Acceleration due to gravity (g)

Drop a pebble and a brick together from the same height, and — if you could remove air resistance — they would hit the ground at exactly the same instant. That surprising fact is the doorway into one of the most-used numbers in school physics: the acceleration due to gravity, $g$ . Every object near the Earth's surface speeds up by about $9.8$ metres per second, every single second it falls, regardless of how heavy it is. This article derives $g = GM/R^2$ from Newton's universal law of gravitation, works through the standard numerical calculation that gives $g \approx 9.8\ \text{m/s}^2$ , and shows why $g$ never depends on the mass of the object doing the falling — a fact students routinely mix up with the constant $G$ .

The core insight

The gravitational force $F = GMm/R^2$ produces an acceleration $a = F/m = GM/R^2$ in which the falling object's own mass $m$ cancels out, so every object near Earth's surface accelerates downward at the same rate, $g \approx 9.8\ \text{m/s}^2$ .

The core idea

Consider an object of mass $m$ resting on or falling near the Earth's surface, where the Earth (mass $M$ , radius $R$ ) is the attracting body. Newton's universal law of gravitation gives the force between them:

F = \frac{GMm}{R^2}

where: $F$ is the gravitational force (N), $G = 6.674\times10^{-11}\ \text{N}\,\cdot\,\text{m}^2/\text{kg}^2$ is the universal gravitational constant, $M$ is Earth's mass, $m$ is the falling object's mass, and $R$ is Earth's radius, used as the distance between centres because the object sits close to the surface compared with $R$ . By Newton's second law, this force produces an acceleration $a = F/m$ in the object. Substituting the expression for $F$ and cancelling $m$ :

a = \frac{GMm}{R^2 m} = \frac{GM}{R^2}

The falling object's own mass has vanished from the result entirely — this is the single most important feature of gravitational acceleration. This acceleration is given its own symbol, $g$ , the acceleration due to gravity:

g = \frac{GM}{R^2}

Because $g$ depends only on the mass and radius of the attracting body (Earth), and both are essentially constant near the surface, $g$ is treated as a fixed value everywhere in this chapter — always about $9.8\ \text{m/s}^2$ , always directed vertically downward, toward the Earth's centre. This is why a stone and a boulder released together fall side by side: the boulder feels a bigger force, but it also has more mass to accelerate, and the two effects cancel exactly. $g$ is easy to confuse with $G$ , but the two play very different roles. $G$ is the universal gravitational constant — a fixed number, the same everywhere in the universe, that appears in the force law between any two masses. $g$ is the acceleration due to gravity at a particular location, and it depends on the mass and radius of whichever body is doing the attracting — so $g$ on the Moon (smaller mass, smaller radius) works out different from $g$ on Earth, even though $G$ is identical in both calculations.

Figure

Earth of radius R with mass M at its centre, an object of mass m at the surface pulled by force F toward the centre, and the resulting acceleration g shown as a downward arrow of the same length regardless of the object dropped

Derivation ADeriving g = GM/R^2

The goal is to show, starting only from the universal law of gravitation and Newton's second law, that the acceleration of any object falling near Earth's surface reduces to a single expression, $g = GM/R^2$ , that contains no trace of the falling object's own mass and therefore explains why light and heavy objects fall together.

Step 1. Write the gravitational force between the Earth and the falling object using the universal law of gravitation. With Earth's mass $M$ , the object's mass $m$ , and Earth's radius $R$ as the separation, $F = GMm/R^2$ . This force acts along the line joining the two centres, so it points straight down, toward the Earth's centre.

Step 2. Apply Newton's second law, $F = ma$ , to the same object. Since the same force $F$ from Step 1 is what causes the object to accelerate, we can equate the two expressions for $F$ : $ma = GMm/R^2$ .

Step 3. Divide both sides by the object's mass $m$ to isolate the acceleration. $a = \dfrac{GMm}{R^2 m} = \dfrac{GM}{R^2}$ . The $m$ on the right cancels completely, leaving an acceleration that depends only on $G$ , $M$ and $R$ — quantities that describe the Earth, not the falling object.

Step 4. Name the result. This acceleration is $g$ , the acceleration due to gravity: $g = GM/R^2$ . Because $M$ and $R$ are (to a very good approximation) fixed for a given planet, $g$ is a constant for that planet, and it is the same for every object dropped near its surface, however large or small that object's mass.

