The force will change linearly with the distance from the centre.
This is connected with the shell theorem, i.e. there is no net force acting on a body inside an uniformly dense sphere.
As the ball gets deeper under the surface, the layers above it stop exerting gravitational force, and all that matters is the mass underneath.
That mass gets smaller with the third power of distance(volume of a sphere) as the ball goes down, but at the same time it is getting closer to the centre of mass attracting it, which force is inversely proportional to the distance squared(Newton's law of gravity).
[tex]F=\frac{GMm}{r^2}[/tex]
[tex]M=ρV[/tex]
[tex]V=\frac{4}{3}πr^3[/tex]
[tex]F=\frac{\frac{4}{3}πr^3 ρGm}{r^2}[/tex]
[tex]F=\frac{4}{3}πr ρGm[/tex]
[tex]a=r \frac{4}{3} πρG[/tex]
G is constant and we can assume the density of Earth ρ to be constant as well, so we have a linear relationship between acceleration and distance.
So the ball starts falling by being accelerated by g=9,81 m/s2, then the acceleration falls to 0 in the centre of the Earth just as the velocity reaches maximum.
Then it gets slowed down more and more the farther away from the centre it gets. As it reaches the surface on the other side, the velocity is again 0 and the acceleration is again g.