Throwing Stuff Down A Mineshaft

That's the magic gravity gnome that lives at the centre of the Earth. What is it that causes the gravity? And if you are halfway to the centre of the Earth, where is that thing that causes the gravity?

Absolutely. And it all works out beautifully - everything in the shell from the surface down to your radius exactly cancels out. So the stuff that's behind you exactly balance the equivalent stuff on the other side of the planet, leaving just the mass that is below you that is relevant. So if you are 2000km from the centre, your gravitational acceleartion is the same as standing on a planet with radius 2000km (assuming same density). Cool Even better, just think what happens if you are inside a hollow shell of matter. By the above, there is no gravity at all! You are weightless anywhere inside the shell, yet anything outside the shell feels the inward gravitational pull as normal! This is the gravitational analogue of the Faraday Cage effect.

Just to be clear, if I wasn't before - this is wrong. As is You are describing the velocity, not the acceleration.The acceleration decreases all the way to the centre, where it is zero. As you overshoot, the acceleration grows again (linearly with depth) but in the opposite direction.But as Parasomnium explains, your picture of what happens is correct.

Yes, it is - though I hadn't thought of it that way - and it's a state of neutral equilibrium.

Oh, very sure I can take you through the integration if you like...

0.0002 Hz

Only if you are located outside the shell. Once inside the shell, you experience no force whatsoever at any point. There is no "floating" towards the centre.It's straightforward calculus, just integrating hoops from pole to pole.

No, there is no tendancy towards the centre, nor any other point. There is zero gravitational force on the test mass inside the spherical shell. Hmmm... if you have two sine waves of enormous amplitude coincide out of phase, would you say that there are are equal and opposite waves at those points, or that there are no waves at those points

I'm going to cheat and refer you here as I hate trying to typeset maths at EvC No, you are truly weightless, and would be in this state anywhere inside the shell.

No more or less than any other force. Gravity is great because it is the real deal - a fundemental force that we can see up close. E/M we can too, if we have a pair of magnets. But E/M is behind just about every other force we encounter, whether it is lifting weights, pushing doors, trains pulling trucks, etc. The large scale mechanical forces are simply infinitely tangled webs of electromagnetic interactions of which we have no awareness. And fundemental forces simply come down to numbers. What is the force's value at this point in space. At that point, there is no way to tell whether a value of +15 is made of +25 and -10, +1000000000000015 and -1000000000000000, etc.

Relative to the shell, it doesn't matter what the external gravitational field is. The shell will respond to that field identically to the shell occupant, and again the occupant will be entirely weightless within the shell.

Probably not a bounce as it is a very smooth process. You will end up against one side of the tunnel which could introduce some nasty friction burns. Your contact will decelerate your rotational motion as you drop towards the centre and then re-accelerate you back to full rotational speed back on the other side.Or just build the tunnel between the poles...

Obviously, in this case you are no longer simply inside a spherical shell, but a very lopsided mass distribution Not quite what you said, with

Er, what do you mean by not "moving"? How is it not in free-fall? Does it now have engines strapped to the shell which are firing? Edited by cavediver, : No reason given.

Yes, I even started to mention this possibility in my last reply but couldn't be arsed Yes, your initial wording: explicity denies the essential "Faraday Cage"-like nature of the shell. It actually shields you from knowledge of the external gravitational fields, by nature of the shell acting as a point source to external fields, i.e. it responds the same as you do, inside the shell. This is what I am trying to emphasise. The fact that you can introduce other fields/fluids or change the mass distribution is irrelevant.