A famous result by Newton will help. [It involves an integration the technicalities of which I would have to work out, or look up]
Any spherical shell has the same gravitational effect as a mass at its centre if you are outside it, and no effect if you are inside it.
As a ball is a sum of spherical shells, you can break it into two concentric parts and ignore one of them.
I see. so it doesn't change wherever you are inside the body? Interesting!
? doesn't seem right to me. seems to me, if you ignore the outer shell than the gravitational pull of the intersphere would be diminished by half. since the radius would be half, there would be a fourfold increase in force since that's a square function, but since the mass would be 1/8 that being a cube function, and, that would result in an overall decrease of gravity by half. then you have to take into account the fact that the outer shell would be exerting a gravitational field 180 degrees opposite to the to the inner field. No?
ok. On second thought my last comment was not correct. I was thinking of gravity of the inner part of the sphere as if the other part on the opposite side of the center had somehow just magically disappeared. It still exerts a gravitational pull on you inside the body. I guess it would sort of even out, although the mathematics for that are a bit over my head.
...but at some point it has to change. surely if you're at the center of the body then the gravitational pull would be zero, right?
I'm thinking it would be infinity in all directions? All the mass is now surrounding you rather than below you, uniformly. Interesting points though :)
I see. so it doesn't change wherever you are inside the body? Interesting!
Just to check you are not misunderstanding, the parts that are further than you from the centre have no effect. It needs a bit of integration to show it all cancels out.
So, for example if you are at 4000 miles from the Earth's centre, it is as if only the parts of the Earth less than 4000 miles from its centre remain. [ignoring any discrepancies from spherical].
So what would the acceleration be? Again simplifying by assuming the density is constant and the Earth's radius is 8000 miles, the force is proportional to M / r^2, and it turns out that means it is half as strong as on the surface (1/8 of the mass but r^2 is 1/4 as big).
Generalising, in a ball of constant density, the gravitational acceleration goes down proportional to distance from the centre. Which agrees with mrd55's observation that it becomes zero at the centre itself.
The fact that there is no gravitational force inside a spherical shell seems a bit of fluke, but it's just a mathematical fact (not a trivial one).
Ah okay. Yes, I did misunderstand, but now I comprehend.
While I understand that all gravitational forces will cancel out at the centre of the Earth; I don't understand why you wouldn't feel a force in all directions, as opposed to no force.
To me, it seems logical that all the mass (and thus gravity) is on the outside. Assuming you dug a 10x10m spherical hole at the very centre of the Earth, you would be attracted with half the Earth's surface gravity to one side. I guess I'm thinking of it similar to what forces you'd feel if 2 Earths were placed 10m apart and you were floating in between them. Technically, you wouldn't move, but your body would want to go both ways.
(assume equal mass distribution in the body).
If you dug to HALF the RADIUS of the body, what graviational effect would you feel on yourself?
Assume equal mass distribution in the body, that the body's content wouldn't crush you etc