hey doc al, maybe I am just interperting something incorrectly then. I thought that Newtons law states that F = \frac{GmM} {R^2}. Let's do a simple case. Let's say I am at the center of the earth, and there is just a column of Earth above my head and below my feet that all has the same uniform density. The weight that is trying to push down on my head will equal the weigh trying to squash me from me feet, I think we both agree on that point. And that force would just equal the sum of the mass at its distance, times the gravity at that distance, added together over the distance. That would be the force pushing me down from my head, likewise there would be that force equal in magniude present trying to push me from my feet as well.
Now for the gravity. I understand that the NET gravitational force I will feel will be any mass that is below me and is unacounted for above me. What i mean is that if i don't stand directly in the middle of that column of earth, there will be more Earth one one side of me ( either above or below). As a result, there will not be a proper balance of attraction between myself and all the particles of earth, so there will be a net force in the direction where there is the larger amount of earth, and I will feel like I want to fall in that direction.
If I were in the center, there would be equal mass and distance both above and below me, so I would not feel any net gravitational force. You said it does not have to do with net gravitational force, but I do not see how you can avoid it. I thought it is TOTALY dependent on having that net force cancel out. For every particle trying to pull me towards it above my head, there is the same amount of force trying to pull me towards the column of Earth below my feet, and for the reason, the NET force on me is zero, and I go nowhere. But that does not necesairly mean that i would not feel myself trying to be pulled both up and down does it?
Hmm, now after all this typing I kind of see what you mean .haahhaaha. I will post it anyways. (SIGHHHHHHHHHHH)

. If every particle on my body is feeling as thought it is being pulled both up and down, then as a whole, each particle in my body relative to each other, don't want to go anywhere.
I think what I was thinking about would be a gradient across my body. Because I take up space, I could not really be a point masss at the center, therefore, Techincally speaking, I should feel some small component of force from my head down to my toes. The thing was that I was considering the entire space I occupied as the center. In reality the center is a point, so a rigid body like me can't be the center, by definition, some parts of my body will be off center, by 3 feet or so on each side, I am a tall guy :-). So the parts of my body that are off center should feel some force due to the unbalance of force above and below it, but were talking about maybe 3 feet difference at most, I would think this to be a very very very very small force difference, and in affect, so small I would still think that my entire body is weightless. Thanks Doc Al you were right. I did not think about the situation hard enough.
As a side note, Shouldent that single particle feel a force on each side trying to pull it appart like the anlogy I made earlier? Would this force be strong enough to rip electrons off of that point mass, and be a reason why the center of the Earth would have a liquid plasma? Or am I still looking at the situation incorrectly.
Edit: I just made the same exact mistake to this situation. Each electron relative to the atom will not experience a great force tugging at it, thus this cannot be the reason for the plasma. The pressure has to be the only result of the plasma. great pressure = great temperature = energy to rip electrons from their orbits and turn stuff into a plasma state.