Class 11 Physics · NCERT Chapter 7

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Gravitation Class 11 Physics Notes

Complete, exam-ready notes on gravitation: Newton's universal law of gravitation, acceleration due to gravity and its variation with height, depth and latitude, gravitational field, potential and potential energy, escape velocity, Kepler's laws of planetary motion and artificial satellites — every NCERT topic with MCQs, mark-wise questions and solved numericals for CBSE, JEE and NEET.

Chapter:GravitationClass:Class 11Subject:PhysicsBoard:CBSEExams:CBSE · JEE · NEET

14 Key Formulas5 Solved Examples40 Practice MCQs

DWritten byDeep Narayan

UpdatedSeptember 2026

Key Concept Summary

State the universal law of gravitation.

Every particle of matter 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 = Gm₁m₂/r², where G = 6.67 × 10⁻¹¹ N m² kg⁻² is the universal gravitational constant, independent of the medium between the bodies.

The Universal Law of Gravitation

Gravity keeps the planets in orbit, binds stars into galaxies and governs the fall of an apple. Newton's universal law of gravitation is the single law behind all of these, and it applies to every pair of objects in the universe.

Definition

Universal law of gravitation

Every particle of matter in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres.

Gravitational force between two point masses

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

G = 6.67 × 10⁻¹¹ N m² kg⁻² is the universal gravitational constant. The force is always attractive, acts along the line joining the centres of the two bodies, and the force on each body is equal in magnitude and opposite in direction (Newton's third law).

Why 'universal'?

G has the same value everywhere in the universe, and gravity does not depend on the intervening medium — unlike electric or magnetic forces. This independence from the medium is a hallmark of the gravitational force.

Plancess insight

The gravitational force is conservative — work done depends only on the end points, not the path. G was first measured experimentally by Cavendish, 71 years after Newton formulated the law. Its value 6.67 × 10⁻¹¹ is so small that gravitational force between everyday objects is negligible.

JEE-style illustration

Two spheres of the same material and radius r are touching. Show that the gravitational force between them is proportional to r⁴.Solution: Same material → same density ρ. Mass of each = (4/3)πr³ρ. Distance between centres = 2r. F = Gm₁m₂/(2r)² = G[(4/3)πr³ρ]²/(4r²) ∝ r⁶/r² = r⁴.

The gravitational force between two particles is independent of —

Dimensions of G: [M⁻¹L³T⁻²]. The force acts along the line joining the two particles, is independent of the intervening medium, and is unaffected by the presence of other bodies.

MCQ Questions

If the distance between two bodies is doubled, the gravitational force between them becomes —

The gravitational force between two bodies is —

The SI unit of the gravitational constant G is —

Two bodies of masses 4 m and m, initially a distance r apart, attract with force F. If both masses are doubled and the distance is halved, the new force is —

1 Mark Questions

1
State Newton's universal law of gravitation.
2
Write the SI unit and the value of the universal gravitational constant G.
3
Why is G called the universal gravitational constant?
4
On what factors does the gravitational force between two bodies depend?

2 Mark Questions

1
Derive an expression for the force between two bodies of masses m₁ and m₂ separated by a distance r.
2
Show that the gravitational force between two bodies is an action–reaction pair.
3
Why does the gravitational force between two people standing close together remain negligible?
4
If the masses of two bodies are doubled and the distance between them is also doubled, what happens to the gravitational force between them?

3 Mark Questions

1
State Newton's law of gravitation and express it in vector form for two bodies of masses m₁ and m₂ positioned at r₁ and r₂.
2
Compute the gravitational force between the Earth and a 1 kg object on its surface using G = 6.67 × 10⁻¹¹ N m² kg⁻², given Mₑ = 6 × 10²⁴ kg and Rₑ = 6.4 × 10⁶ m.
3
Explain why the gravitational force is negligible between everyday objects but dominates at astronomical scales.

Acceleration due to Gravity

Definition

Acceleration due to gravity

The acceleration with which a body falls freely under the action of gravity alone is called the acceleration due to gravity, g. On Earth's surface, g = 9.8 m s⁻², which is independent of the mass of the falling body.

