Throwing Stuff Down A Mineshaft

I'm not really sure what you're asking, but I'll have a go at it. By "gravitational energy", I'm guessing you mean Potential Energy. That is, the kind of energy an object has because of some other object's gravitational force. Like a book on a shelf has more Potential Energy then a book on the floor.But the "mass of the sphere" is unaffected. The Potential Energy comes from the mass of the sphere. It's not the energy that makes the mass, it's the mass that makes the energy. The book on the shelf has more Potential Energy than the book on the floor, but neither book gains or loses mass, and neither does the Earth (not even a teeny-tiny bit).You are essentially correct that the Potential Energy at the centre of a hollow sphere cancels out*. Think of it as every tiny point in the mass of the sphere is pulling you a tiny bit towards it. (In fact, every tiny point of everything that has mass is actually pulling you a tiny bit towards it, but that's just a side point).On the surface, you have all the tiny points in the mass "below" you, that's why you're pulled "down".Inside the hollow part, you have all the tiny points actually pulling you away towards the surface. Given a perfect sphere, each point on one side has an equal point on the exact opposite side that cancels out. Add in a bit of air resistance and no matter where you are in the centre area, you'll eventually "float" to the centre of the sphere.Now think of a non-perfect hollow sphere (or obloid-type thing I think the word is). With a bit of air resistance, you'll still float to one spot... the centre of mass. You can get the same effect using a perfect hollow sphere with you in the middle and, say, 5-million people all in one country on the outside (and no one else anywhere else). The mass of those 5-million people would actually pull you a bit towards them... right to the centre of mass of the sphere-plus-the-people.Now for an interesting addition... if you're on the surface of an earth that was massive enough, you'ed actually be pulled towards the centre with enough force to crush your bones against the ground.And taking this massive-concept to a hollow sphere... if it were massive enough, it would pull your body apart and splatter you all over the inside walls Also... even though the earth is pulling us to it's centre, we're actually also pulling the earth towards our centres. It's just that the size of the earth is so much that it's extremely negligible.
But, if you want to use exact-precision, you can say that you pull the earth out of orbit everytime you jump up. (Of course, as you push off with your jump you push the earth away, and then pull it back towards you with the force of gravity).
The amount the earth moves is ridiculously small... like milli-milli-micron-milli-half-an-electron type small... but it's calculable.*Note: By "cancels out" we don't mean that the forces go away. It's that the forces are equal-but-opposite. The forces are all still there, and all as strong as they were before, it's just that there's no movement as a result of those forces.Edited by Stile, : Added clarity on what "cancels out" means

(Heat is caused by friction... and you only have friction if you're hitting something... which is what the air resistance is)So, if you remove air resistance, you remove friction and also remove heat-loss.You do still have ambient heat-loss from your body-temperature into the air, but this has no affect on the energy of the motion. It only has an affect on the energy of your body. You'd just die from starvation eventually... then that heat loss stops

I, um.. would like to see the integration on this If you're in a hollow shell (perfect sphere).. and the hollow part has a radius of 100 meters.If you're out by the edge (say, 90m from the centre of mass), you'll be pulled towards the centre of mass until you reach equilibrium (at the centre of mass).Or, perhaps I'm defining "weight" incorrectly?
I'm assuming that you're not "weightless" if you're being pulled towards the centre of mass... you just feel like you have very little weight. But the only place you actually feel truly "weightless" would be at the centre of mass, no?Edited by Stile, : Fixed a confusing typo where I put "fell" instead of "feel"

Ah yes, thanks for the refresher, that's all I needed to see. Of course, there's more mass on "one side", but it's farther away. And the cool properties of a perfect sphere balance this distance-mass relationship perfectly equally.I understand. Yes, even if you're at a spot 90m from the centre (back to my scenario), you wouldn't move due to any gravitational forces, and therefore be weightless.

Perhaps I am simplifying gravity too much. I take it the gravitational forces act via waves, then?I was thinking more like if I have a force of 10N pulling my left arm and 10N pulling my right arm, I don't move (they "cancel out"), but both forces are still there.Am I simplifying gravity too much then, by reducing it's force to a mere number of "x Newtons"?

No problem, but make sure to read my Message 35 in this thread, I'm pretty sure I'm screwing up the "inside the hollow sphere" stuff as cavediver is explaining.I got carried away and forgot the relationship with mass vs distance.I was thinking of gravity as too much of a "linear force" when it's more of a linear force that has a fundamental basis as a wave. In which case, what I said about the forces cancelling out but still being there is totally incorrect, and the forces would actually cancel out so that there is no "feeling" of them at all.This is not true for energy like Kinetic energy (someone actually pulling on my arm)... but I think Potential energy works a bit different and actually does cancel out down to absolute 0 force as opposed to simply 0 net force.See cavediver's reply to my question and I'm pretty sure he's about to confirm that forces due to gravity act as waves.I should be able to confirm that (I found Newtonian physics extremely intuitive in school), but apparently it's been a bit too long since college for me ..which makes me a bit sad, it's only been 9-10 years

Next there'll be a conversation on how much the water level of the ocean changes (if at all) when you throw an anchor overboard...

