# Dont understand Gravitation Inside Earth concept

1. Apr 2, 2009

### barcafan1990

I was reading this very small section in my physics book and cant seem to come to grips with its concept.
It starts off with Newtons shell therem being applied to a particle located inside a uniform shell and that the shell exerts no 'net' gravitational force on a particle inside it.

It then uses the earth as an example and says (if earth's mass were uniformly distributed) for a particle that somehow traveled inward from earths surface that the gravitational force would tend to 1) increase because the particle is moving closer to center of Earth and 2)decrease because the thickening shell of material lying outside the particle's radial position would not exert andy net force on the particle.

*What I'm understanding so far is: By using [F=GMm/r2] case1) is saying that r is decreasing here, thus increasing F and for case2) that M is decreasing, thus decreasing F?

THEN it says for a uniform earth the first case would prevail but for our NON uniform earth the "force on the particle actually increases as it begins to descend, and then the force reaches a maximum at a certain depth, and then decreases as the particle descends farther."

*What? is this just saying that for the earth, we cant apply the general rules, because the earth is special and has some max force. Why does the book confuse me with this fact if this "particle inside the earth" example is impossible anyways? OR am i misunderstanding the last statement completely?

Thank you.

2. Apr 2, 2009

### Nabeshin

Umm, actually in the case of a completely uniform spherical earth, the gravitational force decreases linearly as radius decreases (while inside the shell of the earth). What is meant by the description of the non-uniform earth is that the core of the earth is denser than the crust, and thus as you move through the lighter material, the decreasing radius "wins out" against the decreasing mass and you observe an increase in gravitational force. At some point however you start going through the more dense material and the gravitational force decreases.

For a derivation of the uniform, spherical earth relationship, consider that the mass enclosed by a shell at a given radius is given by the density times volume.

3. Apr 2, 2009

### barcafan1990

I kind of understand. I just like justifying everything with equations, which doesnt seem to be working here.

Yes I think this is what the book goes on to talk about in a sample problem.

Mass of the Earth: M = pVearth = (p4(pi)r3)/3
Thus F = GMm/r2 ...becomes... (4(pi)Gmpr)/3
I dont understand these two equations. If they are the 'same' then why do they both yield different answers if we were to change r.

ie. for the first one if we increase r then F would decrease but
fot the second equation if we increase r then F would increase also

I dont get this. Is the second equation to be used only if we are talking about non uniform spheres?

4. Apr 2, 2009

### Cantab Morgan

I'll take a crack at trying to explain what the author seems to be driving at.

The gravitational acceleration you feel from a point particle of mass M is just GM/r^2, where G is that pesky constant and r is the distance between you and the point particle. Now, suppose you have an extended mass, meaning it's no longer a point particle. Well, happily, it's not much more complicated than that. The acceleration you feel is still the same GM/r^2, but this time the r is the distance between you and the barycenter (or center of mass) of the extended mass.

If you are inside a uniform spherical shell of matter, then it exerts no acceleration on you at all. This is true even if you are not at the center of the shell. This is a mind boggling fact, but it's true.

In your imagination, divide the earth into two parts. A CORE ball of half the radius of the Earth, and the surrounding hollow shell. When you're standing on the surface of the Earth, the acceleration due to the CORE is GM/r^2, where M is the mass of the CORE alone. (You also have an acceleration caused by the shell, of course.)

Now drill yourself down halfway to the center of the earth. You're now standing on the surface of our CORE. But this time, you're only r/2 away from its barycenter. So the gravitational acceleration you feel from the CORE now, is greater! It's 4GM/r^2, because you're closer. That's what the author is saying in point #1. Tendency number one is that your gravitational acceleration due to the core increases as you drill down.

On the other hand, all that mass in the shell no longer accelerates you at all, because you're inside it. This is what the author is saying in point #2. Tendency number two is that your gravitational acceleration due to the outer shell vanishes as you drill down.

If the Earth is uniformly dense everywhere, then tendency #2 wins. You will feel lighter as you drill down.

But, suppose the Earth is not uniformly dense. Let it still be spherically symmetrical, but let the density vary with depth. Suppose the CORE is massive and very very dense, but the upper crust is not dense at all. In such a case, tendency #1 wins, and you will feel heavier as you drill down to the surface of the CORE. (But even in this case, if you drill all the way down to the center, then your weight will vanish.)

5. Apr 3, 2009

### Nabeshin

Cantab Morgan did a good job explaining in words the concepts the author is trying to get across. As far as equations, you're right in deriving the following relationship:

$$F=\frac{4}{3}\pi \rho GMmr$$
This equation is valid within a sphere of uniform density.

The equation when you're outside of the sphere goes back to
$$F=\frac{GMm}{r^2}$$, so the force graph as a function of radius from the center of the earth is actually piece-wise, with the first equation valid for re>r, and the 2nd for r>re. Of course, they yield the same value for r=re, or else one would clearly be wrong.

So I think the confusion arises because equation 1 is only valid while inside the (uniform) sphere. The equations for a non-uniform sphere in density or a uniform density spheroid are much more complicated :). I hope this makes sense to you, and if it doesn't, try drawing a graph of the piece-wise function. That might help.

6. Apr 9, 2009

### raknath

Hi

I am just curious how does this actually work. Do we take the gravity by all the particles inside the earth individually and then take the tensor sum

7. Apr 10, 2009

### arildno

Nope, but the following is:
$$F=\frac{4}{3}\pi \rho Gmr$$