Worked example

Using $G = 6.674\times10^{-11}\ \text{N}\,\cdot\,\text{m}^2/\text{kg}^2$ , Earth's mass $M = 6\times10^{24}$ kg, and Earth's radius $R = 6.4\times10^6$ m, calculate the value of $g$ at the Earth's surface.

Step 1. Write down the formula and substitute the given values: $g = \dfrac{GM}{R^2} = \dfrac{6.674\times10^{-11} \times 6\times10^{24}}{(6.4\times10^6)^2}$ .

Step 2. Work out the numerator: $6.674\times10^{-11} \times 6\times10^{24} = 4.0044\times10^{14}$ .

Step 3. Work out the denominator: $(6.4\times10^6)^2 = 4.096\times10^{13}$ .

Step 4. Divide to get $g$ : $g = \dfrac{4.0044\times10^{14}}{4.096\times10^{13}} = \mathbf{9.78\ m/s^2}$ , which matches closely with the standard value $g \approx 9.8\ \text{m/s}^2$ used throughout this chapter.

Competitive stretch

A stone of mass 2 kg and a ball of mass 8 kg are both dropped from rest from the same height near the Earth's surface. Compare their accelerations, and find the gravitational force (weight) acting on each, taking $g = 9.8\ \text{m/s}^2$ .

Step 1. Since $g$ is independent of the mass of the falling object, both the stone and the ball have the same acceleration, $g = 9.8\ \text{m/s}^2$ , directed downward.

Step 2. Find the force on each using $F = mg$ : for the 2 kg stone, $F = 2 \times 9.8 = 19.6$ N; for the 8 kg ball, $F = 8 \times 9.8 = 78.4$ N. So although the heavier ball experiences four times the force (78.4 N versus 19.6 N), matching its four-times-larger mass, both objects accelerate at the identical 9.8 m/s^2.

Common traps

Confusing $G$ and $g$ . $G$ is the universal gravitational constant ( $6.674\times10^{-11}\ \text{N}\,\cdot\,\text{m}^2/\text{kg}^2$ , the same everywhere); $g$ is the acceleration due to gravity at a specific place ( $\approx 9.8\ \text{m/s}^2$ on Earth's surface, different on the Moon or other planets).
Assuming $g$ depends on the mass of the falling object. It does not — $g = GM/R^2$ depends only on the mass and radius of the attracting body (Earth), never on the object being dropped.
Using a precise decimal like $9.81$ without being told to. CBSE Class 9 problems use $g = 9.8\ \text{m/s}^2$ unless a question specifies a different value.
Forgetting the direction of $g$ . Acceleration due to gravity always points toward the centre of the Earth, which matters when applying sign conventions for objects thrown upward.

Ask the tutor: Want to practice more numericals calculating $g$ for different planets, or see how $g$ feeds into the equations of motion for objects thrown up or dropped? Ask the AI tutor for guided practice.

Why don't heavier objects fall faster if gravity pulls on them harder?

Gravity does pull harder on a heavier object — the force $F = GMm/R^2$ genuinely grows with the object's mass $m$ . But a heavier object also needs a proportionally bigger force to give it the same acceleration, by Newton's second law $F = ma$ . When you divide the bigger force by the bigger mass to get the acceleration, the mass cancels out exactly, leaving $a = GM/R^2$ for every object. That is why a coin and a hammer, dropped together with no air resistance, land at the same instant.

Is g exactly 9.8 m/s squared everywhere on Earth?

Not exactly — $g = GM/R^2$ depends on $R$ , the distance to Earth's centre, and Earth is not a perfect sphere, so $g$ varies slightly with latitude and altitude, and it changes further with height or depth below the surface. For CBSE Class 9 problems, though, this variation is out of scope; you are expected to treat $g \approx 9.8\ \text{m/s}^2$ as a constant everywhere near the surface unless a question explicitly states otherwise.

How is g different on the Moon compared to Earth?

The formula $g = GM/R^2$ still applies, but $M$ and $R$ change to the Moon's mass and radius, both of which are much smaller than Earth's. Working through the numbers gives a lunar $g$ of roughly one-sixth of Earth's value, which is why astronauts on the Moon can jump so much higher for the same effort. $G$ itself never changes — only $M$ and $R$ do, because they describe the attracting body, not the universal constant.

Acceleration due to gravity (g)

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads