Variation of g with Altitude and Depth

Variation with altitude. Consider a point at height $h$ above the Earth's surface, so that its distance from the Earth's centre is $R + h$ rather than $R$. Applying $g = GM/(\text{distance})^2$ at this new distance,

using $g = GM/R^2$ from Section 7.5. This exact form already shows that $g_h < g$ whenever $h > 0$ -- gravity is always weaker at any height above the surface than it is at the surface itself. For the common case where $h \ll R$ (true for any ordinary aircraft or even a low-altitude satellite, since the Earth's radius is about $6400\ \text{km}$), the binomial approximation $(1+x)^{-2} \approx 1 - 2x$ for small $x$ gives the simpler, widely used working formula

Variation with depth. Consider instead a point at depth $d$ below the Earth's surface, so its distance from the Earth's centre is $R - d$. Treating the Earth as a uniform sphere of density $\rho$, a key result used here (proved fully in Exercise 4) is that only the mass enclosed WITHIN radius $R - d$ contributes any net gravitational pull on a point at that depth -- the mass in the spherical shell lying above it, between radius $R-d$ and the full radius $R$, contributes exactly zero net force, its pulls from every direction cancelling out perfectly by symmetry. Since mass scales with volume for a uniform density, the enclosed mass at radius $R - d$ is $M' = M(R-d)^3/R^3$, and so

again using $g = GM/R^2$. Unlike the altitude case, this relation for depth is written here WITHOUT any small-$d$ approximation -- it holds exactly, all the way from the surface ($d=0$, giving back $g_d = g$) down to the very centre of the Earth ($d = R$, giving $g_d = 0$). This last result is a genuinely important physical fact, not simply an algebraic curiosity: at the exact centre of a uniform Earth, the gravitational pulls from every direction cancel perfectly by symmetry, so an object there would be entirely weightless, even though it sits at the very heart of the Earth's enormous mass. …