Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Suppose that I put in the outer space (where gravity from other bodies is negligible) a large, perfectly round sphere totally filled with water. At the bottom (even though "bottom" doesn't make much sense in the deep space) of the sphere there are several marbles, all touching each other.

If I make rotate the sphere, all marbles should stay exactly where they are. Right? Is that true regardless of the force that I apply? And what happens, instead, if I strongly shake the sphere?

share|improve this question
    
if there is gravity close to zero then water will not go to bottom but may reside throughout the volume –  Hussain Akhtar Wahid 'Ghouri' Mar 1 '13 at 9:33
    
Well, I actually said that the sphere is totally filled with water. Marbles are at the bottom, not water. –  AldCer Mar 1 '13 at 23:13

1 Answer 1

up vote 2 down vote accepted

Wrong. You are neglecting the viscosity of the water. Friction from the inner wall of the sphere will move the water, which will in turn move the marbles and they will likely rotate around the sphere, though at a speed slower than the rotation of the sphere.

If your sphere was on Earth, then the only reason the marbles stay at the bottom is gravity pulling them down.

If you shake the sphere (even assuming that the walls do not touch the marbles), then they will move in this situation too due to the force of the water flowing around the sphere.

share|improve this answer
    
Many thanks deadly, I knew that I was neglecting something there. Right: water viscosity is the answer. –  AldCer Feb 27 '13 at 16:39

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.