# Astronaut on a rotating asteroid

This might sound like the silliest question ever, but can someone shed some light on what is actually happening in the following situation:

There is an asteroid in outer space what is rotating with high speed. There is an astronaut standing on its surface.

1. Does the astronaut feel lighter compared to if the asteroid wasn't rotating?

2. Can he jump higher compared to if the asteroid wasn't rotating?

3. What happens if the astronaut makes a big jump vertically? Is he going to land on the exact same place, or the asteroid could "rotate under" the jumping astronaut?

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You're asking very different questions here. One is a repeat: physics.stackexchange.com/questions/1372/… Could you simplify the question to ask just one thing, please? –  Mark Eichenlaub Jun 11 '11 at 0:43
First, ask yourself: in reality, how to land an astronaut on the surface of a rapidly rotating asteroid(say with diameter of 100 m)? –  Martin Gales Jun 11 '11 at 8:45
He was born there :-) –  zsero Jun 11 '11 at 14:18

1. Yes he feels lighter unless he's on the North or South pole, and on most of the asteroid it would feel like partly sideways gravity. Since we can assume the asteroid is rigid, it is distinguished in such a way from my recent [hydrostatic self-gravitation problems][1] question.
2. Yes he can jump higher... unless he's standing on a pole.
3. This part is more complicated, I'll address more detail below

On Equator

If the astronaut jumps from the equator he leaves the ground from a point that has an apparent gravity (gravity combined with rotational acceleration) normal to the surface. For small jumps on a large asteroid, yes, he will make it back to the same spot. But what's the cutoff? I believe it would be the point at where orbital dynamics started to matter.

$$g = \frac{a}{r^2} + b \frac{V_x^2}{r}$$

I believe this would be the relevant equation, since $V_x$, the horizontal velocity in the CM frame, would be the invariant quantity. Say he jumps a distance $d$ upwards, then to the extent that $d (2 a / r^3 + b V_x^2/r^2) \ll a/r^2+b V_x^2/r$, he would land about in the same place. If this is not true he would land in a different place. I'm not entirely sure about this, but it's my best shot.

Poles

Again, if he's on the poles the rotation doesn't matter, and he can jump to infinity and make it back to the exact same spot as an academic exercise. In reality, the jump speed would have to be very carefully produced to go far and make it back without reaching escape velocity.

Elsewhere

Again, the apparent gravity is non-normal. So the astronaut already feels like he's on a slope, and obviously he would have to jump in the direction that felt like "up" for this to make sense at all. But if he did that, similar rules to the Equator would apply.

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"So the astronaut already feels like he's on a slope" You are assuming the asteroid is a sphere. In reality it would probably be an ellipsoid, so that the apparent gravity would again be normal. Unless it were very small and rigid, but then it would probably resemble a potato. –  starblue Jun 11 '11 at 7:19
@starblue There exists exactly one shape that would make the gravity normal in all places given a rotation rate. That would be the solution of the hydrostatic self-gravitation problem. I don't think there is a good case for that here. Even dwarf planets behave as rigid bodies and thus have non-normal gravity. Asteroids themselves can be extremely odd shapes, but the common "potato" is not the shape that balances the forces to make gravity normal everywhere. The shape that does is more like a pancake, or bulging at the equator. –  AlanSE Jun 11 '11 at 14:36