Universal Law Gravitation

8 MCQs1 revision card9-step worked example

Source: NCERT GravitationPYQ coverage: NEET 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026

Lesson

Newton's law of universal gravitation states that every particle attracts every other particle with a force proportional to the product of their masses and inversely proportional to the square of the distance between their centres (NCERT Class 11 Physics Chapter 7, page 4):

F = G m₁ m₂ / r²

Here G = 6.674 × 10⁻¹¹ N·m²/kg² is the universal gravitational constant, measured by Cavendish. The force is always attractive, acts along the line joining the two centres, and obeys Newton's third law — both bodies experience equal and opposite forces regardless of their mass difference.

Key features to lock in:

Inverse-square dependence on distance. The force falls as 1/r², not 1/r. Doubling the centre-to-centre distance reduces the force to one-quarter, not one-half.
Product of masses, not sum. If one mass doubles, the force doubles. If both double, the force quadruples.
Centre-to-centre distance for spheres. For uniform spheres, r is measured from centre to centre (shell theorem), not surface to surface. A common error is using the gap between surfaces instead of the full centre-to-centre separation.
Superposition. The net gravitational force on a body due to multiple masses is the vector sum of individual pairwise forces.
G versus g. G is a universal constant (same everywhere). g (acceleration due to gravity at a planet's surface) is derived from it: g = GM/R². They are not interchangeable.

The law applies to point masses and, by the shell theorem, to any spherically symmetric mass distribution — the condition NEET problems almost always assume. When a problem says "uniform sphere," you can treat it as a point mass at its centre.

Practice MCQs

Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.

MCQ 1Easy RecallPractice

The SI unit of the universal gravitational constant G is:

MCQ 2Easy RecallPractice

Newton's law of gravitation applies exactly (without shell-theorem extension) to:

MCQ 3Easy RecallPractice

The gravitational force between two bodies is always:

MCQ 4Direct ApplicationPractice

Two identical spheres, each of mass m, are separated by a centre-to-centre distance d. The gravitational force between them is F. If the distance is halved to d/2, the new force is:

MCQ 5Direct ApplicationPractice

Two spheres of masses M and 4M are separated by centre-to-centre distance d. The ratio of the gravitational force on M due to 4M to the force on 4M due to M is:

MCQ 6Direct ApplicationPractice

The gravitational force between two uniform spheres is F when their surfaces are in contact. Each sphere has radius R. If the spheres are moved apart so that there is a gap of 2R between their surfaces, the new force is:

MCQ 7Concept TrapPractice

Gravitational force between two bodies depends on the intervening medium:

MCQ 8CalculationPractice

Three identical particles, each of mass m, are placed at the vertices of an equilateral triangle of side a. The magnitude of the net gravitational force on any one particle is:

Quick recall before you leave

Worked Example

Pattern: Direct application of F = Gm₁m₂/r² — force calculation between two bodies. (No topic-specific PYQ pattern survived Rule 7 filtering; all four dossier patterns belong to other topics. This worked example is constructed from the universal gravitation formula and the NCERT law entry.)

1
Given
Two uniform spheres: mass of sphere A = 5.0 × 10⁴ kg, mass of sphere B = 3.0 × 10⁴ kg. Centre-to-centre separation = 2.0 m.
2
Required
Find the gravitational force between the two spheres.
3
Concept
Newton's law of universal gravitation: every pair of point masses (or uniform spheres, by the shell theorem) attracts with force proportional to the product of their masses and inversely proportional to the square of their centre-to-centre distance (NCERT Class 11 Physics Chapter 7, page 4).
4
Formula
F = G m₁ m₂ / r²
5
Substitution
F = (6.674 × 10⁻¹¹) × (5.0 × 10⁴) × (3.0 × 10⁴) / (2.0)²
6
Calculation
Numerator: 6.674 × 10⁻¹¹ × 1.5 × 10⁹ = 6.674 × 1.5 × 10⁻² = 10.011 × 10⁻² = 1.0011 × 10⁻¹ Denominator: 4.0 F = 1.0011 × 10⁻¹ / 4.0 = 2.503 × 10⁻² Rounding to 2 significant figures (limited by the given masses at 2 sig figs each): F ≈ 2.5 × 10⁻² N. **Note on exact values:** The integer 2 in the denominator (2.0)² = 4.0 is an exact-count squaring of the given distance, which itself has 2 sig figs. G is known to 4 sig figs here. The final answer is rounded to 2 sig figs, matching the least precise given quantity.
7
Final answer
F ≈ 2.5 × 10⁻² N (attractive, along the line joining the centres).
8
Common trap
Using surface-to-surface distance instead of centre-to-centre distance. If the spheres each had radius 0.5 m and the problem stated "surfaces are 1.0 m apart," the centre-to-centre distance would be 0.5 + 1.0 + 0.5 = 2.0 m, not 1.0 m. Mixing these up changes the answer by a factor of 4.
9
Similar NEET-style question
Two lead spheres, each of mass 6.0 × 10³ kg, have their centres 5.0 m apart. Calculate the gravitational force between them. (Answer: F = G × (6.0 × 10³)² / (5.0)² ≈ 9.6 × 10⁻⁴ N.) ---

Before solving, remember these

Law

Universal law of gravitation

Every particle attracts every other particle with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between them: F = G m₁ m₂ / r², directed along the line joining them. G ≈ 6.67 × 10⁻¹¹ N·m²/kg².

