Equatorial Gunshot

Due to the conservation of angular momentum, the horizontal component of velocity, $v_x$ of the bullet will remain constant. Fred also has the same value for his $v_x$ . The difference in landing position is due the the change in radius of the circular path undertaken by the bullet.

Angular velocity, $\omega$ is given by

$\omega=\frac{v_x}{r}$ ,

(note that $v_x$ is the tangential velocity)

$\therefore\omega\propto\frac{1}{r}$

The fact that $v_{x bullet}=v_{x Fred}$ lets us look purely at scaling relations. Now

$\theta=\omega t$

$\therefore \theta\propto\frac{1}{r}$

Where $\theta$ is the angular position.

This tells us that as the bullet flies upwards, away from the centre of the Earth, $r_{bullet}$ increases and consequently $\theta_{bullet}$ is less compared to $\theta_{Fred}$ due to his smaller $r_{Fred}$ . Thus if the rotation is in the eastward direction, Fred will advance further over the time interval, and the bullet will be west of him.

$\square$