Acceleration Due to Gravity (g): Meaning and Formula

Physics · Gravitation · NEET

Acceleration due to gravity (g) is the acceleration a body gets only because of Earth's gravity when it falls freely. Its formula is g = GM/R², where M is Earth's mass and R is Earth's radius, and its value at the surface is about 9.8 m/s². Memory hook: "g comes from big G" - small g depends on the planet (M and R), but big G is the same everywhere in the universe.
EarthM, RRmass mgg = GM / R²Falling mass m cancels:mg = GMm/R² → g = GM/R²Value at surface ≈ 9.8 m/s²Direction: towards centre (down)
A free mass m near Earth accelerates downward with g = GM/R². The body's own mass cancels, so g depends only on Earth's mass M and radius R; at the surface g is about 9.8 m/s² and points toward Earth's centre.

Your doubts, answered

Is g the same for a heavy stone and a light stone?

Yes. Both fall with the same g = 9.8 m/s². Earth pulls the heavy stone with more force (F = mg), but the heavy stone also needs more force to accelerate. When you divide force by mass (a = F/m), the mass cancels out, so both get the same acceleration. This is why a coin and a feather fall together in vacuum. NEET loves this idea.

Why does g not depend on the mass of the falling body?

Start from Newton's law: the force on a body of mass m near Earth is F = GMm/R². By Newton's second law F = ma. Setting them equal: ma = GMm/R². The small m cancels on both sides, giving a = g = GM/R². Only Earth's mass M and radius R are left. So g depends on the planet, not on what is falling.

What is the difference between g and the number 9.8?

g is the physical quantity 'acceleration due to gravity'. 9.8 m/s² is just its value on Earth's surface. On the Moon g is about 1.6 m/s², on Jupiter it is larger. So 9.8 is one special value of g, only for Earth's surface. Do not treat 9.8 as a fixed constant like G.

Does g change on a different planet?

Yes. Since g = GM/R², a planet with a different mass M or radius R has a different g. A planet with more mass but the same radius has a larger g. A planet with the same mass but bigger radius has a smaller g. Big G stays the same everywhere; only g changes.

How do I get g = GM/R² step by step?

Write the gravitational force on mass m at Earth's surface: F = GMm/R². This force produces free-fall acceleration, so F = mg. Equate them: mg = GMm/R². Cancel m from both sides to get g = GM/R². This single line connects Newton's universal law to everyday free fall.

⚠️ The NEET trap
Thinking a heavier body has a larger g, so it falls faster (g depends on the falling mass).
g = GM/R² has no term for the falling body's mass. All bodies at the same place fall with the same g. Heavier only means more force (F = mg), not more acceleration.
🧠 The falling body's mass cancels out - g belongs to the planet, not to the object.

Real NEET questions

NEET 2024

The mass of a planet is 1/10th that of the earth and its diameter is half that of the earth. The acceleration due to gravity on that planet is:

A · 9.8 m/s²
B · 4.9 m/s²
C · 3.92 m/s²
D · 19.6 m/s²
Solution: Use g = GM/R². For the planet, M' = M/10 and diameter is half, so radius R' = R/2. Then g' = GM'/R'² = G(M/10)/(R/2)². The (R/2)² in the denominator gives R²/4, so dividing by it multiplies by 4: g' = (4/10)·(GM/R²) = 0.4 g. With g = 9.8 m/s², g' = 0.4 × 9.8 = 3.92 m/s². Answer: C. Key trap: use radius, not diameter, and square it.

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Frequently asked

What is the value of g on Earth's surface?

About 9.8 m/s² (often rounded to 9.81 m/s²). This is how much the speed of a freely falling body increases every second, ignoring air resistance.

What are the units of g?

g is an acceleration, so its SI unit is metre per second squared (m/s²). It can also be written as newton per kilogram (N/kg), because g = force per unit mass; both units are equal.

Is g a vector or a scalar?

g is a vector. It has magnitude (about 9.8 m/s²) and direction (pointing towards the centre of the Earth, i.e. downward).

What is the formula for acceleration due to gravity?

g = GM/R², where G is the universal gravitational constant (6.67 × 10⁻¹¹ N·m²/kg²), M is the mass of the Earth, and R is the radius of the Earth.

Why is g important for NEET?

g is the base of the whole Gravitation chapter. Variation of g with height and depth, weight of a body, escape velocity, and satellite motion all start from g = GM/R². Almost every year NEET asks a numerical that uses this formula.