Say you or some other piece of matter goes down into the center of the earth in some hollow sphere of some sort. What would happen to your weight? Would you weigh less as you go down to the center since the gravity of the stuff above you is pulling you "up"?
I get why there is more pressure and why the SPHERE would weigh more... as it is right in the middle of two huge things driving themselves into each other... but if you were INSIDE the sphere, nothing is really pushing against you, so wouldn't you be seemingly weightless at the very center of the world?

I believe you are correct. In a hollow sphere you are being tugged equally in all directions. Of course this assumes the sphere is of uniform density and thickness throughout.
This has to be, as I know you realise, a theoretical exercise, since the pressures would collapse any sphere you might construct for the purpose.

This has to be, as I know you realise, a theoretical exercise, since the pressures would collapse any sphere you might construct for the purpose.
of course, and even if it somehow managed to bare the pressure, it would probably just melt in a coupule of seconds after entering the mantle

Say you or some other piece of matter goes down into the center of the earth in some hollow sphere of some sort. What would happen to your weight? Would you weigh less as you go down to the center since the gravity of the stuff above you is pulling you "up"?

Your apparent weight would decrease linearly with depth as you went down to the centre of the Earth. At the exact centre, you would be weightless.


I get why there is more pressure and why the SPHERE would weigh more...

No. The sphere would be weightless, too, at the centre of the Earth.

In a uniformly dense sphere, the strength of gravity, according to Newton gravity, changes linearly with respect to distance from the center.

In a sphere of non-uniform density, increasing as distance to the center decreases, such as the Earth, gravitational strength, according to Newton gravity, tends away from linearity and toward an inverse square function.

Not surprisingly at all, at the cetner, with equal amounts of mass in every direction, a non-uniform but symetrical sphere will also provide a weightless situation at the centor.

In a uniformly dense sphere, the strength of gravity, according to Newton gravity, changes linearly with respect to distance from the center.

In a sphere of non-uniform density, increasing as distance to the center decreases, such as the Earth, gravitational strength, according to Newton gravity, tends away from linearity and toward an inverse square function.

Not surprisingly at all, at the cetner, with equal amounts of mass in every direction, a non-uniform but symetrical sphere will also provide a weightless situation at the centor.

If gravity weekens as you get further away from the source would it not tend to get stronger as you get closer to the source?

If gravity weekens as you get further away from the source would it not tend to get stronger as you get closer to the source?It is a question of the geometry. For any inverse-square law the force inside a spherical shell is 0. So, none of the mass "above" you counts, only the mass below. For a uniform density sphere the mass below you is a function of r^3 so r^3/r^2 gives you r, the linear term that James mentioned.

-Dale

It is a question of the geometry. For any inverse-square law the force inside a spherical shell is 0. So, none of the mass "above" you counts, only the mass below. For a uniform density sphere the mass below you is a function of r^3 so r^3/r^2 gives you r, the linear term that James mentioned.

-Dale

Yes
Of course it all has to do with location in regard to the mass.

Sometimes the obvious eludes me.