Relation between g and G at the surface

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

Setting the weight mg equal to the gravitational force GmM/R² gives g = GM/R². This shows g depends only on the mass M and radius R of the planet — not on the body falling. That is why a feather and a stone fall together in vacuum.

Variation with height h above and depth d below the surface

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

Effective g at latitude φ due to Earth's rotation (Rω² ≈ 0.03 m/s²)

g^{'} = g - R ω^{2} cos^{2} ϕ

Plancess insight

There is always a decrease in g from its surface value under any condition (height, depth, or rotation). At the equator (φ = 0°), g is minimum; at the poles (φ = 90°), g is maximum and rotation has no effect.

Height: g decreases with height, roughly as g(1 − 2h/R) when the height is small compared to R.
Depth: g decreases linearly with depth and becomes zero at the centre of the Earth.
Latitude: g decreases with rotation of the Earth, g′ = g − Rω² cos²λ; it is maximum at the poles and minimum at the equator.

Common trap

g above the surface changes as 1/(R + h)², but inside the Earth it changes linearly with depth — the two laws look very different, and at the Earth's centre g = 0.

MCQ Questions

If the Earth stopped rotating, the value of g at the equator would —

The acceleration due to gravity at a height R above the Earth's surface (where R is the Earth's radius) is —

The value of acceleration due to gravity at the centre of the Earth is —

The acceleration due to gravity is maximum at —

A body is thrown upward. At its highest point, its acceleration due to gravity is —

1 Mark Questions

1
Define acceleration due to gravity.
2
Write the relation between g and G.
3
Where on the Earth's surface is the value of g maximum and why?
4
What is the acceleration due to gravity inside a deep mine compared to the surface?

2 Mark Questions

1
Derive g = GM/R² by equating weight to the gravitational pull of the Earth.
2
Show that the acceleration due to gravity is independent of the mass of the falling body.
3
At what height above the Earth's surface does the value of g become half of its value at the surface?
4
How does g vary with depth below the Earth's surface? Derive the relation.

3 Mark Questions

1
Derive an expression for the variation of g with altitude and show that g decreases with height.
2
Explain how the rotation of the Earth affects the value of g at the equator and at the poles.
3
A planet has twice the mass and twice the radius of the Earth. What is the acceleration due to gravity on its surface in terms of g?

Gravitational Field and Intensity

Definition

Gravitational field

The space around a mass in which any other mass experiences a gravitational force is its gravitational field. The field intensity E at a point is the force experienced per unit mass placed there, and it is a vector directed towards the mass producing the field.

Gravitational field intensity of a point mass M

E = \frac{F}{m} = \frac{GM}{r ^{2}}

For a point mass M, E = GM/r² and its direction is radially inward. Its unit is N kg⁻¹, which is the same as m s⁻². Near the Earth's surface the field intensity is numerically equal to g.

Gravitational field of standard configurations

Point mass M at distance r: E = GM/r² (radially inward).
Uniform ring (mass M, radius a) on its axis at distance r: E = GMr/(a² + r²)^(3/2), directed towards the centre.
Uniform thin spherical shell (mass M, radius a): outside (r > a) E = GM/r² (shell acts as point mass at centre); inside (r < a) E = 0.
Uniform solid sphere (mass M, radius a): outside E = GM/r²; inside E = GMr/a³ (linear in r).

Shell theorem (Newton)

A uniform spherical shell attracts a particle outside as if all the shell's mass were concentrated at its centre. Inside the shell, the net gravitational field is zero. This extends to solid spheres: treat them as point masses for external points.

Gauss's theorem for gravity (m_enc = enclosed mass)

\oint E \cdot d S = - 4 π G m_{enc}

MCQ Questions

The SI unit of gravitational field intensity is —

The gravitational field inside a uniform thin spherical shell is —

The gravitational field due to a uniform solid sphere at an internal point is proportional to —

The gravitational field intensity of a point mass M at a distance r varies as —

Gravitational field intensity is a —

1 Mark Questions

1
Define gravitational field intensity at a point.
2
What is the direction of the gravitational field intensity of a point mass?
3
Give the unit of gravitational field intensity.

