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 \(a_{1}\) and draw lines from the vertices of the square through the point \(O\) (we are interested in the weight at those points), then on the opposite side of the shell we obtain a small square of a different size with side \(a_{2}\) (Figure 1). The forces of attraction exerted at the point \(O\) by the two small squares are oppositely directed and proportional according to the law of gravitation to \(m_{1}/r_{1}^{2}\) and \(m_{2}/r_{2}^{2}\). But the masses of the squares, \(m_{1}\) and \(m_{2}\) , are proportional to their areas. Therefore the gravitational forces are proportional to the expressions \(a_{1}^{2}/r_{1}^{2}\) and \(a_{2}^{2}/r_{2}^{2}\).

I suggest that the reader prove that these ratios are the same, that is, that the forces of attraction at point \(O\) acting from the two small squares balance.

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 \(H\) under the Earth. A point situated at depth \(H\) only experiences the attraction exerted by the internal layers of the Earth. The formula for the acceleration due to gravity, \(g = GM/R^{2}\) , also applies to this case, where \(M\) and \(R\) are the mass and radius not of the entire Earth but of its “internal” part with respect to this point. If the Earth had the same density in all its layers, the formula for \(g\) would assume the following form: \[g = G \, \dfrac{\dfrac{4}{3} \pi \rho \, (R - H)^{3}}{(R - H)^{2}} = \dfrac{4}{3} \, \pi \, G \rho \, (R - H)\] where \(\rho\) is the density, and \(R\) is the Earth’s radius.

This implies that \(g\) would be directly proportional to \((R - H)\): the greater the depth \(H\), the smaller would be \(g\).

But as a matter of fact, the behaviour of \(g\) near the Earth’s surface—we are able to observe it up to a depth of \(5\,\mathrm{km}\) (below sea level)—does not obey this law at all. Experiments show that \(g\), on the contrary, increases with depth within these layers. The lack of agreement between the experiments and our formula is explained by the fact that the difference in density at various depths was not taken into account.

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 \(g\) increases.