Gravity at the centre of the earth

In response to a question that I have received at LP, I have decided to put a justification up for why gravity, will decrease as you move towards the centre of the earth. As usual in a physics problem we will make a couple of assumptions to get the answer, but I don’t think they are terribly bad ones. If someone wants a mathematical derivation of this they should read the Wikipedia entry on the Shell Theorem, particular the section on a solid sphere.

First I want to clear up a couple of confusions. Newtonian gravity is governed by an inverse square law. The magnitude of the force of gravity being equal to g = GM/r^2, where M is the mass of the object we are being attracted to. Now naively you may assume that this means that gravity goes to infinity as you approach the centre of the earth. Well it would if the earth’s Mass was all concentrated in the centre, but its not which is the key point. When we are outside a sphere its acceptable to treat the whole thing as a point mass, but not once we are inside.

Now consider a spherical shell, like a tennis ball. The bellow diagram shows an arrow pointing to an arbitrary point inside that sphere’s cross section, the dashed line is meant to go through the arbitrary point and the centre of the circle (it doesn’t quite because I drew it badly, but just imagine it does). The other two sectors are symmetrical about this dashed centre line.

Now for a circle we would have the length of the two arcs (l and L in the diagram) given by l= r *phi, and L = R* phi. For a sphere if we rotate that arc around to give us a capped section on either side, we get the area_l = r^2 * phi, and area_L = R^2 * phi. Now since this is a thin shell the mass of each of those is just the mass = density * area * thickness. Most importantly the mass is proportional to the area or rather mass_l = k * l^2, mass_L = k*L^2

So when we put those two masses into the Newtonian gravity formula, we get the mass term on the top line proportional to r^2, which cancels with that same term on the bottom line, and we find, miraculously that the strength of gravity from the two different arcs is the same in exactly opposite directions. ie it cancels. So we can work around the circle find each pair of opposite sections will cancel and this works for any point INSIDE this shell.

So if we want to find the effect of gravity we never have to worry about the effect of a shell that we are inside. So as we approach the centre of the earth its only the mass closer to the centre than us that matters.

Now assuming the earth is fairly uniformly dense the mass inside us will be = density * 4 * pi * r^3 /3. So if we put that into our Newtonian gravity formula we get when we are a distance R from the centre:

g = 4* M * density * R^3

    3 * R^2


g = (4/3) * density * R

So as we go towards the centre the force of gravity goes to zero.

ps. If anyone knows a good way to render maths in WordPress I’ll clean this up and make it easier to read.


5 Responses to Gravity at the centre of the earth

  1. Sacha says:

    Sorry Steve I don’t know how to make it look good. Maybe a pdf attachment?

    Did people really ask about this? I think I forget how much people don’t know about basic physics.

  2. Steve says:

    Yes they did ask. Check the link. Actually its an old question Rod Clarke asked me years ago. He was using it as a way of identifying himself to me as asking again, but GMB also took some interest.

  3. Steve says:

    I thought about doing it as word equations and trying to save them as images but it is too much trouble.

  4. james says:

    havn’t really used this since mathematica automatically produces mathml web pages.

  5. Rod Clarke says:

    Thanks very much Steve,

    Now I understand it.

  6. tigtog says:

    This brings HSC physics back to me vividly. I’d forgotten about this one entirely (having only done physics for biomechanics since). Ta.

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