For the brave of heart, here is what I'm getting for the complete equations of motion. I won't describe how I got them in detail, hopefully it's right :-).
Coordinates:
\theta is zero at the equator, and pi/2 at the north pole. It's similar to a latitude.
\phi is zero on the Grenwich meridian, similar to a longitude.
r is of course the distance from the center of the Earth.
Differentiation with respect to time is indicated by a dot above the variable, i.e. \dot{r} = dr/dt\ddot{r} - r \, \dot{\theta}^2 - r \, sin^2(\theta) \, \dot{\phi}^2 = -gM/r
\ddot{\theta} + \frac{2}{r}\,\dot{\theta}\dot{r} - \frac{1}{2} \, sin(2\theta) \, \dot{\phi}^2 = 0
\ddot{\phi} + \frac{2}{r} \, \dot{\phi}\dot{r} + 2 \, tan(\theta) \, \dot{\theta}\dot{\phi}= 0
What this means is that if someone is standing still on the equator, \dot{\phi} is 2 pi radians / 24 hours due to the Earth's rotation while \dot{\theta} is zero. This gives us
\ddot{\phi} + \frac{2}{r} \, \dot{\phi}\dot{r} = 0
which means that if someone throws a ball up, making \dot{r} nonzero (positive), d^2 \phi / dt^2 becomes nonzero (negative). The magnitude of the force is proportional to the radial velocity, as one expects of a coriolis force. Thus the ball accelerates west immediately when it is thrown up, stops accelerating at the peak of its trajectory, and accelerates east on its downward trip, eventually returning to an east/west velocity of zero. However, while the east/west velocity is zero, the east/west position is negative (west).
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A trouble spot - the above equations are for a circular Earth. It can be seen that the second equation doesn't have a stationary solution for \theta :-(, thus it's not quite right.
The equations faithfully say that an object on a spinning sphere would tend to experience a force moving them towards the equator if they are not there already. But the Earth is not a sphere.
