How g Changes

Climb a mountain or descend a mine and you weigh a little less either way.

11 minUpdated 21 October 2026acceleration due to gravityheightdepthlatitudemass of the Earth

Lesson at a glanceShow the key points

At height h: g' = g/(1 + h/R)² ≈ g(1 − 2h/R) for h ≪ R.
At depth d: g' = g(1 − d/R), zero at the centre.
g is largest at the surface and falls off both above and below it.
Earth's rotation reduces apparent g most at the equator: g' = g − ω²R there, unchanged at the poles.
The Earth is flattened (equatorial radius ≈ 21 km larger) and uneven in density, which also changes local g.
Weighing the Earth: M = gR²/G ≈ 6 × 10²⁴ kg.

We usually treat g as a constant 9.8 m/s². It isn’t. It changes with height, depth, latitude and even what lies under your feet.

With height

At a height h above the surface of the Earth (mass M, radius R):

g' = \frac{GM}{(R+h)^2} = \frac{g}{(1 + h/R)^2}

where $g = GM/R^2$ is the surface value. For h much smaller than R, the binomial approximation gives

Small heights

g' \approx g\left(1 - \frac{2h}{R}\right)

With depth

In a mine a depth d below the surface you are inside the Earth. Treating the Earth as uniform, only the ball beneath you pulls (the shell above gives zero field), so from the solid-sphere result:

At depth d (uniform Earth)

g' = \frac{GM(R - d)}{R^3} = g\left(1 - \frac dR\right)

exactly linear, falling to zero at the centre

Worked example

Up a balloon, down a mine

Taking R = 6400 km and g = 9.80 m/s² at the surface, find g (a) 32 km up, (b) 16 km down.

Interactive

Climb or dig: g falls either way

Treat the Earth as a uniform sphere. Above the surface g falls as 1/r²; below it, only the mass deeper than you pulls, so g falls in proportion to r. For heights small compared with the radius, g(1 − 2h/R) is a good shortcut. Separately, the Earth's spin trims the apparent g most at the equator.

height400km

g (uniform sphere)8.681m/s²

shortcut g(1 − 2h/R)8.575m/s²

apparent g at latitude 0° (sea level)9.766m/s²

Height (+) or depth (−)

h

400km

Latitude

\lambda

0°

Other effects

The Earth spins

In the rotating frame of the Earth, a body at colatitude θ (angle from the north pole) also feels a centrifugal force $m\omega^2R\sin\theta$ outward from the axis. At the poles this vanishes; at the equator it points straight up, opposing gravity:

g'_{\text{equator}} = g - \omega^2 R \approx g - 0.034\ \text{m/s}^2

We met this in the chapter on circular motion: apparent g is least at the equator, greatest at the poles.

The Earth isn’t a sphere

The spin has also flattened the Earth: the equatorial radius is about 21 km larger than the polar radius. Being further from the centre at the equator lowers g there too, reinforcing the rotation effect.

The Earth isn’t uniform

Mountains, ocean basins, dense ore bodies and oil-bearing rock all nudge g locally. Geologists map these tiny variations (gravimetry) to find what lies underground.

Weighing the Earth

“Weight of the Earth” is meaningless (weight is the Earth’s pull on something), but its mass follows from g:

M = \frac{gR^2}{G} = \frac{(9.8)(6.4\times10^6)^2}{6.67\times10^{-11}} \approx 6.0\times10^{24}\ \text{kg}

What to remember

1Height: $g' = g/(1+h/R)^2 \approx g(1 - 2h/R)$ .
2Depth (uniform Earth): $g' = g(1 - d/R)$ , zero at the centre.
3Spin and flattening make apparent g smallest at the equator, largest at the poles.
4Local density variations change g slightly; M(Earth) = gR²/G.

Check yourself

At what height does g fall to a quarter of its surface value?

Where would a spring balance show the greatest reading for the same object?

Keep exploring

Hand-picked videos, simulations and readings that explain this idea from a different angle.

WatchAcceleration due to gravity at the space stationKhan Academy ReadThe value of gThe Physics Classroom ReadGravitation near Earth's surfaceOpenStax University Physics §13.2 ReadGravity of EarthWikipedia

PreviousThe Gravitational Field Next Satellites, Orbits & Kepler's Laws

How g Changes

Climb a mountain or descend a mine and you weigh a little less either way.

11 minUpdated 21 October 2026acceleration due to gravityheightdepthlatitudemass of the Earth

Lesson at a glanceShow the key points

At height h: g' = g/(1 + h/R)² ≈ g(1 − 2h/R) for h ≪ R.
At depth d: g' = g(1 − d/R), zero at the centre.
g is largest at the surface and falls off both above and below it.
Earth's rotation reduces apparent g most at the equator: g' = g − ω²R there, unchanged at the poles.
The Earth is flattened (equatorial radius ≈ 21 km larger) and uneven in density, which also changes local g.
Weighing the Earth: M = gR²/G ≈ 6 × 10²⁴ kg.

We usually treat g as a constant 9.8 m/s². It isn’t. It changes with height, depth, latitude and even what lies under your feet.

With height

At a height h above the surface of the Earth (mass M, radius R):

g' = \frac{GM}{(R+h)^2} = \frac{g}{(1 + h/R)^2}

where $g = GM/R^2$ is the surface value. For h much smaller than R, the binomial approximation gives

Small heights

g' \approx g\left(1 - \frac{2h}{R}\right)

With depth

In a mine a depth d below the surface you are inside the Earth. Treating the Earth as uniform, only the ball beneath you pulls (the shell above gives zero field), so from the solid-sphere result:

At depth d (uniform Earth)

g' = \frac{GM(R - d)}{R^3} = g\left(1 - \frac dR\right)

exactly linear, falling to zero at the centre

Worked example

Up a balloon, down a mine

Taking R = 6400 km and g = 9.80 m/s² at the surface, find g (a) 32 km up, (b) 16 km down.

Interactive

Climb or dig: g falls either way

height400km

g (uniform sphere)8.681m/s²

shortcut g(1 − 2h/R)8.575m/s²

apparent g at latitude 0° (sea level)9.766m/s²

Height (+) or depth (−)

h

400km

Latitude

\lambda

0°

Other effects

The Earth spins

g'_{\text{equator}} = g - \omega^2 R \approx g - 0.034\ \text{m/s}^2

We met this in the chapter on circular motion: apparent g is least at the equator, greatest at the poles.

The Earth isn’t a sphere

The Earth isn’t uniform

Mountains, ocean basins, dense ore bodies and oil-bearing rock all nudge g locally. Geologists map these tiny variations (gravimetry) to find what lies underground.

Weighing the Earth

“Weight of the Earth” is meaningless (weight is the Earth’s pull on something), but its mass follows from g:

M = \frac{gR^2}{G} = \frac{(9.8)(6.4\times10^6)^2}{6.67\times10^{-11}} \approx 6.0\times10^{24}\ \text{kg}

What to remember

1Height: $g' = g/(1+h/R)^2 \approx g(1 - 2h/R)$ .
2Depth (uniform Earth): $g' = g(1 - d/R)$ , zero at the centre.
3Spin and flattening make apparent g smallest at the equator, largest at the poles.
4Local density variations change g slightly; M(Earth) = gR²/G.

Check yourself

At what height does g fall to a quarter of its surface value?

Where would a spring balance show the greatest reading for the same object?

Keep exploring

Hand-picked videos, simulations and readings that explain this idea from a different angle.

PreviousThe Gravitational Field Next Satellites, Orbits & Kepler's Laws