JEE Mains Physics · Gravitation

Acceleration due to Gravity: Surface and Height

On the surface g = GM/R² = (4/3)πGρR; at height h it falls as g/(1 + h/R)², which is close to g(1 − 2h/R) only when h is small.

All Gravitation notes JEE Mains Physics strategy

Why this matters

Sixteen PYQs, fifteen of them multiple choice, and one from 2026. Six compare g on planets of different mass, radius or density; ten find g or a weight at some height above the surface. All of them are ratio questions: write g as a fraction of its surface value and the arithmetic is one line.

Concept 1 of 2: g on the surface: mass, radius and density

Surface gravity depends on how much mass there is and how far the surface is from the centre. If you are told the mass, use

GM/R^{2}

. If you are told the density, replace M by

\tfrac{4}{3}\pi R^{3}\rho

and g becomes proportional to

\rho R

. Which form to use depends only on which quantity the question holds fixed.

Definition

$g = \dfrac{GM}{R^{2}} = \dfrac{4}{3}\pi G\rho R$ .
Same mass: $g \propto \dfrac{1}{R^{2}}$ . Halving the diameter (or the radius) makes g four times larger.
Same density: $g \propto R$ , and since $M \propto R^{3}$ , $g \propto M^{1/3}$ .
Two planets in general: $\dfrac{g_1}{g_2} = \dfrac{\rho_1R_1}{\rho_2R_2}$ .
A small change with mass fixed: $\dfrac{\Delta g}{g} = -2\dfrac{\Delta R}{R}$ . The radius shrinking by 1% raises g by 2%.
Weight mg changes from planet to planet; mass does not.

Surface gravity

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

Worked example

A planet has three times the earth's radius and two thirds of the earth's average density. Find g on its surface. Take g on earth as 9.8 m/s².

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 13 April 2023 · Q102Moderate

Example 1 · Gravitation · Acceleration due to Gravity: Surface and Height

Two planets

A

and

B

of radii

R

and

1.5R

have densities

\rho

and

\rho/2

respectively. The ratio of acceleration due to gravity at the surface of

B

A

is:

Same density is not same mass

With density fixed, a bigger planet has MORE gravity (g ∝ R). The 1/R² rule holds only when the mass is fixed. Read which quantity the question keeps constant.

A diameter changes in the same ratio as the radius

'The diameter is reduced to one third' means R becomes R/3, so with the same mass g becomes 9g. Do not halve the ratio because the word was diameter.

Drill 5 more on g on the surface: mass, radius and density

Concept 2 of 2: g at a height above the surface

Above the surface the whole earth still acts as a point mass at its centre, so g falls as the inverse square of the distance from the CENTRE. Write that distance as

R + h

and compare it with R. Only when h is a small fraction of R may you use the straight-line approximation.

Definition

$g_h = \dfrac{GM}{(R + h)^{2}} = \dfrac{g}{(1 + h/R)^{2}}$ .
Useful values: $h = R/4 \to \tfrac{16}{25}g$ ; $h = R/2 \to \tfrac{4}{9}g$ ; $h = R \to \tfrac{1}{4}g$ ; $h = 2R \to \tfrac{1}{9}g$ ; $h = 9R \to \tfrac{1}{100}g$ .
To make g fall to $g/n$ : $1 + h/R = \sqrt{n}$ , so $h = (\sqrt{n} - 1)R$ .
Small heights, $h \ll R$ : $g_h \approx g\left(1 - \dfrac{2h}{R}\right)$ ; the percentage fall is $\dfrac{2h}{R} \times 100$ .
A distance 'from the centre' is r; a height 'above the surface' is $h = r - R$ .

g at height h

g_h = \frac{g}{(1 + h/R)^{2}} \approx g\left(1 - \frac{2h}{R}\right)\ \ (h \ll R)

Worked example

A body weighs 72 N on the earth's surface. What does it weigh at a height equal to half the earth's radius?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 4 Apr 2026 Shift 2 · Q5Moderate

Example 2 · Gravitation · Acceleration due to Gravity: Surface and Height

The height in terms of radius of the earth

(R)

, at which the acceleration due to gravity becomes

\frac{g}{9}

, where

g

is acceleration due to gravity on earth's surface, is

\_\_\_\_

From the centre or from the surface?

A body '2R from the surface' is 3R from the centre, so g is g/9. A body '2R from the centre' is at height R, so g is g/4. Read the phrase before you write r.

A height of one diameter is 2R

A point whose height equals the earth's diameter is 3R from the centre, so g there is g/9, not g/4.

The 2h/R rule fails for large heights

At h = R/2 the approximation g(1 − 2h/R) gives zero, which is absurd. Use g/(1 + h/R)² unless h is a few percent of R.

Drill 9 more on g at a height above the surface

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

g on the surface: mass, radius and density
Surface gravity
$g = \frac{GM}{R^{2}} = \frac{4}{3}\pi G\rho R$
g at a height above the surface
g at height h
$g_h = \frac{g}{(1 + h/R)^{2}} \approx g\left(1 - \frac{2h}{R}\right)\ \ (h \ll R)$

Watch out for (5)

Same density is not same mass
→ g on the surface: mass, radius and density
A diameter changes in the same ratio as the radius
→ g on the surface: mass, radius and density
From the centre or from the surface?
→ g at a height above the surface
A height of one diameter is 2R
→ g at a height above the surface
The 2h/R rule fails for large heights
→ g at a height above the surface

Test yourself on Gravitation

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Gravitation chapter test Drill all 16 questions