NDA Physics · Gravitation

Gravitational Field and Potential

The acceleration due to gravity at a planet's surface is g = GM/R² (equivalently g = (4/3)πGρR); gravitational potential measures energy per unit mass, and a region of equal potential does no work on a body moved through it.

All Gravitation notes NDA Physics strategy

Why this matters

This is the chapter's busiest subtopic — seven PYQs, including its hardest item. The recurring engine is g = GM/R² and its density form g = (4/3)πGρR: scale a planet's mass, radius or density and read off the new g. The bank also tests the deeper ideas — that equal potential means zero work, that weightlessness in orbit means zero normal reaction (not zero gravity), and that in vacuum every body falls with the same g. Master the two forms of g and the field-versus-potential distinction and the marks fall out.

Concept 1 of 6

Surface gravity — g = GM/R²

Intuition

Drop a small mass near a planet's surface and Newton's law of gravitation gives it an acceleration GM/R². This is g — it grows with the planet's mass and shrinks with the square of its radius. To compare two planets, scale M linearly and R as an inverse square.

Definition

The acceleration due to gravity at the surface of a planet of mass $M$ and radius $R$ is $g = \dfrac{GM}{R^2}$ . It follows from equating the weight $mg$ of a surface mass with the gravitational force $\dfrac{GMm}{R^2}$ ; the test mass $m$ cancels, so g is the same for all bodies at that surface. To compare planets, multiply by the mass factor and divide by the square of the radius factor.

Surface gravity

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

gacceleration due to gravity at the surface
Mmass of the planet
Rradius of the planet

Worked example

Planet X has the same mass as Earth but half its radius. How does the surface gravity on X compare with Earth's g?

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1GravitationMODERATE

A planet has a mass

M_1

and radius

R_1

. The value of acceleration due to gravity on its surface is

g_1

. There is another planet 2, whose mass and radius both are two times that of the first planet. Which one of the following is the acceleration due to gravity on the surface of planet 2?

[Q95 · Sep · 2018]

Radius enters as a square in g, just like in F

g = GM/R². When both mass and radius double, g does NOT stay the same — the mass factor 2 is divided by the radius factor squared (2² = 4), giving g/2. Always square the radius factor.

Concept 2 of 6

Surface gravity from density — g = (4/3)πGρR

Intuition

If a problem gives a planet's density instead of its mass, rewrite M as density × volume. The R³ from the volume cancels two of the three R's in the radius-squared denominator, leaving g proportional to ρ and to R. So for two planets of the SAME density, the bigger one has the stronger surface gravity.

Definition

Writing the mass as $M = \rho \cdot \dfrac{4}{3}\pi R^3$ and substituting into $g = \dfrac{GM}{R^2}$ gives

g = \dfrac{4}{3}\pi G \rho R.

So at fixed density,

g \propto R

: a larger planet of the same material has a larger surface gravity. More generally

g \propto \rho R

Surface gravity from density

g = \dfrac{4}{3}\pi G \rho R

$\rho$ mean density of the planet
Rradius of the planet
gsurface gravity (∝ ρR)

Worked example

Two planets are made of the same material (same density). One has twice the radius of the other. Compare their surface gravities.

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2GravitationMODERATE

Suppose there are two planets, 1 and 2, having the same density but their radii are

R_1

and

R_2

respectively, where

R_1 > R_2

. The accelerations due to gravity on the surface of these planets are related as

[Q110 · Apr · 2019]

Same density does NOT mean same gravity

When two planets share a density, g = (4/3)πGρR makes g proportional to R, so the larger planet has the stronger surface gravity. Don't assume equal density gives equal g — only same density AND same radius would.

Concept 3 of 6

Average density of a composite body

Intuition

When a body is built from parts of different densities — a solid core wrapped in a shell — its average density is the total mass divided by the total volume, not the average of the two densities. Find each part's mass (density × volume), add them, and divide by the whole volume.

Definition

The average (mean) density of a composite body is

\bar{\rho} = \dfrac{\text{total mass}}{\text{total volume}} = \dfrac{\sum_i \rho_i V_i}{\sum_i V_i}.