I think I understand... let me ramble a bit and if you could confirm what I'm saying it would be greatly appreciated.All forces act in the same manner.
Forces seemingly cancel each other out as waves would do when they act on the same "spot".
Forces due to Gravity (Potential energy) act "everywhere" (in their mass vs distance relationship).
Forces due to Kinetic energy act locally on the spot they interact.I think that was my confusion. The force of gravity being "everywhere" vs. local forces due to something's mass running into something else's mass.All forces act the same, but if I have a planet's gravity pulling on my left and an equal planet's gravity pulling on my right... I don't get ripped apart between the two because the "everywhere" force due to their gravities actually cancels out.All forces act the same, but if I have a truck pulling on my left and an equal truck pulling on my right... I get ripped apart because this Kinetic energy is focused locally on my arms, which itself comes down to fields interacting. However, the interactions of these fields are more local and not so pervasively "everywhere" such as the interaction due to a gravitational force.Something like that?

Disregarding all energy loses due to friction/air resistance... yes.Well, not exactly, you wouldn't really "land". As in, you wouldn't shoot out the end of the hole, move to your side, and then come back down and "land" on the ground. With no lateral force, there would be nothing to move you away from being right over the hole.If you dropped into a hole that went through the centre of the earth (disregarding air resistance), you would end up popping out the other end (feet first, so you'd feel "upsidedown") exactly the same height beyond the surface as you started... this "height" is measured from your centre of mass... not the top of your head or bottom of your feet or something like that. And then you'd start your drop all over again (head first this time) back to the other end and you'd end up hovering just above the hole, with your centre of mass at the exact same position you started in. This would repeat over and over again forever.If the hole was sized just perfectly enough to fit you, and say you could spread your legs so that you didn't fall into it, and then brought your legs together to "jump in", you'd get through on the other side enough to spread your arms out on the surface and stop yourself (remember you'd be upsidedown at this point). Or you could not do anything, drop back in head-first, and you'd be able to regain your original position and spread your legs to stop yourself from doing another full oscillation.

I'm not positive about this, but I think that the rotation of the Earth wouldn't have any effect. That is, the goat is on the Earth already (that's where it jumps from to get into the hole). So that would mean the goat has all the rotational momentum and motion of the Earth. Now, the goat can't really run up to the hole and jump in... it would eventually hit the far side with it's own additional momentum. It would really have to straddle the hole (or be held over it somehow) and drop into it so that it can't reach the sides.I do know that if the goat hits the sides on the way down, each hit will lose a bunch of energy... this will ensure that the goat won't make it all the way back up the other side. The poor banging goat will still oscillate a bit, but it would be similar to air resistance and the goat would eventually end up at the centre of the earth. Rrahin's explanation is much better and more in-depth. I'm just going to say that you're half-right. Stuff that is closer to you most certainly does pull you stronger. The part that you're half-wrong is that in a perfect sphere, if you're closer to one side, there's less "stuff" pulling stronger on you, and there's more "stuff" pulling weaker on you. The properties of a perfect sphere are such that these effects perfectly balance... lots of weak stuff = less strong stuff. Therefore you go nowhere.Of course, this only works in an isolated, perfect sphere. In a non-perfect sphere, or a different shape, or a perfect sphere with something else on one side of it... you'll be pulled towards those imperfections depending on how significant they are.Added by Edit: Oh... if anyone's wondering, this "Dyson" sphere stuff is a perfect sphere with another perfectly spherical cavity inside it, that is located at the exact centre.Any deviation from that, like if you have a perfect sphere cavity in another perfect sphere, but the cavity is off-set to one side somewhat... would not produce the same effect. You would then be drawn towards the imperfection somewhat. Edited by Stile, : Added Dyson Sphere explanation

Well, I have to admit I don't really know what's going on here anyway. And even when I did think I knew what was going on I got parts wrong (like the inside of the perfect sphere).But... this is what I think (for whatever it's worth).I agree with you that if the goat bounces of a side, he's likely to bounce of multiple sides and the shit goes to hell (..literally? ).However, I'm still not sure that the goat hits the side at all. Rotational speed is strange... it depends on how far you are from the centre. Yes, the relative speed of the Earth will drop (since we're getting close to the centre... each specific area of the Earth will be moving slower and slower). However, I'm not convinced that the relative rotational speed of the goat stays constant. As the goat moves closer to the centre of the Earth, I'm guessing that the goat's rotational velocity reduces equally with respect to how close the goat is to the centre of the earth.Now, I may just be flat wrong about that and I don't have any hard facts to back myself up with this one, and I hope someone more knowledgeable chimes in to set us straight. But for now, I'm standing on the side that the goat's rotational velocity reduces itself as the poor fella falls so that the goat never touches the sides anyway. That is, neglecting air resistance and stuff like that. Long live the free falling goat!I suppose my real answer is: *shrug*