Take the 2-minute tour ×

Assume that a space habitat is shaped like a can with the "top" open. lets say we attach the can's top to a string with an astronaut standing on the base. We then rotate the can in deep space using the string (somewhat like a hula hoop) at a constant rate so that the force exerted on his feet by the base of the can is equal to his weight on earth.

Will the astronaut feel the rotation of the habitat (the can) or would he simply feel he is on the surface of the earth? If he detects a rotation, would he be able to tell the direction of the rotation?

share|improve this question
All of these question depend on the length of the rotational radius and on the equipment available for testing---i.e. just your own senses, a repeatable rubber band catapult, high precision instruments. And no one has tested them in person, so for the "just you own senses" option there is some uncertainty. –  dmckee May 4 '12 at 23:53
I should add that he only have his senses to rely on. –  Jus12 May 5 '12 at 0:02
How big is the hoop? The astronaut will feel pushed sideways while jumping. –  Ron Maimon May 5 '12 at 2:04
It is the weak equivalence principle which says that he will not know the difference as long as he is on the floor of the can. en.wikipedia.org/wiki/Equivalence_principle . If he jumps he will see the floor moving under him, a coriolis effect en.wikipedia.org/wiki/Coriolis_effect . –  anna v May 5 '12 at 13:11

1 Answer 1

up vote 4 down vote accepted

No one else has taken a crack at it, so I'll just point you in the direction of the answer.

He won't notice unless the pseudo-forces due to rotation (centrafugal, Coriolis and Euler) are large enough to notice. So take a human being to have a height of 2 meter.

Using the conventions in the wikipedia pages and assuming that the angular velocity of the can is unaffected by anything our experimenter does we write to "centrifugal force" as $$ F_{centrifugal} = m \omega \times (\omega \times r) = m \frac{v^2}{r} $$ always pointing out (i.e. down from the tester's perspective).

Two parts of the testers body at different radii will experience different "gravity". It is not clear how big a difference between head and feet will be noticed, but lets take it to be 10% of g. He notices if $$ 0.1 g \ge \omega^2 r_{out} - \omega^2 r_{in} = \omega^2 \Delta r$$ where $\Delta r$ is 2 meters (the height of our tester).

The "Coriolis force" depends on velocities as well as radii, which is to say it might be noticed during movements not parallel to the axis of rotation. $$ F_{Coriolis} = -2 \omega \times v $$ Standing up suddenly, for instance will generate a "sideways" force. I personally guess that that will be much easier to notice than the differential centripetal acceleration, so I'm going to set this limit a a few percent of a gee. Assume that "standing suddenly" is a 1 meter motion in one quarter of a second. He notices if $$ 0.03 g \ge \left| 2 \omega \cdot (4 \text{ m/s})\right| $$ Similarly pouring liquids can generate speeds on order of meters per second.

The Euler force is not present due to our assumption on constant angular velocity.

So, stick your design parameters into these expressions, diddle the limits if you don't like my choices and draw your conclusions.

share|improve this answer

Your Answer


