# Gravitational Force at the Center of the Earth?

1. Mar 2, 2007

### Kaoi

Okay, this was just a curious question that's been bobbing around in my head recently.

The formula for gravitational force between two masses is

$$F_{g} = G \frac{m_{1}m_{2}}{r^2}$$

right?

So, gravitational force increases as you approach the center of the earth. I was wondering, what is the gravitational force at the exact center of the earth? I thought it could be approximated by

$$\lim_{r \rightarrow 0} G \frac{m_{1}m_{2}}{r^2} = \infty$$

But I know that's wrong. I just don't know why. Why wouldn't there be infinite force at the center of the earth, and if there was, why wouldn't we collapse on ourself? :rofl:

2. Mar 3, 2007

### haiha

Your second formula is ok for a black hole.
For normal plannet like earth, using Gauss law we can have the folowing equation :

F(r) = G(4p*r*rho*m)/3

with rho is the density of the earth,
m : mass of the thing going to the earth center.
r : distance to the center.

The key point is that when you go into the earth center, the mass behind you will not have any gravitational force on you, it cancels out. So at the earth center, the gravitational force is zero.

3. Mar 3, 2007

### ZapperZ

Staff Emeritus
This equation is correct only for a location OUTSIDE the surface of the earth, i.e. r>R where R is the radius of the earth. For inside the earth, if one assumes the earth being a sphere with a uniform density, then haiha has given the proper form to use.

Zz.

4. Mar 3, 2007

### arildno

Within the classical domain, Newton's general law of gravitation holds ABSOLUTELY, as long as you apply it CORRECTLY.

Your paradox with the limiting process arises because you haven't applied the law correctly.
The net result of a correct usage in this case is given in haiha's post.

5. Mar 3, 2007

### ObsessiveMathsFreak

The formula for gravity beneath the surface of the earth is very interesting. If you perform the calculations, you will find that gravity is zero at the earth's center, increases proportially to to the radius r as you travel up to the surface, and then falls off as 1/r^2 after you leave the surface.

6. Mar 3, 2007

### arunma

As others have already said, the formula you provided is only valid for regions of space outside the earth. Newton's theory of gravity works just fine inside the earth as well (otherwise Newton would have noticed an obvious error!). But you need to use the integral form of Gauss' Law for gravity (you can use Newton's law itself, but the calculation is a bit more tricky). Simply put, Gauss' Law says that the closed surface integral of the gravitational field across a closed surface is equal to $$4 \pi G$$ times the mass enclosed by the surface.

I strongly recommend that you do this calculation yourself, because it only takes about 5-10 minutes, and will likely answer all of your questions. One very interesting result you'll notice is that until you get outside the surface of the earth, the gravitational force is proportional to the distance from the center of the earth. Whenever force is proportional to distance, this should remind you of simple harmonic motion. Indeed, gravitational force inside the surface of a planet is identical to the force that a stretched spring exerts on a mass. It actually turns out that if you could drill a hole from one side of the moon to the other and drop a ball through, it would go back and forth in simple harmonic motion. If you did this on the earth, then air resistance would cause the ball to experience damped harmonic motion.

7. Sep 12, 2010

### magujabi

What you haven't taken into account, though, is that as the radius is decreased, so is the mass. Therefore, you're not dealing with a case of a finite number divided by zero, but with zero divided by zero.

8. Sep 12, 2010

### Staff: Mentor

Mass is not concentrated at the very center. When you are below surface you are attracted not only by what is below you, but also by what is left above.

9. Sep 12, 2010

### Staff: Mentor

In the simplified case considered here, where we assume a spherically symmetric mass distribution, the net attraction from the mass above you is zero.

10. Sep 12, 2010

### magujabi

But, what this says then, is that, at the center of the earth, if you would be attracted by what is above, then the net force would be zero if we integrate all the continuous outward radial force.

11. Sep 12, 2010

### ZapperZ

Staff Emeritus
Isn't that what everyone has been trying to say? It is even stated explicitly in msg. #5 from more than 3 years ago!

Zz.

12. Sep 12, 2010

### magujabi

Well, I just joined this forum this morning. Forgive me if I haven't read all of the messages. Since you're being so polite, though, sure, I agree now that that's what everybody has been trying to say.

13. Sep 12, 2010

### Staff: Mentor

I know, I was just pointing to thing that was obviously missed by the OP.

Three years ago.

I need new glasses.

14. Sep 13, 2010

### einstor

consider this, since r>R (radius of earth) r reducing causes magnitude of force increasing, but if r<R when r reduces so the total mass of the earth included to attract reduces too, until r=0 we'll get the total mass of the earth covered is 0, so the total force is 0 (analogy of Gauss Law in Electric Field of an uniform distributed sphere charge say with radius R and density p(rho)..

Last edited: Sep 13, 2010
15. Sep 13, 2010

### magujabi

Sure, this makes perfect sense. This problem unfortunately, though, doesn't lend itself to a convenient algebraic solution. Looking at the force below the surface of the earth, we can use 4/3(pi)rmp, using p for rho, the mass density of the earth. The attraction from above then, would be 4/3(pi)(R-r)mp for the attraction from above. Combining these, and since they are opposite in direction, we get F = 4/3(pi)(2r-R)mp, which says that the force is zero half way to the center. The two dimensional approach just doesn't seem give a correct answer.

16. Sep 13, 2010

### Staff: Mentor

This is correct, and is the formula given in post #2 by haiha. It's the gravitational force on a mass 'm' that is a distance 'r' from the center of the earth. It includes the gravitational attraction from all parts of the earth.
No. The attraction 'from above' is zero.

17. Sep 13, 2010

### Staff: Mentor

Correct me if I am wrong, but isn't it a matter of how we define 'above'? If we assume 'above' means everything that is further from the center than we are, attraction from above is zero, as it already canceled out. But if 'above' is understood as 'whatever we have have over our heads' and 'below' means everything that we can see when we look down (that is, if we take into account our intuitive feeling of direction), then we are attracted by whatever is above AND attracted by whatever is below - and these attractions cancel out.

Or am I missing something? I am walking on a thin ice :uhh:

18. Sep 13, 2010

### magujabi

I am new to this forum, and sorry if I have just barged in and joined discussions, but, could you expalin to me what would be the gravitational force on a ball bearing, sitting on a bench, in straight alignment and in between, two other ball bearings, one of less mass, and the other of greater mass, irrespective of the gravitational attraction of the earth of course.

19. Sep 13, 2010

### Staff: Mentor

Well, the statement "The attraction 'from above' is zero" certainly depends on what we mean by 'above'. (That's why I put it in quotes.) From the context of this discussion, I'd say that it meant your first example: everything at a greater distance from the center. That's certainly what I meant.

I have no idea how to sharply define your second meaning, 'whatever we have have over our heads'. In any case, why would you think above and below would 'cancel out'? Wouldn't that imply gravity = 0?

In any case, the formula given in post #2 includes the gravitational attraction from all the earth's mass. (Given the simplifying assumption of a uniform distribution of mass.)

20. Sep 13, 2010

### Staff: Mentor

As usual, lack of precision on my side

I am somewhere inside. Earth mass consist of inner and outer part (which are what you defined as above and below). Outer shell has part that is below and above (my definition of above/below), and their attractions cancel out. Above and below are separated by the plane (marked in red). Inner sphere is what is left attracting.

It is a little bit convoluted, but it better fits my feeling about what is above/below. In the end it yields the same result.

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