Gravity at Centre of Earth: Infinite?

[Ralph Spencer]

g = G*m1*m2/r^2
What will be the g faced by a point particle at the exactly centre of earth?
g= (6.67*10^-11*1*5.97*10^24)/0^2
g= Infinite ?

 

[Doc Al]

No. Taking the simple model of a spherically symmetric earth, for any distance from the center only the mass within that distance contributes to g at that point. So at r = 0 (at the center of the earth) g = 0.

 

[arildno]

1. g at the surface g=GM/r^2, where M is the mass of the Earth, and r is the radius out to the surface.

2. g=0 at the Earth centre.

3. How to explain 1 and 2?
Without doing the maths, I merely state the reuslt:

At any given point within the Earth, the "local g"-value will only be affected by the mass of the ball with the Earth's center as its center, and upon which you are placed on the surface. The massive shell around contributes nothing at all!

4. Thus, for an ARBITRARY r-value, we substitute "M" in the formula with \rho\frac{4}{3}\pi{r}^{3}, where \rho is the (constant) density, and you recognize in the other parts the volume for a sphere of radius r.

Inserting this into the formula, we gain the following expression for local g:
g=\frac{4}{3}G\rho\pi{r}[/itex], i.e, the local g decreases linearly to zero as we proceed towards r=0

 

[D H]

g=\frac{4}{3}G\rho\pi{r}

Note well: arildno's result is predicated upon a constant density. The Earth's density is anything but constant. Most notably, the Earth's core is mostly iron and is considerably more dense than the rock that forms the mantle and crust. A more realistic model is that gravitational acceleration is roughly constant down to the core/mantle boundary and only proceeds to decrease toward zero with depth after crossing this boundary. In fact, gravitational acceleration is slightly greater at the core/mantle boundary than it is at the surface.

 

[arildno]

Note well: arildno's result is predicated upon a constant density. The Earth's density is anything but constant. Most notably, the Earth's core is mostly iron and is considerably more dense than the rock that forms the mantle and crust. A more realistic model is that gravitational acceleration is roughly constant down to the core/mantle boundary and only proceeds to decrease toward zero with depth after crossing this boundary. In fact, gravitational acceleration is slightly greater at the core/mantle boundary than it is at the surface.

Indeed.

A simplified model, even to the point of being erronous, can, however, yield some important qualitative insights, even though it fails miserably in quantitive prediction.

 

[magpies]

If gravity at the center of the Earth was really infinite it would also mean gravity at the center of just about every thing was infinite. Unless there was a force stoping the infinite gravity from pulling the rest of the object into it... this wouldn't work.

 

[mishrashubham]

In physics dividing by the denominator zero is often taken as infinity while actually it is a quantity that is not defined in mathematics. Infinity and "not defined" are two distinct things. So g is definitely not infinity at the centre of the earth.

 

[Vanadium 50]

First, this is a year old. Second, there is no dividing by zero involved. Third, the question has already been answered correctly in message #2.

 

[mishrashubham]

1. Sorry I didn't check the dates.
2. Actually there is dividing by zero. force=GMm/r^2, and the asker asked what would happen when r=0. then GMm/r^2=GMm/(0)^2=GMm/0

I was not trying to give a solution to the problem I was trying to state that the assertion that GMm/0=infinity is false

3. As I said above, question had already been answered and I was not giving a solution.

 

[ZapperZ]

1. Sorry I didn't check the dates.
2. Actually there is dividing by zero. force=GMm/r^2, and the asker asked what would happen when r=0. then GMm/r^2=GMm/(0)^2=GMm/0

There isn't! The expression being used is only valid for r>R, where R is the radius of the earth. For r<R, it is a completely different expression, so the issue of dividing by zero doesn't exist!

Zz.

 

[Idoubt]

Without getting into math the best way to explain is that there is mass all around you pulling at you from every direction at the centre of the earth, and since the Earth is nearly symmetric the forces cancel each other out.

on the surface of the Earth all the mass is on one side and pull in one directions and hence u have max gravitational force

Actually because of the asymmetry of the earth, g is higher at the poles, this is because at the equator all the mass of Earth is a bit more spread out to both sides while at the poles they are closer together ( ie if you draw straight lines between you and every atom on the earth, the lines will be more closely spaced from the poles )