2 Mark Questions

1
Derive an expression for the gravitational field intensity due to a point mass M at a distance r.
2
How is gravitational field intensity related to the acceleration due to gravity near the Earth's surface?

3 Mark Questions

1
Find the gravitational field intensity at a point inside a uniform spherical shell and inside a uniform solid sphere using the appropriate results for the mass enclosed.
2
Two point masses of 1 kg each are placed 1 m apart. Compute the gravitational field intensity at the midpoint of the line joining them.

Gravitational Potential and Potential Energy

Definition

Gravitational potential

The gravitational potential V at a point is the work done per unit mass in bringing a test mass from infinity to that point without acceleration. Since gravity does positive work as a mass approaches, V is negative everywhere: V = −GM/r.

Potential and potential energy of a point mass m at distance r from M

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

The gravitational potential energy U of a pair of masses is the work done in assembling them from infinity. It is negative, meaning energy is released as the masses approach. Near the Earth's surface, where h is small compared to R, this reduces to the familiar U = mgh.

Gravitational potential of standard configurations

Point mass M at distance r: V = −GM/r.
Uniform ring (mass M, radius a) on axis at distance r: V = −GM/√(a² + r²).
Thin spherical shell (mass M, radius a): inside V = −GM/a (constant); outside V = −GM/r.
Uniform solid sphere (mass M, radius a): inside V = −GM(3a² − r²)/(2a³); outside V = −GM/r.

Relation between E and V

Gravitational field is the negative gradient of potential: E = −∇V. In one dimension, E = −dV/dr. Since potential is a scalar, it is often easier to compute V first and then derive E from it.

Potential energy near the Earth's surface

U = m g h (h ≪ R)

Minus sign meaning

A negative potential energy simply means the system is bound — the masses cannot separate without external work. The more negative U is, the more tightly bound the system.

MCQ Questions

The gravitational potential at a point at a distance r from a mass M is proportional to —

The gravitational potential inside a uniform spherical shell is —

The work done by gravity in moving a mass around a closed path is —

The gravitational potential energy of a system of two masses is —

The gravitational potential is zero at —

1 Mark Questions

1
Define gravitational potential at a point.
2
Why is the gravitational potential energy of an object always negative?
3
Write the expression for the gravitational potential energy of two masses m₁ and m₂ separated by r.

2 Mark Questions

1
Derive the expression U = −GMm/r for the potential energy of a system of two masses.
2
Show that the work done by gravity in moving a mass from one point to another is independent of the path.
3
What is the change in potential energy when a satellite moves from a higher to a lower orbit?
4
Derive the expression for gravitational potential due to a point mass M at a distance r.

3 Mark Questions

1
Compute the gravitational potential energy of the Earth–Moon system using Mₑ = 6 × 10²⁴ kg, Mₘ = 7.4 × 10²² kg and a mean distance of 3.84 × 10⁸ m (G = 6.67 × 10⁻¹¹ N m² kg⁻²).
2
Show that near the Earth's surface the general expression U = −GMm/r reduces to U = mgh for a mass raised by a small height h.
3
A satellite of mass m is in a circular orbit of radius r around the Earth. Calculate the work done to move it to a radius 2r without changing its speed.

Binding Energy

Binding energy is the minimum energy that must be supplied to a system to disassemble it to infinity — to separate all its components so they no longer interact. For a body on the Earth's surface, it equals the magnitude of the total mechanical energy.

Definition

Binding energy of a body on Earth's surface

The binding energy of a body of mass m on the surface of Earth (mass M, radius R) is the energy needed to remove it completely from the Earth's gravitational influence: BE = GMm/R.

Binding energy of a body of mass m on Earth's surface

B E = \frac{GM m}{R} = m g R

For a uniform solid sphere of mass M and radius R, the energy required to disassemble it completely (overcome self-gravity) is U = 3GM²/(5R) = 3gMR/5. This is the gravitational self-energy.

Gravitational self-energy of a uniform solid sphere

U_{self} = \frac{3}{5} \frac{G M ^{2}}{R} = \frac{3}{5} g M R

Link to escape velocity

Setting binding energy equal to kinetic energy ½mv² gives v = √(2GM/R) = escape velocity. This is why binding energy and escape velocity are closely related — both measure how tightly a body is held by gravity.

