Re: rotating space station

[John Mallinckrodt <ajmallinckro@CSUPOMONA.EDU>]

On Thu, 28 Sep 2000, Rick Tarara wrote:

> From: "Kilmer, Skip" <kilmers@GREENHILL.ORG>
>
> > But the ball, being closer to the axis, must be going slower than the
> > astronaut's feet. Wouldn't it land behind her?
> > Skip
>
> But the straight line path of the ball is shorter than the curved path of
> the edge of the station.  Bet it all works out so that the ball hits right
> where it should to fool the space station resident into thinking/feeling
> that gravity is at work just as if on earth.

Think about what happens in the limit that the release point
approaches the center of the station.  In that limit, the ball
never hits the "ground."  Now, unless you think that limit is
approached discontinuously, the implication is that the ball must
fall further and further behind the astronaut's feet as the height
of the release point increases.

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.

In the rotating frame, the astronaut sees the ball begin at rest
and start moving in the "downward" direction due to the
centrifugal force.  However, as the "downward" speed increases, so
does the the Coriolis force which is directed "to the rear"
(assuming that the astronaut is facing in the direction of her
motion as seen by inertial observers.)  Thus the ball hits the
"ground" "behind" the astronaut's feet.

If the drop is from a large enough altitude, things get a little
more complex and the ball will undergo a spiraling motion with one
or more full spirals centered on the axis of rotation and,
depending on the details, *can* hit the "ground" "in front" of the
astronaut (but only if you are willing to call what is really
falling 350 degrees *behind*, falling "10 degrees ahead.")

This is a great problem for Interactive Physics.  In fact, I have
a closely related IP simulation on my IP web page at

  <http://www.csupomona.edu/~ajm/myweb/index.ip.html>

John Mallinckrodt      mailto:ajm@csupomona.edu
Cal Poly Pomona          http://www.csupomona.edu/~ajm