Acceleration Due to Gravity Formula: g = GM/R² Explained

The acceleration due to gravity formula is $g = \frac{GM}{R^2}$ : the gravitational constant times the planet's mass, divided by the square of its radius. For Earth this gives about 9.8 m/s², the value used in most school problems (9.81 m/s² is also common). It is the acceleration of any freely falling body near the surface, whatever its mass.

The formula follows from Newton's law of gravitation. The pull on a mass m at the surface is $\frac{GMm}{R^2}$ , and that pull is also its weight, mg. Set them equal and m cancels. Students also meet this as the gravitational acceleration formula or the gravity acceleration formula; all three names mean the same g.

Quick facts

Formula
$g = \frac{GM}{R^2}$
Earth value
9.8 m/s² (9.81 m/s² also used)
SI unit
m/s², equivalently N/kg
Dimensional formula
[M⁰ L¹ T⁻²]
At height h
$g_h = \frac{g}{(1 + h/R)^2} \approx g\left(1 - \frac{2h}{R}\right)$
At depth d
$g_d = g\left(1 - \frac{d}{R}\right)$
Latitude φ
$g' = g - \omega^2 R\cos^2\phi$

Deriving the acceleration due to gravity formula

Force on a mass m at Earth's surface: $F = \frac{GMm}{R^2}$ .
That same force is the weight: $F = mg$ .
Equate: $mg = \frac{GMm}{R^2}$ .
Cancel m: $g = \frac{GM}{R^2}$ .

Because m drops out, a hammer and a feather fall with the same acceleration once air resistance is removed. The result also gives a handy identity used across the gravitation chapter: GM = gR².

Worked examples

Example 1: Earth's surface gravity

Find g at Earth's surface from G, M = 5.97 × 10²⁴ kg and R = 6.37 × 10⁶ m.

Top: $GM \approx 3.98 \times 10^{14}\ \text{N m}^2/\text{kg}$ . Bottom: $R^2 \approx 4.06 \times 10^{13}\ \text{m}^2$ .

$g = \frac{3.98 \times 10^{14}}{4.06 \times 10^{13}} \approx 9.82\ \text{m/s}^2$

Close to the measured 9.8 m/s²; the small gap comes from rounding and Earth's spin.

Example 2: gravity on Mars

Mars has mass 6.42 × 10²³ kg and radius 3390 km. Find g there.

$g = \frac{6.674 \times 10^{-11} \times 6.42 \times 10^{23}}{(3.39 \times 10^6)^2} \approx 3.73\ \text{m/s}^2$

About 38 percent of Earth's value.

Example 3: up a mountain and down a mine

Find g at a height of 32 km and at a depth of 3185 km (half Earth's radius). Start from 9.8 m/s² at the surface and R = 6370 km.

Height (h much less than R): $g_h \approx 9.8\left(1 - \frac{2 \times 32}{6370}\right) \approx 9.70\ \text{m/s}^2$ .

Depth: $g_d = 9.8\left(1 - \frac{3185}{6370}\right) = 9.8 \times 0.5 = 4.9\ \text{m/s}^2$ .

Variation of g: height, depth, latitude and shape

How g changes
Change	Formula	Trend
Height h above the surface	$g_h = g\frac{R^2}{(R + h)^2}$	Falls as the inverse square of distance from the centre
Height small next to the radius	$g_h \approx g(1 - 2h/R)$	Falls linearly
Depth d below the surface	$g_d = g(1 - d/R)$	Falls linearly, zero at the centre
Latitude φ, from Earth's spin	$g' = g - \omega^2R\cos^2\phi$	Least at the equator, greatest at the poles

At the equator the spin term is $\omega^2 R \approx 0.034\ \text{m/s}^2$ , the same outward effect described on the centrifugal force page. Earth's slight bulge at the equator lowers g there a little more, because the surface is farther from the centre.

For the same small distance, g falls twice as fast going up as going down: compare 2h/R with d/R.

Units, dimensions and g versus G

m/s² and N/kg are the same unit, because a newton is a kilogram metre per second squared. The dimensional formula of acceleration due to gravity is that of any acceleration, a length over time squared.

Do not mix up g and G. Little g is an acceleration that changes from place to place, roughly 9.8 m/s² here and 1.6 m/s² on the Moon. Big G is the constant in Newton's law of gravitation, fixed at 6.674 × 10⁻¹¹ in SI units on every planet and in every lab.

Mistakes that cost marks

Using the radius in km. R must be in metres before squaring.
Using the height h in place of the distance R + h in the exact formula.
Applying the small height formula 1 − 2h/R when h is comparable to R. Use the exact form instead.
Writing g = GM/R without the square.
Saying g is zero in orbit. At 400 km up it is still about 8.7 m/s²; astronauts float because they fall freely with their craft.

Escape and orbit problems both start from this g; see the escape velocity page.

Frequently asked questions

Is g a vector quantity?

Yes. Near Earth its size is about 9.8 m/s², and its direction is what we call down, along the line to Earth's centre.

What is the formula for acceleration due to gravity?

Divide G times the planet's mass by the square of its radius: g = GM / R². Earth's numbers give close to 9.8 m/s².

What is the value of g on the Moon?

About 1.6 m/s², roughly one sixth of Earth's value, because the Moon's mass is much smaller relative to its radius squared.

Does acceleration due to gravity depend on the mass of the falling body?

No. The body's own mass cancels in the derivation, so all bodies fall with the same g in a vacuum.

What is the difference between g and G?

g is the local free fall acceleration in m/s² and varies with place. Capital G never changes: it is 6.674 × 10⁻¹¹ in SI units wherever you are.