Magnitude of Coriolis force given an east-west velocity I'm trying to make sure I understand how the Coriolis effect works for velocity vectors that are not purely north-south (with regard to the surface of the earth). So consider an eastbound (again with regard to the surface of the earth) object at some fixed speed $s$. Its velocity vector $\vec{v}$ has no poleward component, so is perpendicular to the earth's angular velocity vector $\vec{\omega}$, and therefore the angle between them is $\pi/2$. It follows that, independent of the latitude of the object, the magnitude of the Coriolis force is constant, at $2ms||\vec{\omega}||$. Is that true -- that the Coriolis effect for an east-west bound object is independent of the object's latitude? (related: this answer)
 A: First, it's not the Earth's angular momentum that matters, it's the angular velocity, $\vec{\omega}$.
The Coriolis Force for a mass $m$ moving with velocity $\vec v$ is:
$$ \vec F = -2m\vec{\omega} \times \vec v  $$
So that is independent of latitude, longitude, and altitude. Of course, at different latitudes, the resulting force has different direction relative to the local vertical. Likewise for $\vec v$, which has different local interpretations as you move it around the globe.
A: The answer is yes, in the sense that the magnitude of the force doesn't change for an east/westward bound object with constant speed $\vec{v}$ (although the direction certainly does). 
You can verify it by creating a spherical coordinate system with coordinates "east", "north", and "up", noting the angular velocity $\vec{\omega}$ points "up", and using the Coriolis formula $\vec{F} = 2m\vec{\omega}\times\vec{v}$ for any velocity pointing "east".
This is as a result of the fact, when approximating the Earth as a sphere, the "east" coordinate is always perpendicular to the angular velocity vector of the Earth $\vec{\omega}$ (which points "up"), unlike the "north" coordinate. (You've all pointed this out as well; just confirming it's valid.)
