Gravity Is Strongest Right Where You Stand, Not Deep Underground

We usually take $g=9.8$ m/s${}^2$ as fixed. But climb far above the ground or go deep below it, and **$g$ changes** — in different ways in each direction.

All of it follows from Newton's law of gravitation and one idea about which part of the Earth actually pulls on you.

This part covers the formula for $g$, its variation with height, its variation with depth, and how the two compare. Throughout, take $R=6400$ km and $g=9.8$ m/s${}^2$ at the surface.

**Equating the weight $mg$ of a body at the surface with the gravitational force $\frac{GMm}{R^2}$ gives $g=\frac{GM}{R^2}$, which does not depend on the body's own mass.**

Writing $M=\frac{4}{3}\pi R^3\rho$ also gives $g=\frac{4}{3}\pi G\rho R$.

Worked example 1 — checking the value.

$$g=\frac{6.67\times 10^{-11}\times 6\times 10^{24}}{(6.4\times 10^6{)}^2}\approx 9.77{\text{ m/s}}^2$$

Worked example 2 — Earth's mean density.

$$\rho =\frac{3g}{4\pi GR}=\frac{3\times 9.8}{4\pi \times 6.67\times 10^{-11}\times 6.4\times 10^6}\approx 5.5\times 10^3{\text{ kg/m}}^3$$

Worked example 3 — another planet. A planet with twice Earth's mass and twice its radius has

$$g^{\prime }=\frac{G(2M)}{(2R{)}^2}=\frac{g}{2}=4.9{\text{ m/s}}^2$$

An everyday example. A bathroom weighing scale measures the force $mg$; a $60$ kg person presses on it with about $588$ N.

The substance. **A heavy stone and a light pebble fall with the same $g$**, because $m$ cancels out.

**At height $h$, the distance from Earth's centre is $R+h$, so $g_h=g{\left(\frac{R}{R+h}\right)}^2$; for heights much smaller than $R$ this is approximately $g_h\approx g\left(1-\frac{2h}{R}\right)$.

Derivation of the approximation.**

$$g_h=g{\left(1+\frac{h}{R}\right)}^{-2}\approx g\left(1-\frac{2h}{R}\right)\text{for }h\ll R$$

**Worked example 1 — at $h=R$.** $g_h=g{\left(\frac{1}{2}\right)}^2=\frac{9.8}{4}=2.45$ m/s${}^2$.

Worked example 2 — a small height. At $h=32$ km:

$$g_h\approx 9.8\left(1-\frac{64}{6400}\right)=9.8\times 0.99\approx 9.70{\text{ m/s}}^2$$

Worked example 3 — finding the height. Where is $g$ only $64\%$ of its surface value?

$${\left(\frac{R}{R+h}\right)}^2=0.64\Rightarrow \frac{R}{R+h}=0.8\Rightarrow h=\frac{R}{4}=1600\text{ km}$$

Worked example 4 — a low orbit. At $400$ km up, $g_h=9.8{\left(\frac{6400}{6800}\right)}^2\approx 8.7$ m/s${}^2$.

An everyday example. A climber on a very high mountain peak weighs very slightly less than at the base camp far below.

The substance. Astronauts in a low orbit are not beyond gravity — $g$ there is still about $8.7$ m/s${}^2$; they feel weightless because they are falling freely.

**At depth $d$, only the inner sphere of radius $R-d$ pulls on a body, because the outer shell exerts no net force inside it; for uniform density this gives $g_d=g\left(1-\frac{d}{R}\right)$.

Derivation.** The inner sphere's mass is $M\frac{(R-d{)}^3}{R^3}$, so

$$g_d=\frac{G}{(R-d{)}^2}\cdot \frac{M(R-d{)}^3}{R^3}=\frac{GM}{R^2}\cdot \frac{R-d}{R}=g\left(1-\frac{d}{R}\right)$$

Worked example 1 — halfway down. At $d=\frac{R}{2}$, $g_d=\frac{g}{2}=4.9$ m/s${}^2$.

Worked example 2 — a deep mine. At $d=3.2$ km:

$$g_d=9.8\left(1-\frac{3.2}{6400}\right)=9.8\times 0.9995\approx 9.795{\text{ m/s}}^2$$

Worked example 3 — finding the depth. Where is $g$ $1\%$ less than at the surface?

$$1-\frac{d}{R}=0.99\Rightarrow d=0.01R=64\text{ km}$$

Worked example 4 — the centre. At $d=R$, $g_d=0$.

An everyday example. A miner working far below ground experiences a very slightly smaller $g$ than someone at the pithead.

The substance. This formula assumes uniform density — the real Earth is denser towards the centre, so $g$ actually changes differently at first.

**$g$ is greatest at the surface; it falls with height as $\frac{1}{(R+h{)}^2}$ and approaches zero only very far away, while it falls linearly with depth and becomes exactly zero at the Earth's centre.

- Above the surface**: $g\propto \frac{1}{r^2}$ — never zero at any finite height
- Below the surface: $g\propto r$ — straight-line decrease to zero at $r=0$
- Small distances: going up by $x$ reduces $g$ by $\frac{2x}{R}$ of its value; going down by $x$ reduces it by $\frac{x}{R}$

Worked example 1 — same distance. At $32$ km up, $g\approx 9.70$ m/s${}^2$; at $32$ km down,

$$g_d=9.8\left(1-\frac{32}{6400}\right)\approx 9.75{\text{ m/s}}^2$$

**Height reduces $g$ about twice as fast as depth, for small distances.

Worked example 2 — matching values.** At depth $\frac{R}{2}$, $g=\frac{g_0}{2}$. The height with the same $g$ satisfies

$${\left(\frac{R}{R+h}\right)}^2=\frac{1}{2}\Rightarrow h=(\sqrt{2}-1)R\approx 0.414R\approx 2650\text{ km}$$

An everyday example. Going up a mountain or down a mine both reduce $g$ slightly, but for the same small distance, the mountain wins.

The substance. **At the centre $g$ is zero, but that does not mean gravitational potential is zero** — that idea comes in the next part.