{array}{l}
\ds \mbox{\boldmath $\omega$} = -\omega \cos \lambda {\bf i} +
\omega \sin \lambda {\bf k}\\
\textrm{Coriolis Force }-2m \mbox{\boldmath $\omega$} \times {{\bf v}}
\end{array}$
\end{center}

\begin{description}
\item[(i)]
\begin{eqnarray*} {\bf v} = -V {\bf i}\ \  {\rm \ \ Coriolis\ Force} & = &
-2m\omega \times - V {\bf i} \\ & = & 2m\omega V(-\cos \lambda
{\bf i} + \sin\lambda {\bf k}) \times {\bf i} \\ & = & 2m\omega
V\sin \lambda {\bf k} \times {\bf i} \\ & = & 2m\omega V\sin
\lambda {\bf j}
\end{eqnarray*}

\item[(ii)]
\begin{eqnarray*} {\bf v} = -V {\bf k}\ \  {\rm \ \ Coriolis\ Force} & = &
2m\omega V(-\cos \lambda {\bf i} + \sin\lambda {\bf k}) \times
{\bf k}
\\ & = & - 2m\omega V\cos \lambda {\bf j}
\end{eqnarray*}
\end{description}



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