I need to know why on Earth (ha ha..!) this behaviour happens when a body jumps in the Moon (a celestial body without any atmosphere at all).
Video explanation: https://youtu.be/Qgs5E5gKO48
The steps I followed using Kerbal Space Program:
Start in position A
Perform a clean jump on the Anti-radial vector
Wait some time
Slow myself to not impact the ground and kill myself.
End up in position B
I know this is basic, but, should not the object end up in position A again?
Should not the object end up in position A again?
No, it shouldn't, because of the Coriolis effect (assuming the moon is rotating about an axis with respect to the stars).
From the perspective of a frame rotating with the moon, the coriolis acceleration is $-2\,\vec\omega\times \vec v$, where $\vec w$ is the moon's angular velocity with respect to inertial space and $\vec v$ is the velocity of the jumping Kerbal.
For simplicity, I'll assume the jump is performed at the moon's equator, at initial coordinates $\vec r = r\hat x$. The jump, in moon-fixed coordinates, gives the Kerbal an initial velocity of $\vec v = v_0\hat x$. The dominant acceleration is the downward acceleration due to gravity, $-g\hat x$ (not 9.80665 m/s2). There's also a much smaller acceleration in the $-\hat y$ direction due to the Coriolis effect. I'll focus on that and ignore that this slight drift changes the direction of the gravitational acceleration vector. (This effect is very small.)
With this assumption, the x component of velocity is $v_x(t) = v_0 - gt$. This means that the y component of acceleration is $\ddot y(t) = -2\omega v_x(t) = -2\omega(v_0-gt)$. Integrating twice yields $y(t) = -\omega v_0 t^2 + \frac 1 3 \omega g t^3$. Substituting $t = t_f = 2\frac{v_0^2}g$, the time at which the jumping Kerbal lands, yields $y_f = -\frac 4 3 \omega \frac{v_0^3}{g^2}$ (in other words, slightly to the west of the initial position).
