# How much pressure does the earth's central atom experience?

If mass is distributed evenly about it: "central atom", then it should be weightless right? If reasonably so, does it still experience the pressure?

I would like to calculate, but I'm not so proficient in math to my own assumptions.

And then what about the atom at the center of gravity in every object?

The main reason I pose such an oppositional proposition is in essence because I reasoned, that within a tunnel beneath a mountain, I ought to weigh less as the mountain pulls me upwards with its mass; so I figure, what if the mass was evenly distributed around you, how much would you weigh? It seems to me the same effect as flat space, like 360 degrees of waves cancelling out at the convergence point.

Unfortunately, gravity would then have to be a "surface phenomenon", unlike Newton's experimentally confirmed understanding of earth's pressure gradient.

You are correct that as you get very close to the center of the earth, the value of $g$ can become arbitrarily low. If you could somehow create a space there, you could potentially float in it because you would not be pulled in any particular direction with respect to the earth.

But while gravity is not strong there, it is strong in other places (like your mountain), and those things are all pushing down. So the pressure at the center of the planet is high. It can only be estimated, but is more than $300 GPa$.

While you can pick certain points where the sum of gravitational forces cancel out, I'm not sure why you use the term 'surface phenomenon'.