It remains for us to throw light on another interesting question. How will the force of gravity change if we go deep underground?
The weight of an object is the result of the tension in, so to say, invisible “threads” reaching out to this object from every piece of matter in the Earth. Weight is the resultant force, the result of the addition of the elementary forces exerted on the object by the Earth’s particles. All these forces, even though directed at different angles, pull a body “down”—towards the centre of the Earth.

But what will be the weight of an object in an underground laboratory? Forces of attraction will be exerted on it both by the internal and external layers of the Earth.
Consider the gravitational forces exerted at a point lying inside the Earth by an external layer. If we break up this layer into thin shells, cut out in one of them a small square with side
I suggest that the reader prove that these ratios are the same, that is, that the forces of attraction at point
Having broken up a thin shell into pairs of “opposite” similar squares, we established a remarkable fact: a thin homogeneous spherical shell does not act on a point within it. But this is true for all the thin shells into which we broke up the spherical layer lying above the underground point we are interested in.
Hence, the layer of the Earth lying above the body might just as well be absent. The action of its individual parts on the body is neutralized, and the resultant force of attraction exerted by the external layer is equal to zero.
Of course, throughout this reasoning we have assumed the Earth’s density to be constant within each shell.
The result of our reasoning permits us to easily obtain a formula for the gravitational force exerted at any depth
This implies that
But as a matter of fact, the behaviour of
The average density of the Earth is easily found by dividing its mass by its volume. This yields a value of 5.52. At the same time, the density of the surface bedrocks is much smaller—it is equal to 2.75. The density of the Earth’s layers increases with depth. Within the surface layers of the Earth, this effect dominates the ideal decrease which follows from the formula just derived, and so the value of