Variation of g

Stand on a mountain top and you weigh very slightly less than at sea level; descend into a deep mine and you weigh less again, for a different reason. The acceleration due to gravity, $g$ , is not a fixed $9.8\ \text{m/s}^2$ everywhere — it is the local strength of Earth's pull, and it changes with how far you are from the planet's centre and with how much of the planet's mass still lies "outside" you. Starting from the surface relation $g = GM/R^2$ , this article derives how $g$ falls off with altitude above the surface and with depth below it, and states — without derivation — why $g$ also depends on latitude because of Earth's rotation. The same handful of ideas explains why a pendulum clock runs slow on a mountain, why $g \to 0$ at Earth's centre, and why the equator is the "lightest" place to stand on the planet.

The core insight

Since $g=GM/R^2$ depends on distance from Earth's centre and on the mass enclosed within that distance, $g$ decreases (approximately linearly, $g_h\approx g(1-2h/R)$ ) with altitude $h\ll R$ , decreases linearly ( $g_d=g(1-d/R)$ ) with depth $d$ down to zero at the centre, and is minimum at the equator and maximum at the poles due to Earth's rotation.

The core idea

Altitude. Treat Earth as a uniform sphere of mass $M$ and radius $R$ , so at the surface $g = GM/R^2$ . At height $h$ above the surface, the distance from Earth's centre becomes $R+h$ , so gravity there is

g_h = \dfrac{GM}{(R+h)^2} = g\left(1+\dfrac{h}{R}\right)^{-2}

where: $g_h$ is the value of $g$ at height $h$ , and $g$ is the surface value. For $h \ll R$ , the binomial approximation $(1+x)^{-2}\approx 1-2x$ turns this into a simple linear drop,

g_h \approx g\left(1-\dfrac{2h}{R}\right)

so $g$ decreases with altitude, and — for small $h$ — nearly linearly in $h$ .

Depth. Below the surface, only the mass enclosed within radius $(R-d)$ pulls on you; a uniform spherical shell of mass lying above your depth exerts zero net gravitational force on points inside it (an exact result for a uniform shell). Taking Earth's density $\rho$ as uniform, the enclosed mass at depth $d$ is $M' = \tfrac{4}{3}\pi(R-d)^3\rho$ , so

g_d = \dfrac{GM'}{(R-d)^2} = \dfrac{4}{3}\pi G\rho(R-d) = g\left(1-\dfrac{d}{R}\right)

where: $g_d$ is gravity at depth $d$ , using $g = \tfrac{4}{3}\pi G\rho R$ at the surface. Unlike the altitude case, this decrease is exactly linear (no approximation needed), and $g_d \to 0$ as $d \to R$ , i.e. at Earth's centre.

Figure

Line graph of g against distance from Earths centre showing g rising linearly from zero at the centre to a peak at the surface then falling off as inverse square with height above the surface

Latitude (mention only, no derivation required). Earth's rotation means part of the gravitational pull at any point on the surface is used up supplying the centripetal force needed to keep that point moving in a circle about the rotation axis. This effect is largest at the equator (largest radius of rotation) and vanishes at the poles (on the axis itself), so the effective $g$ you would measure is minimum at the equator and maximum at the poles. No formula for the exact latitude dependence is expected at this level — only this qualitative trend.

Derivation Ag falls linearly with depth because only the enclosed mass pulls on you

The goal of this derivation is to show, in a few algebraic steps starting from nothing more than $g=GM/R^2$ and the volume of a sphere, exactly why gravity below the surface drops off in a straight line rather than curving the way it does above the surface, and why that line must reach zero precisely at the Earth's centre.

Step 1. Model Earth as a uniform sphere of radius $R$ and (uniform) density $\rho$ , so its total mass is $M = \tfrac{4}{3}\pi R^3\rho$ , and its surface gravity is $g = GM/R^2 = \tfrac{4}{3}\pi G\rho R$ .

Step 2. At depth $d$ below the surface, you are at distance $r = R-d$ from the centre. Only the mass inside radius $r$ contributes to gravity there, because a uniform shell of mass outside radius $r$ pulls equally from all sides and cancels out — a standard shell-theorem result assumed here without re-proving it. That enclosed mass is $M' = \tfrac{4}{3}\pi r^3\rho = \tfrac{4}{3}\pi(R-d)^3\rho$ .

Step 3. Apply $g_d = GM'/r^2$ with $r=R-d$ : $g_d = \dfrac{G\cdot\tfrac{4}{3}\pi(R-d)^3\rho}{(R-d)^2} = \tfrac{4}{3}\pi G\rho(R-d)$ . One power of $(R-d)$ survives the division, which is exactly why this expression is linear in $(R-d)$ rather than an inverse-square law.

