Question

How far would you have to drill into the Earth, to reach a point where your weight is reduced by 7.5% ? Approximate the Earth as a uniform sphere.

Answer 1

Let F1 = weight on the surface at radius r1
Let F2 = weight in the hole at radius r2
Let m1 = mass of person
Let m2 = mass of entire earth (of radius r1)
Let m3 = mass of earth (of radius r2)

Look up in your physics book. Somewhere it should derive the fact that outer shells exert a net zero force on an inner body. It might be derived regarding charged spheres, but applies as well to gravity.

So the concept is, down in the hole both the distance between the two masses and the mass of the earth have been reduced. The mass of the earth is proportional to the volume, which is proporational to the cube of the radius.

On the surface of the earth:
F1 = Gm1m2/(r1)^2 (Eqn 1)

In the hole:
F2 = Gm1m3/(r2)^2 (Eqn 2)
and
m3 = m2 x (r2/r1)^3 (Eqn 3)

Subsitute Eqn 3 into 2:
F2 = Gm1m2(r2/r1)^3 / (r2)^2
F2 = Gm1m2r2/(r1)^3 (Eqn 4)

But it was given:
F2 = (100% - 7.5%)F1
F2 = 0.925F1 (Eqn 5)

Substitute Eqn 4 and 1 into Eqn 5:
Gm1m2r2/(r1)^3 = 0.925 x Gm1m2/(r1)^2

Cancel Gm1m2:
r2/(r1)^3 = 0.925 / (r1)^2

Multiply each side by (r1)^2
r2/r1 = 0.925

Rearrange:
r2 = 0.925 x r1

Answer: Drill down a distance equal to 7.5% of the earth's radius.