GeologyRocks

it would still stop in the core, regardless of gravity, because atmospheric pressure would be equal from both sides of the hole at that point.

Here's what one person said...I'm not too sure myself

~Jenny~

Definately stop in the middle if you ask me....The gravity would balance.

Jon

I've been playing with a similar idea these last few weeks. I'm beginning to think it is a Quantum Mechanics problem. Way over my head.

Keeping an open hole all the way through is a bit different, but does this make sense?

Any point within the Earth has mass above and beneath, and mass side to side.

Here at the surface, the Gravitational Constant of 32 ft/sec² should decrease and approach a value of zero with increased depth, and ideally arrive at a relative value of zero ft/sec² when measured at the Earth's centre. (There could really be an Omni-directional force of gravity, roughly 16ft/sec², pulling that centre point, making the Net Effect of gravity = zero.)

I've been looking for some input on this for a while but it still isn't sorted out yet.

Mike

So do you think the ball would just get to the middle and stop, or do you think it would go past the middle then come back (oscillating for a while) and eventually stop?

~Jenny~

The last one: everything oscillates for a bit...

That was my suggestion, only in place of the word "oscillate" I think I said "wobble a bit" and used overdramatised arm movements to illustrate my point.

This post is going to map out my thinking of this problem, from beginning to end, so you'll just be following my train of thought.

so i was thinking about this last night, and how my previous answer wouldn't make sense since in general things move from areas of high value to the area of lower value. pressure inside the hole would be greater than on the surface, so it wouldn't push the ball into the hole. Rather, it would seem to make it rise if it had an affect on the ball.

so i'm changing my guess to the ball rising and being pushed into space.

at least i think. an airplane wing works on the principle of low pressure above the wing and high pressure below the wing when air is forced past. This doesn't totally apply because this is a sphere, so air blowing over it will not inherently cause areas of low and high pressures above or below the ball. So the pressure difference has to be already there, which it is. even if it's just a tiny amount, any increase in elevation, no matter how small, will result in a lower pressure (atmospheric anomolies aside).

well.........this reminds me of something. bouyancy. things float in water if they're less dense than the water, sink of not. So maybe this is a question of density of the ball. is it greater or less than density of the surrounding air? so this could be like.....a balloon. if the balloon is filled with helium, it flies away. filled with exhaled air from lungs, it sinks to the floor.

so could this just be a question of how dense the ball is compared to the surrounding air?

I think we can assume that the ball is more dense than the surrounding air, as just about all solids are. So, we know from experience what it's going to do at the start: big hole, small hole, it accelerates down with gravity.
It all gets a bit more theoretical when the ball starts to fall far enough for it to be a reasonable proportion of the way to the earth's centre of mass. I remember reading that g is the same any depth below the surface of a sphere of uniform density; in effect, the extra gravitational field strength derived from being closer to the centre is cancelled out by the upward gravitational attraction of the part of the (planet) that is now above our theoretical ball.
That leaves a few unanswered questions, even if it is true.
1: as the core of the earth is probably more dense than the rest of it, will g increase as the ball drops?
2: what about the angular momentum? Might the ball, if the hole were the right shape, maybe even tend to go into an orbit around the core?

It seems intuitive that if the hole were the right shape (or if we just forget the hole and imagine a "ghost ball" that can move freely through the ground) and if, as always in this sort of thing, we neglect friction (so a vacuum hole or a super-ghostly ball) the ball would probably convert its potential energy into kinetic, then rise back up to the surface in australia, and carry on oscillating like that; but the angular momentum issue still seems a problem. I think, when I imagine how one changes a space probe's orbit, we might end up with our ghost ball in a very small, very fast orbit around the earth's core; or maybe an extremely elliptical orbit, returning it to about where it came from at the apogee but never getting very close to australia.

Hmmm. Well, that didn't come very close to an answer, but it's a few thoughts. I am pretty confident that, in the real, with-friction big hole paradigm, the ball will end up hovering in the dead centre.

if the hole was super smooth couldn't the ball just flow into space due to centrifugal force?

Almost certainly not. If the initial direction of the stone were going to be up, then with or without a hole everything -- including the ground and the atmosphere -- would go up. The only way I can imagine it ending up higher than it started is if it were to undergo a gravitational slingshot effect. Anyone have more info on these?
In the real scenario, air resistance would keep the terminal velocity very low, and lower in the very dense air near the core. So it would be moving slowly on getting there, and then stop there.

I think.......