It is a volume-weighted average of the part densities — never a simple arithmetic mean of

\rho_1

and

\rho_2

. Compute each part's mass separately, sum, then divide by the total volume.

Average density of a composite body

\bar{\rho} = \dfrac{m_1 + m_2}{V_{\text{total}}} = \dfrac{\rho_1 V_1 + \rho_2 V_2}{V_1 + V_2}

$\rho_i, V_i$ density and volume of part i
$\bar{\rho}$ average density of the whole body

Dense core (ρ) inside a lighter shell (ρ/2). The average density is total mass ÷ total volume, a volume-weighted blend — here 9ρ/16, not the mid-value 3ρ/4.

Worked example

A solid sphere of radius R and density ρ is surrounded by a thin coating that doubles its volume, the coating having density 2ρ. Find the average density of the combined body.

Practice this conceptself-check · 3 quick reps

From the bank · past-year question

Example 3GravitationHARD

A spherical shell of outer radius R and inner radius R/2 contains a solid sphere of radius R/2. The density of the solid sphere is

\rho

and that of the shell is

\rho/2

. What is the average mass density of the larger sphere thus formed ?

[Q56 · Apr · 2024]

Average density is volume-weighted, not the mean of densities

For the shell-and-core sphere, averaging ρ and ρ/2 to get 3ρ/4 is wrong — that ignores that the denser core is the smaller part. The correct answer 9ρ/16 comes from total mass ÷ total volume, where the shell (the larger volume) pulls the average down.

Concept 4 of 6

Gravitational field versus potential

Intuition

The gravitational field is the force per unit mass (a vector, the local 'pull'); the potential is the potential energy per unit mass (a scalar, the 'height' in the energy landscape). Two points can sit at the same potential — the same energy height — while the field (the slope) differs between them. Moving a mass between two equal-potential points does no work.

Definition

The gravitational field $\vec{E} = \vec{F}/m$ is the force per unit mass (a vector). The gravitational potential $V$ is the potential energy per unit mass (a scalar). The work done by gravity moving a mass $m$ from A to B is $W = -\Delta U = -m(V_B - V_A)$ . If the potential is equal at A and B, then $V_B - V_A = 0$ , so $W = 0$ — even though the field may differ at the two points. Gravity is conservative, so this holds regardless of the path.

Work and equal potential

W = -\,m\,(V_B - V_A); \qquad V_A = V_B \Rightarrow W = 0

Wwork done by gravity, A → B
V_A, V_Bgravitational potential at A and B
mmass moved

Field lines (red) give the local pull; dashed circles are equipotentials. A and B sit on the same equipotential, so gravity does zero work moving a mass between them — even though the field strength differs.

Worked example

A mass is carried from one point to another along an equipotential surface (potential the same everywhere on it). How much work does gravity do?

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 4GravitationMODERATE

For an object in a gravitational field, the gravitational potential is the same at two points

A

and

B

, but the gravitational field is not the same at these two points. Which one of the following statements is correct?

[Q56 · Apr · 2026]

Equal potential, not equal field, decides the work

Work by gravity depends on the potential DIFFERENCE, not on the field. If A and B are at the same potential, the work is zero even when the field strength differs between them. A different field never makes gravity non-conservative.

Concept 5 of 6

Weightlessness in orbit — zero normal reaction

Intuition

An astronaut on the space station feels weightless not because gravity has switched off, but because the station and the astronaut are both falling freely around the Earth together. With nothing pushing up on them, the floor exerts zero normal reaction — and 'apparent weight' is exactly that contact force.

Definition

Weightlessness in orbit means the normal (contact) reaction is zero, not that gravity is absent. The astronaut and station are in free fall — both accelerating toward the Earth at the local g — so there is no contact force between the astronaut and the floor. Gravity is very much still acting (it is the centripetal force keeping the orbit); the astronaut's acceleration is not zero; and there is no real 'centrifugal' push.

Apparent weight = normal reaction

W_{\text{apparent}} = N = 0 \quad (\text{free fall}); \qquad F_{\text{gravity}} \neq 0

Nnormal (contact) reaction from the floor
$W_{\text{apparent}}$ apparent weight (= N)
$F_{\text{gravity}}$ actual gravitational pull (non-zero)

Worked example

Why does a person standing inside a lift whose cable has snapped (free fall) feel weightless, even on Earth?

