What should happen when an astronaut drops a moon-pebble into a hole that was drilled right through the moon?

Question

Will it pop up at the opposite continent, and return to the austronaut again. In other words, will the pebble describe a wave-movement? When so, is it still reasonable to define gravity as a force?

Answer 1

theoretically i suppose, if the moon was perfectly spherical and its density was even throughout.

and...theoretically, i suppose the pebble will keep going through the hole with a steady decrease in amplitude, till it stops in the middle, due to drag.

Answer 2

Provided the moon is perfectly spherical and its mass is evenly distributed, the pebble should accelerate toward the center of the moon (no drag in space) and slow as it moves toward the opposite "continent" it will continue this pattern witha decreasing amplitude untill it finally settles at the center of the moon.

Gravity is both the force of the acceleration and the deceleration.

Answer 3

Yes, it will oscillate back and forth. And with no drag, it will continue forever. It's the same as being in orbit.

Other answers that assume it will slow down and stop are wrong.

Don't understand your last sentence, why should this change anything?  Pebble is converting potential energy to kinetic energy and then back to potential, etc, over and over again.

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Answer 4

I have heard this problem several times and there are some things people fail to take account of. Remember that the the moon makes one full revolution every time it passes around the earth. This is the reason that we always see the"man in the moon" and never what is know as the "dark side of the moon".

The moon rotates around the Earth once every 27ish days and in this time it also rotates about its own axis once. So if you want to find its angular velocity.

w = 2 * pi * 1 / (2700 * 24 * 60 * 60)

w = 2.69E-6 rad/s

So if you are on the surface of the moon that means that you have a tangential velocity of this angular velocity times the radius of the moon (1780km).

Vt1 = w * R

Vt1 = 4.794m/s

Which is roughly 17km/h. Now think of it this way, when you are on the surface of the moon you are travelling quite fast because you are far away from the center, like a ball on the end of a long string. Now when you start to the center the radius changes. Say you are halfway to the center of the moon, your tangential velocity will have to be:

Vt2 = 2.39 m/s or 8.63 km/h

So you can see that our tangential velocity has changed. This is what is know as coriolis acceleration. This is often not considered in analysis and many people find it hard to grasp what is happening. If you are having trouble understanding it let me know.

Now we know from Newton that objects in motion tend to stay in motion until they are acted on by and external force. Now there is only one thing that can push the pebble in the tangential direction and that is the wall of the hole. This will also impose a radial force on the pebble and slow it decent down. This means that it will lose a lot of energy on the trip to the other side, won't make it all the way to the other side and will never have enough energy to come back to the surface. It will eventually settle in the middle after a number of oscillations.

So in order for your problem to work, the astronaut would have to be standing exactly on one of the two poles of the moon. This way there is no tangential velocity, and he/she should have to drop it quite straight.