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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?

David Z
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hyperknot
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2 Answers2

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Because of the centrifugal force countering gravity of the rotating asteroid, the astronaut would weigh less than on a non-rotating body. Also, the astronaut would be able to jump higher since his muscular strength would be unaffected. Finally, if the astronaut jumped vertically, he would not land on the same spot due to Coriolis force effects.

Michael Luciuk
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  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.

Alan Rominger
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