-- NCERT Class 11 Physics, Ch. 7, p. 4

Formulas

8 formulas — click to collapse

Escape velocity from a body's surface

v_{e} = \frac{2 GM}{R} = 2 g R

Minimum speed for an object to escape gravity to infinity from radius R. Earth: ~11.2 km/s.

Symbol	Quantity	SI Unit
v_e	escape velocity	m/s
M	planet mass	kg
R	planet radius	m
g	surface gravity	m/s^2

Valid when

Launched from surface
No air drag
Body treated as point/sphere

Gravitational potential energy (point masses)

U = - \frac{GM m}{r}

PE of two-body system; negative because gravity is attractive (work to separate them is positive).

Symbol	Quantity	SI Unit
U	grav PE	J
M, m	two masses	kg
r	separation	m

Valid when

Reference U=0 at r=infinity
Point or spherical masses

g variation with altitude

g_{h} = g (\frac{R}{R + h})^{2} \approx g (1 - 2 h / R)

Gravitational acceleration decreases with altitude above Earth's surface.

Symbol	Quantity	SI Unit
g_h	g at height h	m/s^2
g	surface g	m/s^2
R	Earth radius	m
h	altitude	m

Valid when

Static observer at altitude
Earth treated as uniform sphere

g variation with depth

g_{d} = g (1 - d / R)

Inside Earth (uniform density), g decreases linearly with depth, vanishing at centre.

Symbol	Quantity	SI Unit
g_d	g at depth	m/s^2
g	surface g	m/s^2
R	Earth radius	m
d	depth	m

Valid when

Earth treated as uniform density sphere

Kepler's third law

T^{2} = \frac{4 π ^{2}}{GM} a^{3}

Square of orbital period proportional to cube of semi-major axis. Holds for elliptic orbits about a central mass.

Symbol	Quantity	SI Unit
T	orbital period	s
a	semi-major axis	m
M	central mass	kg

Valid when

Two-body system with central mass M >> orbiting mass
Bound orbit

Orbital velocity for circular orbit

v = \frac{GM}{R + h}

Speed of circular orbit at altitude h above body of mass M, radius R.

Symbol	Quantity	SI Unit
v	orbital speed	m/s
M	central mass	kg
R+h	orbit radius	m

Valid when

Circular orbit
M >> orbiting mass

Satellite total mechanical energy

E = - \frac{GM m}{2 ( R + h )}

Total energy = KE + PE = -KE (virial). Always negative for bound orbit; E -> 0 at infinity.

Symbol	Quantity	SI Unit
E	total energy	J
M, m	central mass and satellite mass	kg
R+h	orbit radius	m

Valid when

Circular orbit
Bound (E < 0)

Newton's law of gravitation

F = \frac{G m _{1} m _{2}}{r ^{2}}

Attractive force between any two masses. Inverse-square central force.

Symbol	Quantity	SI Unit
F	force	N
G	grav constant = 6.674e-11	N*m^2/kg^2
m1, m2	masses	kg
r	centre-to-centre distance	m

Valid when

Point masses or spherically symmetric distributions
r > sum of body radii (else use shell theorem)

Exam Traps & Common Mistakes

These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.

4 items — click to collapse

Trap

g at altitude: linear vs inverse-square

Category: Similar Terms

Student treats g(h) as linear in h. Actual: g(R/(R+h))² (inverse-square).

When it triggers

Question asks for g at significant altitude (e.g. R/2 above surface).

How to avoid

Use g_h = g (R/(R+h))². Linear approximation g(1 - 2h/R) only valid for h << R.

Trap

Kepler's third law: linear vs 3/2 power

Category: Similar Terms

Student treats T proportional to a (linear) instead of a^(3/2).

When it triggers

Question gives change in semi-major axis and asks for new period.

How to avoid

T² ∝ a³, so T ∝ a^(3/2). Doubling a multiplies T by 2^(3/2) ≈ 2.83.

Mistake

Treats g(h) as linear in h for significant heights.

Root cause: formula misuse

Correction

Use g_h = g(R/(R+h))² (inverse-square). Linear approximation g(1-2h/R) is only valid for h << R. For h = R/2, the exact formula gives g_h = (2/3)² g = 4g/9, not g(1-1) = 0.

Mistake