Why does the acceleration due to gravity not depend on the mass of the falling body?

The gravitational force on a body is F = GMm/R², proportional to its own mass m. By Newton's second law the acceleration is g = F/m = GM/R², in which m cancels out. Because the larger pull on a heavier body is exactly offset by its larger inertia, every body accelerates at the same g. In a vacuum a feather and a stone fall together — a fact Galileo demonstrated and astronauts have confirmed on the airless Moon.

How is the mass of the Earth estimated from g?

Rearranging g = GM/R² gives M = gR²/G. Acceleration g is readily measured, the radius R is known, and G comes from Cavendish's experiment. Substituting g = 9.81 m s⁻², R = 6.37 × 10⁶ m and G = 6.67 × 10⁻¹¹ N m² kg⁻² gives M ≈ 5.97 × 10²⁴ kg. This is the basis of the popular statement that 'Cavendish weighed the Earth'.

What is the difference between weight and mass?

Mass m is the amount of matter in a body — an intrinsic property that stays constant wherever the body is taken. Weight W = mg is the gravitational force pulling the body toward the Earth, measured in newtons. Because g varies from place to place, weight changes with location while mass does not. Your weight on the Moon is about one-sixth of your weight on Earth, but your mass is unchanged.

How does the surface value of g relate to Earth's mean density?

Writing the Earth's mass as M = (4/3)πR³ρ for a sphere of uniform density ρ and substituting into g = GM/R² gives g = (4/3)πGρR. So at the surface g is proportional to both the radius and the mean density of the body. A larger or denser planet has a larger surface g.

Why is g slightly larger at the poles than at the equator?

The Earth is not a perfect sphere; its radius is smaller at the poles (about 6357 km) than at the equator (about 6378 km). Since g = GM/R² decreases with larger R, g is greater at the poles and smaller at the equator. Weight, being mg, is therefore maximum at the poles and minimum at the equator. This is a surface-value effect; variation with altitude and depth is treated separately.

Does a spring balance measure mass or weight?

A spring balance measures weight — the force W = mg stretching its spring. Manufacturers calibrate its scale in kilograms for everyday convenience by dividing by the standard g, but the underlying quantity is force in newtons. Take the same spring balance to the Moon and its reading falls to about one-sixth, even though the body's mass is unchanged. A two-pan beam balance, by contrast, compares masses and reads the same everywhere.

Acceleration Due to Gravity (Surface) — NEET Physics

Q: What is the difference between g and G?

G is the universal gravitational constant — the same everywhere in the universe, with value 6.67 × 10⁻¹¹ N m² kg⁻² and dimensions [M⁻¹L³T⁻²]. g is the acceleration due to gravity, a local quantity that changes from place to place, with value about 9.8 m s⁻² on Earth's surface and dimensions [LT⁻²]. They are linked at the surface by g = GM/R², but they are different physical quantities.

Deriving g = GM/R² at the surface

The Earth can be imagined as a sphere built of a large number of concentric spherical shells, the smallest at the centre and the largest at the surface. A point on the surface lies outside every one of these shells, so each shell attracts a body there exactly as if the shell's mass were concentrated at the common centre. Summing over all the shells, the entire mass of the Earth behaves, for a body on the surface, as if it were a single point mass $M$ at the centre.

Take a body of mass $m$ resting on the surface, at distance $R$ (the Earth's radius) from the centre. The gravitational force on it follows from the universal law of gravitation:

$$F = \frac{G\,M\,m}{R^{2}}$$

By Newton's second law this force produces an acceleration $g$ given by $F = mg$. Equating the two expressions for $F$ and cancelling the body's mass $m$:

$$g = \frac{F}{m} = \frac{G M}{R^{2}}$$

At the surface the whole Earth acts as a point mass M at its centre, distance R away. The body's own mass m cancels, so every body falls with the same g.

The crucial feature of this result is that the body's own mass $m$ has cancelled. The acceleration due to gravity depends only on $G$, the Earth's mass $M$ and its radius $R$ — not on the body being dropped. A heavy ball and a light ball, released together in the absence of air, reach the ground at the same instant. This is precisely what Galileo demonstrated, and what astronauts later confirmed on the airless Moon, where a feather and a stone fall side by side.

