So I was in the shower today...

[Radical Raccoon]

Aga:

Actually, if you look at the formula, you will realise I am right. At a distance of 0, you are dividing by zero, which is mathematically defined to be infinity.

Since the average center of mass of all these particles making up Earth is the Earth's center, I stand by my argument.

The formula is correct, but you are using it incorrectly.

If a ball is resting at the surface of the earth (m1) with a mass of 10, and the center of the earth (m2) has a mass of 1000 (just to make things easy) with a radius of 10 then:

pull = (m1-m2)/(d^2)
pull = (1000-10)/(10^2) (leaving out G to simplify things)
pull = 990/100 = 9.9

Now move the ball (m1) to the center of the earth and now you have:

pull = (1000-10)/(0^2)
pull = 990/0 = ERROR! Infinity!

But this is wrong. You are only taking 2 masses into consideration when there are actually 3.

Basically, if the ball is in the center of the earth then you have 3 masses to take into consideration: the ball, the half of the earth behind the ball, and the half of the earth infront of the ball.

Now if the ball (m1) is in the center of the earth, you have 2 halfs of the earth pulling on it. 2 center of masses in opposite directions that would be about 1/2 between the ball and the surface (500 units away).

pull1 = (500-10)/(5^2)
pull1 = 490/25 = 19.6

pull2 = (500-10)/(5^2)
pull2 = 490/25 = 19.6

So now you have a force of 19.6 pulling in both directions, resulting in a net zero force. Conclusion: you would feel no effect of the earth's gravitational pull if you were to reside in the center.