Step 4. Compare with the surface value from Step 1. Since $g = \tfrac{4}{3}\pi G\rho R$ , dividing gives $\dfrac{g_d}{g} = \dfrac{R-d}{R} = 1-\dfrac{d}{R}$ , so $g_d = g\left(1-\dfrac{d}{R}\right)$ . At $d=0$ this correctly gives $g_d=g$ ; at $d=R$ (the centre) it gives $g_d=0$ , matching the physical fact that at the exact centre the pull from all directions cancels symmetrically.

Worked example

Find $g$ at a height of $32\ \text{km}$ above Earth's surface, and separately, the depth at which $g$ falls to half its surface value. Take $g = 9.8\ \text{m/s}^2$ and $R = 6400\ \text{km}$ .

Step 1. Since $h = 32\ \text{km} \ll R = 6400\ \text{km}$ , use the linear altitude approximation: $g_h \approx g\left(1-\dfrac{2h}{R}\right) = 9.8\left(1-\dfrac{2\times32}{6400}\right) = 9.8(1-0.01) = 9.8\times0.99$ .

Step 2. This gives $g_h = 9.702\ \text{m/s}^2$ , a drop of about $0.098\ \text{m/s}^2$ , roughly $1\%$ , at this altitude.

Step 3. Now use the depth relation $g_d = g(1-d/R)$ , setting $g_d = g/2$ : $\dfrac{1}{2} = 1-\dfrac{d}{R}$ , so $\dfrac{d}{R}=\dfrac{1}{2}$ , giving $d = R/2 = 6400/2$ .

Step 4. So $g_h = 9.702\ \text{m/s}^2$ at $h=32\ \text{km}$ , and $g$ falls to half its surface value at a depth of $d = 3200\ \text{km}$ — about halfway to the centre, confirming that the depth relation needs no factor of $2$ , unlike the altitude one.

Common traps

Using the exact inverse-square formula $g_h = GM/(R+h)^2$ when the linear approximation $g_h\approx g(1-2h/R)$ is expected — that approximation is valid only for $h\ll R$ , so check the sizes before approximating.
Mixing up the two factors: altitude carries a factor of $2$ (from expanding a squared term with the binomial approximation), depth carries no factor of $2$ — it is exactly linear in $d/R$ .
Assuming $g$ decreases the same way going down as going up — outside Earth $g$ falls off as $1/r^2$ (curved), inside Earth it falls off linearly, reaching exactly zero at the centre.
Thinking the latitude variation is mainly due to a big change in distance to Earth's centre — within this syllabus the dominant cause is the rotational (centripetal) effect, not radius change.
Extrapolating the altitude approximation $g_h\approx g(1-2h/R)$ to large $h$ comparable to $R$ — it silently breaks down there; only the exact $g_h=GM/(R+h)^2$ remains valid, and that case is not usually asked at this level.

Ask the tutor: Want to see the shell-theorem argument proved in more depth, or practice a problem that combines altitude, depth and latitude in one question?

Why does the depth formula not have the factor of 2 that the altitude formula has?

The factor of $2$ in the altitude formula comes from approximating $(1+h/R)^{-2}\approx 1-2h/R$ using the binomial expansion of a squared, inverse-square-law term — that " $2$ " is the exponent from $(1+x)^{-2}$ , valid only when $h\ll R$ . The depth formula is different in kind: it is not an approximation at all. Because only the enclosed mass $\propto (R-d)^3$ pulls on you, and gravity itself scales as $1/(R-d)^2$ , the two powers combine to leave gravity proportional to just $(R-d)^1$ — an exactly linear relation with no factor of $2$ anywhere.

Does g really become exactly zero at the centre of the Earth?

Yes, in this idealised uniform-density model. At the very centre, $d=R$ , so $g_d = g(1-d/R) = g(1-1) = 0$ . Physically this makes sense too: at the exact centre, mass pulls on you equally in every direction, and all those pulls cancel by symmetry, leaving zero net gravitational force. Real Earth is not perfectly uniform (density increases toward the core), so the true profile deviates slightly from a straight line, but it still falls to zero at the centre.

If g is smaller at the equator, does that mean I would weigh less there than at the poles?

Yes, very slightly. Since weight is $mg$ , a smaller local $g$ means a smaller measured weight for the same mass, even though the mass itself has not changed. The effect is small — a few tenths of a percent — because most of the pull toward Earth's centre is far larger than the small centripetal share diverted by rotation. This is why sensitive gravity measurements and even some sporting records account for the location's latitude, but ordinary weighing at home never needs to.

Variation of g

The core idea

Worked example

Common traps

Before this

Where this leads