The typical value g ≈ 9.8 m s⁻²

Substituting the Earth's data into $g = GM/R^{2}$ returns the familiar number. With $M = 6.0\times10^{24}\ \text{kg}$, mean radius $R = 6.4\times10^{6}\ \text{m}$ and $G = 6.67\times10^{-11}\ \text{N m}^{2}\,\text{kg}^{-2}$, the surface value comes out close to the experimentally observed $g \approx 9.8\ \text{m s}^{-2}$. NCERT records that Galileo, by experiments with bodies rolling down inclined planes, arrived at a value of $g$ close to this modern figure.

Because $g$ is directed toward the centre of the Earth, that direction is exactly what we call vertical. For small heights the value of $g$ hardly changes, so a body released from rest covers $h = \tfrac{1}{2}gt^{2}$ — the standard kinematic equations $v = u + gt$, $v^{2} = u^{2} + 2gs$ all use this same constant $g$. On Earth's surface the working value is taken as $9.8\ \text{m s}^{-2}$ (sometimes $9.81\ \text{m s}^{-2}$ for precise work, or $10\ \text{m s}^{-2}$ to ease arithmetic).

The surface value is not perfectly uniform across the globe. Because the Earth is not a perfect sphere — its radius is about $6357\ \text{km}$ at the poles and $6378\ \text{km}$ at the equator — and since $g \propto 1/R^{2}$, the value of $g$ is slightly larger at the poles than at the equator. This page treats only the surface value and the relations that flow from it; how $g$ falls off as you climb above the surface or descend into a mine is developed separately in variation of g with altitude and depth.

g versus G — local vs universal

Two symbols that look almost alike stand for entirely different quantities, and NEET exploits the confusion relentlessly. $G$ is the universal gravitational constant: a fixed number, the same everywhere in the universe, with value $6.67\times10^{-11}\ \text{N m}^{2}\,\text{kg}^{-2}$. $g$ is the acceleration due to gravity: a local quantity that changes from place to place and from planet to planet. The relation $g = GM/R^{2}$ links them at a surface, but they are not interchangeable.

Property	G (constant)	g (acceleration)
Nature	Universal constant of gravitation	Acceleration produced by gravity
Value	`6.67 × 10⁻¹¹ N m² kg⁻²`	≈ `9.8 m s⁻²` on Earth's surface
Dimensions	`[M⁻¹L³T⁻²]`	`[LT⁻²]`
Varies with place?	No — same everywhere	Yes — changes with location and body
Depends on the body?	No	On the attracting body (M, R), not the falling body

Weighing the Earth: M = gR²/G

The relation $g = GM/R^{2}$ can be inverted to find the most famous unknown of all — the mass of the Earth. Solving for $M$:

$$M = \frac{g R^{2}}{G}$$

Every quantity on the right is independently knowable. The acceleration $g$ is readily measured (for instance, with a simple pendulum). The radius $R$ is a known geometric quantity. And $G$ was measured directly by Cavendish's torsion-balance experiment. With these three numbers in hand, $M$ follows. This is why the achievement is popularly summarised as "Cavendish weighed the Earth" — strictly, he measured $G$, which is the missing piece that unlocks $M$.

Prerequisite

The value of $G$ in $M = gR^{2}/G$ comes from a delicate torsion-balance measurement — see the gravitational constant for Cavendish's method and its NEET dimension questions.

g and the planet's mean density

There is a second, equally useful way to write the surface value. Treating the Earth as a sphere of uniform density $\rho$, its mass is the density times its volume:

$$M = \frac{4}{3}\pi R^{3}\rho$$

Substituting this into $g = GM/R^{2}$ and cancelling one power of $R$:

$$g = \frac{G}{R^{2}}\cdot\frac{4}{3}\pi R^{3}\rho = \frac{4}{3}\pi G \rho R$$

This form shows that the surface value of $g$ is proportional to both the radius $R$ and the mean density $\rho$ of the body. A planet that is larger or denser has a stronger surface gravity. The density form is the quickest route for comparison problems where two planets are specified by their radius and density rather than their mass.

Weight versus mass

The force with which a body is pulled toward the Earth is its weight. If $m$ is the body's mass, its weight is

$$W = mg$$

Weight is a force, so its SI unit is the newton. A person of mass $50\ \text{kg}$ has weight $50\ \text{kg}\times 9.8\ \text{m s}^{-2} = 490\ \text{N}$. Because $g$ varies from place to place, the weight of a body changes from place to place — it is greatest at the poles (smallest $R$) and least at the equator, and it decreases at higher altitudes or inside the Earth. Mass, by contrast, is an intrinsic property of the body: it measures the amount of matter and stays constant wherever the body is taken.

