\documentclass[a4paper,12pt]{article}
\usepackage{epsfig}
\newcommand{\un}{\underline}
\begin{document}
\parindent=0pt

\begin{center}
\textbf{Vector Functions and Curves}

\textit{\textbf{Applications}}
\end{center}

\textbf{Question}

A satellite in a low (radius of orbit is approximately the radius of
the Earth) circular orbit passes over both poles. It takes the
satellite two hours to make one revolution.

If an observer stands on the equator, what would they make the
approximate value of the Coriolis force on the satellite as it passes
overhead, heading south?

What direction would the satellite seem to moving to the observer as
it passes overhead?


\textbf{Answer}


Assume that the observer is standing at the origin , with $\un{i}$
pointed to the east and $\un{j}$ pointed to the north.

The Earth has angular velocity $2\pi /24$ radians per hour
northward:
$$\un{\Omega} = \frac{\pi}{12}\un{j}.$$

As the Earth is rotating west to east, the actual north to south
velocity of the satellite will appear to be moved to the west by $\pi
R/12$ km/h, given that $R$ is the radius of the Earth (in km).

As the satellite circles the Earth at a rate of $\pi$ radians/h, the
observer at the origin would take its velocity to be
$$\un{v}_R = -\pi R\un{j} -\frac{\pi R}{12} \un{i}.$$
With $\un{v}_R$ making an angle with the southward direction of 
$$\tan^{-1} \left ( \frac{\pi R/12}{\pi R} \right ) = \tan^{-1}(1/12)
\approx 4.76^{circ}$$

Therefore the satellite appears to be moving a direction which is
$4.67^{\circ}$ west of south.

And so the apparent Coriolis force is
\begin{eqnarray*}
-2\un{\Omega} \times \un{v}_R & = & -\frac{2\pi}{12}\un{j} \times
 \left ( -\pi R\un{j} - \frac{\pi R}{12}\un{i} \right )\\
& = & -\frac{\pi^2 R}{72}\un{k}
\end{eqnarray*}
This points towards the ground.



\end{document}













