Vector differentiation and rotating frames I'm working with a math problem from the chapter "Applications of Vector Differentitation" from the book "Calculus A Complete Course". The problem includes rotating frames and the coriolis force. The problem is the following:
A satellite is in a low, circular, polar orbit around the earth (i.e., passing over the north and south poles). It makes one revolution every two hours. An observer standing on the earth at the equator sees the satellite pass directly overhead. In what direction does it seem to the observer to be moving? From the observer's point of view, what is the approximate value of the Coriolis force acting on the satellite?
I have a hard time understanding how to solve this problem, and even where to begin. My book have some pages taking about rotating frames and the coriolis force with an example, but I'm still stuck. An older version of my textbook can be found here: http://eng.usc.ac.ir/files/1508848447251.pdf. On page 630-632 they talk about rotating frames, but I still don't get this problem above.
 A: The satellite has a speed of $v_s=(40~000~\mathrm{km})/(2~\mathrm{h})$ in polar direction, while the observer has a speed of $v_o=(40~000~\mathrm{km})/(24~\mathrm{h})$ in equatorial direction. Their velocities are orthogonal in the frame of Earth. In the frame of the observer the direction of the satellite will deviate from north-south with an angle $\phi$ given by
$$
\tan\phi = \frac{v_o}{v_s} = \frac{(40~000~\mathrm{km})/(24~\mathrm{h})}{(40~000~\mathrm{km})/(2~\mathrm{h})} = \frac{1}{12}.
$$
I leave it to you to try to determine whether it is in north-east or north-west.
The Coriolis force is given by $\vec{F}_{\text{Coriolis}}=-2m_s \vec{\omega}_e\times\vec{v}_{so},$ where $m_s$ is the satellite mass, $\vec{\omega}_e$ is the vector angular velocity of Earth, and $\vec{v}_{so}$ is the velocity of the satellite in the frame of the observer. The size of $\vec{F}_{\text{Coriolis}}$ is $|\vec{F}_{\text{Coriolis}}|=2m_s |\vec{\omega}_e||\vec{v}_{so}|\sin\phi.$ Here, $|\vec{\omega}_e|=2\pi/(24~\mathrm{h})$ and $|\vec{v}_{so}|=\sqrt{v_o^2+v_s^2}.$ The value of $\sin\phi$ you can get from $\tan\phi=1/12$; you can even take $\sin\phi=1/12$ since $\sin\approx\tan$ for small angles.
A: I use $s$ for satellite, $e$ for earth, and $o$ for observer. Say that the satellite revolve around the earth with relative angular velocity $\vec{\omega}_{s}$ while earth rotates with angular velocity $\vec{\omega}_{e}$. The velocity of satellite relative to the observer is given by the following equation:
$$
\frac{d}{dt}\vec{r}_{so}=\left(\vec{\omega}_{s}+\vec{\omega}_{e}\right)\times\vec{r}_{s}-\vec{\omega}_{e}\times\vec{r}_{o}
$$
Differentiate once with respect to time to obtain the relative acceleration:
$$
\begin{align}
\frac{d^{2}}{dt^{2}}\vec{r}_{so}&=\left(\frac{d}{dt}\vec{\omega}_{s}+\frac{d}{dt}\vec{\omega}_{e}\right)\times\vec{r}_{s}+\left(\vec{\omega}_{s}+\vec{\omega}_{e}\right)\times\left(\frac{d}{dt}\vec{r}_{s}\right)-\vec{\omega}_{e}\times\frac{d}{dt}\vec{r}_{o}\\
\\
&=\vec{\omega}_{e}\times\vec{\omega}_{s}\times\vec{r}_{s}+\left(\vec{\omega}_{s}+\vec{\omega}_{e}\right)\times\left(\vec{\omega}_{s}+\vec{\omega}_{e}\right)\times\vec{r}_{s}-\vec{\omega}_{e}\times\vec{\omega}_{e}\times\vec{r}_{o}\\
\\
&=2\vec{\omega}_{e}\times\vec{\omega}_{s}\times\vec{r}_{s}+\vec{\omega}_{s}\times\vec{\omega}_{s}\times\vec{r}_{s}+\vec{\omega}_{e}\times\vec{\omega}_{e}\times\vec{r}_{s}-\vec{\omega}_{e}\times\vec{\omega}_{e}\times\vec{r}_{o}
\end{align}
$$
The first term on the right is the Coriolis acceleration. Notice that at equator, the three vectors $\vec{\omega}_{e},\vec{\omega}_{s},\vec{r}_{s}$ are mutually perpendicular so the Coriolis acceleration equals zero.