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 5GravitationMODERATE

An astronaut whose weight on the Earth is 600 N experiences weightlessness on International Space Station orbiting around the Earth. It means that

[Q65 · Sep · 2024]

Weightless does NOT mean gravity-free

The common wrong choice says the gravitational pull on the astronaut is zero. It isn't — that pull is precisely what keeps the station in orbit. Weightlessness means the NORMAL REACTION is zero because everything is in free fall together.

Concept 6 of 6

g is the same for all bodies — free fall and weight

Intuition

Because the test mass cancels in g = GM/R², every object near a planet's surface falls with the same acceleration g, regardless of its own mass or shape — in a vacuum a feather and a coin land together. The same g sets an object's weight mg, so on a body with smaller g (the Moon) the weight, and any spring stretch it causes, shrinks in the same proportion.

Definition

The acceleration due to gravity $g$ is independent of the falling body's mass: in equating $mg = \dfrac{GMm}{R^2}$ , the test mass $m$ cancels. So in a vacuum all bodies fall with the same $g$ and take equal time to fall a given height. Weight is $W = mg$ ; a spring's extension $x = \dfrac{mg}{k} \propto g$ , so on a world with smaller $g$ the same hanging mass produces a proportionally smaller extension.

Weight and spring extension scale with g

W = mg, \qquad x = \dfrac{mg}{k} \propto g

gacceleration due to gravity (independent of the body's mass)
Wweight of the body
xspring extension; k = spring constant

Worked example

A spring stretches 12 cm when a mass hangs from it on Earth. If the same mass hangs from the same spring on a planet where g is half of Earth's, find the new extension.

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 6GravitationEASY

In a vacuum, a five-rupee coin, a feather of a sparrow bird and a mango are dropped simultaneously from the same height. The time taken by them to reach the bottom is

t_1

t_2

and

t_3

respectively. In this situation, we will observe that

[Q72 · Sep · 2017]

In vacuum, mass and shape don't change the fall time

Without air resistance a coin, a feather and a mango fall with the SAME g and reach the bottom together (t₁ = t₂ = t₃). The 'heavier falls faster' intuition only holds when air drag is present.

Spring extension follows g, not just the mass

Extension x = mg/k is proportional to g. The same hanging mass stretches the spring less on the Moon (g/6 → extension/6). Don't leave the extension unchanged just because the mass is unchanged.

Drill 1 more on g is the same for all bodies — free fall and weight

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 (6)

Surface gravity — g = GM/R²
Surface gravity
$g = \dfrac{GM}{R^2}$
Surface gravity from density — g = (4/3)πGρR
Surface gravity from density
$g = \dfrac{4}{3}\pi G \rho R$
Average density of a composite body
Average density of a composite body
$\bar{\rho} = \dfrac{m_1 + m_2}{V_{\text{total}}} = \dfrac{\rho_1 V_1 + \rho_2 V_2}{V_1 + V_2}$
Gravitational field versus potential
Work and equal potential
$W = -\,m\,(V_B - V_A); \qquad V_A = V_B \Rightarrow W = 0$
Weightlessness in orbit — zero normal reaction
Apparent weight = normal reaction
$W_{\text{apparent}} = N = 0 \quad (\text{free fall}); \qquad F_{\text{gravity}} \neq 0$
g is the same for all bodies — free fall and weight
Weight and spring extension scale with g
$W = mg, \qquad x = \dfrac{mg}{k} \propto g$

Watch out for (7)

Radius enters as a square in g, just like in F
→ Surface gravity — g = GM/R²
Same density does NOT mean same gravity
→ Surface gravity from density — g = (4/3)πGρR
Average density is volume-weighted, not the mean of densities
→ Average density of a composite body
Equal potential, not equal field, decides the work
→ Gravitational field versus potential
Weightless does NOT mean gravity-free
→ Weightlessness in orbit — zero normal reaction
In vacuum, mass and shape don't change the fall time
→ g is the same for all bodies — free fall and weight
Spring extension follows g, not just the mass
→ g is the same for all bodies — free fall and weight

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