An aeroplane flies due east along the equator with a speed of 300 m/s. Determine the magnitude and direction of Coriolis acceleration.

Hello!

Coriolis acceleration and Coriolis force appear in a rotating and therefore non-inertial frame of reference. Actually there is "no such force" (no body causes it), but it is convenient to describe motion in a rotating frame of reference. The centrifugal force has the same character.

The frame of reference connected with Earth is a rotating one. Its speed of rotation is small (one revolution per day), therefore Coriolis force is weak enough.

The formula for Coriolis acceleration is `-2 Omega xx V` , where `Omega` is the angular velocity vector, V is the velocity of the  in rotating frame and `xx` is for cross-product. `Omega` has the direction of the rotation axis.

In our case the direction of the flight is the same as the direction of rotation and the direction of Coriolis acceleration is upward (out of the axis). Its magnitude is `2*(2 pi)/(24*3600)*300` (we convert one revolution per day to radians per second). Numerically it is about `0.044 (m/s^2).`