In the interior of a sphere of uniform density, the gravitational strength is proportional to the radius, exactly like Hooke's law for an ideal spring. That's why the object in the bore hole undergoes harmonic motion.
Why is the graviational strength proportional to the radius?
Firstly, you have to know that the field strength is zero inside a hollow sphere. This is part of that shell theorem.
So for a point at a given depth inside the sphere, we can divide the sphere into a hollow sphere consisting of everything less deep, and a solid sphere consisting of everything deeper. Only the deeper sphere matters; we can ignore the hollow sphere.
So as we progress toward the centre, the attraction is due to a smaller and smaller sphere, whose mass is proportional to r^3. The radius is shrinking though, which has the effect of increasing gravity: the gravitational field strength is proportional to 1/r^2. Wen we combine these factors, we get r.
Why is the graviational strength proportional to the radius?
Firstly, you have to know that the field strength is zero inside a hollow sphere. This is part of that shell theorem.
So for a point at a given depth inside the sphere, we can divide the sphere into a hollow sphere consisting of everything less deep, and a solid sphere consisting of everything deeper. Only the deeper sphere matters; we can ignore the hollow sphere.
So as we progress toward the centre, the attraction is due to a smaller and smaller sphere, whose mass is proportional to r^3. The radius is shrinking though, which has the effect of increasing gravity: the gravitational field strength is proportional to 1/r^2. Wen we combine these factors, we get r.