So I was in the shower today...

[Agamemnus]

Alright, SJ Zero, I have no freaking clue what your integration notations stand for, but integrate this, if you can:

Code:

m1 = mass of the earth
v1 = volume of the earth
d1 = density of the earth
D = diameter of the earth
r = D/2
n = depth of our one kilogram mass (n=r means the particle in question is in the center of Earth)
m1 = mass of small attracted object = 1kg
m2 = mass of circle slice

Circle area = pi * R^2
R in the circle at distance x from the particle in question:
x^2 + R^2 = r^2
R = (r^2 - x^2)^.5



component 1 = integration from x=0 to x=n of (integration from b=0 to b=R of (b/(n^2+b^2)))


Similarly,

component 2 = integration from x=0 to x=(2r-n) of (integration from b=0 to b=R of (b/((2r-n)^2+b^2))))

F = (pi*G*m1*m2) * (component 1 - component 2)

So what I did is take the earth as a sphere and divide it into infinitely thin circle slices. Then apply the gravitation formula to each slice, assuming we are dealing with a 1kg mass x meters in depth. First integrate the circles above and then integrate the circles below the point, and subtract those gravitational forces to get the net force. Any angular forces are of course cancelled out such that the center of gravity is always in the center of the circle.

We assume that the density is uniform, which isn't correct but good enough for now.

So, what is the sum of the forces?