(B) Variation in g with Depth

The Earth can be modelled as built up from many thin, uniform, concentric spherical shells, one inside another, with the total mass M being the sum of all their individual masses. At the surface itself, by the shell theorem (section 5.3), the entire mass acts as though concentrated at the centre, giving the familiar $g=\dfrac{GM}{R^2}$; assuming the Earth has uniform density $\rho=\dfrac{M}{V}=\dfrac{M}{\frac{4}{3}\pi R^3}$, this can be rewritten as $g=\dfrac{4}{3}\pi R\rho G.$

Now consider a point P at depth d below the surface (Fig. 5.7), so that its distance from the centre is $(R-d)$. The Earth at this depth naturally splits into two parts: an inner solid sphere of radius $(R-d)$, and an outer spherical SHELL of thickness d surrounding it. By the shell theorem, the outer shell -- since P lies INSIDE it -- contributes exactly ZERO net force at P; only the mass $M'$ of the inner sphere of radius $(R-d)$ matters, and (again by the shell theorem, now applied to this smaller sphere as seen from outside it) that mass acts as though concentrated at the centre. With uniform density, $M'=\dfrac{4}{3}\pi(R-d)^3\rho$, so the acceleration due to gravity at depth d is $g_d=\dfrac{GM'}{(R-d)^2}=\dfrac{4}{3}\pi(R-d)\rho G.$

Dividing this by the surface expression $g=\frac{4}{3}\pi R\rho G$ cancels the common factors $\frac{4}{3}\pi\rho G$, giving the very clean linear result $\dfrac{g_d}{g}=\dfrac{R-d}{R}\quad\Longrightarrow\quad g_d=g\left(1-\dfrac{d}{R}\right).$ Unlike the altitude case (which falls off as an inverse square, only approximately linear for small h), this depth relation is EXACTLY linear in d for a uniform-density Earth, all the way down to the very centre. At $d=R$ (the Earth's centre), $g_d=0$ exactly -- a body at the very centre of the Earth experiences no net gravitational pull at all, by the same shell-cancellation symmetry that makes the force zero anywhere inside a hollow shell. …