Gravitational field intensity at the center of the earth

1. Dec 14, 2010

I Like Pi

1. The problem statement, all variables and given/known data

what is the gravitational field intensity at the center of the earth

2. Relevant equations

3. The attempt at a solution

i was told that it was zero because you have the earth's mass surrounding it, and graivty moves to the center of its mass (center of radius) therefore the net force would be zero? is this correct?

also if you sub in 0 in the gravitational field intensity formula for the radius, you get a value over zero, meaning it is undefined... so i can't really use that to support why it is zero at the center......

thanks

2. Dec 14, 2010

NobodySpecial

Yes - in simple terms, ignoring the moon and tidal effects

You also have zero mass inside that radius so it's zero/zero.
In fact you should be thinking about 'r' as the distance to each bit of mass above you - and since the earth is all around you there is an equal force in every direction out from the centre - hence zero overall field

Last edited: Dec 14, 2010
3. Dec 14, 2010

Staff: Mentor

You presumably are asking about gravitational acceleration, and yes, this is correct. A reasonably good model for the Earth is that it is a sphere whose density is a function of distance from the center of the Earth (and nothing else). With this model, the gravitation acceleration at some point inside the surface of the Earth depends only on that portion of the Earth that is closer to the center of the Earth than is the point in question. Imagine dividing the Earth into two parts, one part being the portion of the Earth closer to the center of the Earth than the point in question and the other part being the portion of the Earth from from the center of the Earth than the point in question. The gravitational acceleration at the point in question is the sum of the gravitational acceleration toward the inner part plus the gravitational acceleration toward the outer part. The contribution of the outer portion is identically zero because the gravitational acceleration anywhere inside a spherical shell of mass is identically zero. All that remains is the contribution from the inner portion,

$$a(r) = \frac{GM_{\text{inner}}}{r^2}$$

where Minner is the mass of that inner portion.

That's because you are using Newton's law incorrectly. The correct usage is to only use the mass at or below the point in question.

If the density of the Earth is finite at every point inside the Earth, including at the center, then as r approaches zero, the inner mass will approach zero much faster than will r2. That in turn means that a(r)→0 as r→0.

4. Dec 14, 2010

I Like Pi

Thank you very much! You have both helped! I really appreciate it!