Mass is intrinsic and constant; weight is the force mg and tracks the local value of g. The same body weighs about one-sixth as much on the Moon, though its mass is unchanged.

This distinction explains the behaviour of weighing instruments. A spring balance responds to the stretching force $mg$, so it truly measures weight; its scale is merely calibrated in kilograms for everyday convenience. Take the same spring balance to the Moon and its reading drops to roughly one-sixth, because $g$ there is about one-sixth of Earth's. A two-pan beam balance, on the other hand, compares the unknown against standard masses, so both sides scale with $g$ equally and it reads the same mass everywhere.

Worked examples

Example 1 — g from M and R

Compute the acceleration due to gravity at the Earth's surface, given mass $M = 6.0\times10^{24}\ \text{kg}$, mean radius $R = 6.4\times10^{6}\ \text{m}$ and $G = 6.67\times10^{-11}\ \text{N m}^{2}\,\text{kg}^{-2}$.

Apply the surface relation. $g = \dfrac{GM}{R^{2}}$.

Numerator: $GM = (6.67\times10^{-11})(6.0\times10^{24}) = 4.00\times10^{14}$.

Denominator: $R^{2} = (6.4\times10^{6})^{2} = 4.10\times10^{13}$.

Divide: $g = \dfrac{4.00\times10^{14}}{4.10\times10^{13}} \approx 9.8\ \text{m s}^{-2}$, the standard surface value.

Example 2 — Weighing the Earth

(NCERT Example 7.6 style) Estimate the mass of the Earth from $g = 9.81\ \text{m s}^{-2}$, $R = 6.37\times10^{6}\ \text{m}$ and $G = 6.67\times10^{-11}\ \text{N m}^{2}\,\text{kg}^{-2}$.

Invert the surface relation. $M = \dfrac{g R^{2}}{G}$.

Substitute: $M = \dfrac{(9.81)\,(6.37\times10^{6})^{2}}{6.67\times10^{-11}}$.

Evaluate: $(6.37\times10^{6})^{2} = 4.06\times10^{13}$; numerator $= 9.81\times4.06\times10^{13} = 3.98\times10^{14}$; dividing by $6.67\times10^{-11}$ gives $M \approx 5.97\times10^{24}\ \text{kg}$ — the accepted mass of the Earth, exactly as NCERT obtains.

Example 3 — Weight on another body

A body has mass $m = 60\ \text{kg}$ and weighs $588\ \text{N}$ on Earth ($g = 9.8\ \text{m s}^{-2}$). Find its weight on the Moon, whose mass is $M_{\text{moon}} = 7.3\times10^{22}\ \text{kg}$ and radius $R_{\text{moon}} = 1.74\times10^{6}\ \text{m}$. Take $G = 6.67\times10^{-11}\ \text{N m}^{2}\,\text{kg}^{-2}$.

Step 1 — surface g on the Moon. $g_{\text{moon}} = \dfrac{G M_{\text{moon}}}{R_{\text{moon}}^{2}} = \dfrac{(6.67\times10^{-11})(7.3\times10^{22})}{(1.74\times10^{6})^{2}}$.

Numerator $= 4.87\times10^{12}$; denominator $= 3.03\times10^{12}$; so $g_{\text{moon}} \approx 1.6\ \text{m s}^{-2}$ — about one-sixth of Earth's $g$.

Step 2 — weight on the Moon. The mass is unchanged at $60\ \text{kg}$. Weight $W_{\text{moon}} = m\,g_{\text{moon}} = 60\times1.6 \approx 96\ \text{N}$, roughly one-sixth of the $588\ \text{N}$ on Earth. The body's mass stayed the same; only its weight changed because $g$ changed.

Quick recap

Surface g in one breath

Surface value: $g = GM/R^{2}$; the falling body's mass cancels, so every body falls with the same $g$.
Typical value on Earth: $g \approx 9.8\ \text{m s}^{-2}$, directed toward the centre (the vertical).
Invert to weigh the Earth: $M = gR^{2}/G$ — "Cavendish weighed the Earth" via his measurement of $G$.
Density form: $g = \tfrac{4}{3}\pi G \rho R$; same density ⇒ $g \propto R$, same mass ⇒ $g \propto 1/R^{2}$.
$g$ is local ($[LT^{-2}]$, changes with place); $G$ is universal ($[M^{-1}L^{3}T^{-2}]$, fixed everywhere).
Weight $W = mg$ (newtons) varies with $g$; mass $m$ (kilograms) is intrinsic and constant.

Acceleration Due to Gravity (Surface)