How do you add velocities of objects with multiple frame-of-reference corrections?

Question

On a spherical planet of equatorial circumference 36,000 km, which rotates once/24 hr (west->east, like the Earth), a train is traveling due east (relative to the planet surface) at 200 km/hr at latitude 30°N. A person on the train throws an object due west at a speed of 150 km/hr relative to the train. Relative to an observer stationary in space (not rotating with the planet), how fast is the thrown object moving directly below?

Answer 1

Strategy: Assemble the sequential speeds, and add them up (as vectors).

Tactics: First, obtain the proper motion of the planet at that point. The mention of 30°N should trigger a geometry fact: indeed, there exists a 30-60-90 right triangle drawn inside the planet, with vertices at the center of the planet, at the location of the train, and at the point of projection of the location of the train perpendicularly onto the equatorial plane of the planet. This triangle has hypotenuse r (from the center of the planet to the train), and a leg lying in the equatorial plane of the planet, of length (distance from the planetary axis of rotation) of √3r/2. Then the rotational speed of the planet at the location of the train is 1500 km/hr * √3/2, to the east (the equatorial velocity is reduced proportional to the diameter of the slice through the planet at the latitude of the train; there is no vector math required for a direction decomposition, since the velocity is due east). That is the velocity of the planet surface relative to the frame of reference of the observer. The train is adding 200 km/hr east (relative to the frame of reference of the planet surface), and the object is adding 150 km/hr west (relative to the frame of reference of the train). Add up all the velocities to the east, subtract the object speed; that's about 1349 km/hr east in all.