Question

Centrifugal force.
Suppose you have a rock on a rotating asteroid. Suppose the asteroid rotates so fast that the rock is lifted from the surface. Is the centrifugal force still acting on the rock?

Answer 1

(Note this is a simplification - assuming circular motion and smooth asteroid, rocks etc).

Answer 2

There is no such thing as centrifugal force. If you have been taught that there is you should get a GOOD text book and learn for yourself.

The force that causes the rock to stay on the surface of the asteroid is due to gravity - it acts TOWARDS the centre of the asteroid and is generally known as centripetal force.

It causes the rock to follow the circular path which keeps it on the surface. The force required to keep a body moving in a circular path is
\[mr \omega ^{2}\]

If the asteroid is somehow 'sped up' to be rotating faster (i.e. omega increases), then at some point the gravitational force will not be sufficient to hold the rock in the path with radius r

The gravity will continue to act, but no further friction on the rock will occur to further increase its rotational speed - it will continue to 'orbit' with the speed at which it left the surface

Answer 3

Thank you for your answer.
I was speaking of centrifugal force because I find it simpler to consider the rotating system of the asteroid (but I know that it is a fictitious force).
I'd like to understand a little more in detail what happens when the rock leaves the surface: now, as you say, there is no more friction and thus the rock keeps its velocity with which it left the surface. This means that in the rotating system of the asteroid you should see that the rock moves only radially with respect to the surface. How can I determine now the motion of the rock?
1)Will it return to the surface or will it simply stay there and orbit around the asteroid?
2) From the point of view of the non inertial frame of the lifted rock, is there any (fictitious) centrifugal force still acting on it? Afterall, the rock is still moving, but I would say that if the initial angular velocity was that of the asteroid \[\omega=v/R\] where R is the radius of the asteroid, now the angular velocity of the rock, which has the same velocity but is now placed at R'>R should be \[\omega'=v/R′\] However I'm not very convinced, because this is true only istantaneously and furthermore the velocity now has no more only one tangent component with respect to the surface, but two components, one radial and one tangent to the spherical surface of radius R'. So if v is actually
\[v=\sqrt{v _{t}^2+v_{r}^2}\]
then I'm afraid that my reasoning cannot be applied...

Answer 4

My interpretation is that you have the situation where the asteroid is slowly increasing in angular velocity.

The force required to keep the rock in circular motion at the surface r is mrW^2

The force acting is gravity (GMm/r^2)

Whilst the rock is touching the surface then there is friction, and any increase in the surface speed of the asteroid will be transferred to the rock but the focre required to keep it in the circular track will increase.

Once they are equal then the rock has no more contact force - i.e it is 'hovering' at the surface, and any increase in speed of the asteroid will NOT be transferred.
The tangential velocity will be JUST enough to keep it at the radius of the asteroid - i.e. it will orbit at height = 0 above the surface.

IF however it were given a little extra velocity (but the same miracle that is speeding up the asteroid perhaps) then it would move into a higher orbit - BUT the gravitational force at the new height will be less - so the rock will eventually escape from the gravity.

All assuming 'perfect assumptions etc.etc.

Quite an interesting problem - I'm happy to be told I've got it wrong - but htat seems to be the situation to me...