Why is the value of "g' Zero at the centre of the Earth?

Question

Note: Here "g" is the acceleration due to gravity at the particular place.

Answer 1

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
or

g = (4/3) * density * R

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

Answer 2

looking at how g was derived, the equation is g=GM/r, where G is the universal gravitational constant, 6.67*10^-11, M is the mass of the earth, and r is the distance from the object to the earth's center. Since r in this case would be zero, the accel of gravity would be undefined, or in easier terms, 0

Answer 3

The acceleration is due to the attraction of every little bit of Earth around you, added up. If you're at the exact center, then for every little bit off in one direction there's a corresponding little bit off in the exact other direction, so those add to zero. add up all the pairs of little bits, each zero, to get the complete acceleration: 0.

Answer 4

If I had to guess, I would say 9.8 (meters/seconds Squ.) But, the force would be in all directions since you are being pulled toward mass. So, in turn it would simulate 0 gravity, but in reality it's just a cancellation of vectors.