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Indeed, good explanations of this fact exist for both the inertial
and rotating observer. In the inertial frame, the distance
traveled by the ball is sqrt(2Rh+h^2) where h is the height of the
release and R is the radius of the space station and the speed of
the ball is omega*(R-h) so the time it takes the ball to fall is
srqt[ 2*(h/R) + (h/R)^2 ]
t_b = -------------------------
omega*(1 - h/R)
(Notice what happens as h/R --> 1.)
To reach the place where the ball will fall, the astronaut's feet
must move through an angular displacement of arccos(1 - h/R), so
the time it takes the feet to get there is
arccos(1 - h/R)
t_f = ---------------
omega
Now it is not *too* difficult to show that t_f < t_b for all h >
0. Perhaps the easiest way to see this is simply to plot
srqt[ 2*(h/R) + (h/R)^2 ]
t_b/t_f = -------------------------
(1 - h/R)*arccos(1 - h/R)
as a function of h/R and note that it is a monotonically
increasing function. Thus, the ball lands behind the feet.