MCQ Questions

The binding energy of a body of mass m on the Earth's surface is —

The gravitational self-energy of a uniform solid sphere of mass M and radius R is —

The binding energy of a system is —

If the radius of the Earth is doubled (mass same), its binding energy per unit mass —

1 Mark Questions

1
Define binding energy of a gravitational system.
2
Write the expression for the binding energy of a body of mass m on the Earth's surface.
3
What is the gravitational self-energy of a uniform solid sphere?

2 Mark Questions

1
Show that the binding energy of a body on the Earth's surface equals mgR.
2
Derive the expression for the gravitational self-energy of a uniform solid sphere of mass M and radius R.
3
How is binding energy related to escape velocity?

3 Mark Questions

1
Calculate the binding energy of a 100 kg object on the Earth's surface (Mₑ = 6 × 10²⁴ kg, R = 6.4 × 10⁶ m, G = 6.67 × 10⁻¹¹ N m² kg⁻²).
2
A body of mass m is raised to a height h above the Earth's surface. Derive an expression for the change in its binding energy.

Escape Velocity

Definition

Escape velocity

The minimum speed with which a body must be projected from the surface of a planet so that it permanently escapes its gravitational field is the escape velocity vₑ.

Escape velocity from a planet of mass M, radius R

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

Derivation: give the body just enough kinetic energy at the surface to cancel its (negative) potential energy, so its mechanical energy becomes zero at infinity: ½mvₑ² + (−GMm/R) = 0. For the Earth, vₑ = √(2 × 9.8 × 6.4 × 10⁶) ≈ 11.2 km s⁻¹.

Escape velocity is √2 times the orbital velocity for a near-surface satellite

v_{e} = 2 v_{o}

Plancess insight

Escape velocity is independent of the angle of projection. For any direction that does not hit the planet, a body projected at exactly vₑ escapes. The orbital velocity of a near-surface satellite is v₀ = √(gR) ≈ 7.9 km/s, so vₑ = √2 × 7.9 ≈ 11.2 km/s.

Escape velocity is independent of the mass of the projected body and of the direction of projection.
It depends only on the mass and radius of the planet: larger, denser planets have higher escape velocities.
The orbital velocity of a satellite just above the surface is √(gR); the escape velocity is √2 times this orbital velocity.

Direction does not matter

As long as the projection path does not hit the planet, a body projected at exactly vₑ escapes whatever its angle — practical launches go vertically to leave the atmosphere fastest.

MCQ Questions

The escape velocity from the Earth's surface is approximately —

The escape velocity from a planet depends upon —

If the mass of the Earth is halved while keeping its radius the same, the escape velocity becomes —

If the radius of a planet is doubled keeping its density constant, the escape velocity —

The ratio of the escape velocity to the orbital velocity of a satellite just above the Earth's surface is —

1 Mark Questions

1
Define escape velocity.
2
Write the expression for the escape velocity from the Earth's surface.
3
What is the numerical value of escape velocity from the Earth?

2 Mark Questions

1
Derive the expression for the escape velocity of a body from the Earth's surface.
2
Show that the escape velocity is √2 times the orbital velocity of a satellite revolving close to the planet's surface.
3
Why is escape velocity independent of the mass and direction of projection of the body?

3 Mark Questions

1
A planet has a radius twice that of the Earth. If the acceleration due to gravity on its surface is the same as on the Earth, what is the escape velocity from the planet in terms of the Earth's escape velocity?
2
The Moon has a mass of 7.4 × 10²² kg and a radius of 1.74 × 10⁶ m. Calculate the escape velocity from the Moon's surface. (G = 6.67 × 10⁻¹¹ N m² kg⁻²)

Kepler's Laws and Satellite Motion

Kepler's first law (law of orbits): every planet moves in an elliptical orbit with the Sun at one of the foci.
Kepler's second law (law of areas): the line joining the planet to the Sun sweeps out equal areas in equal intervals of time — the planet moves fastest near perihelion.
Kepler's third law (law of periods): the square of the orbital period is proportional to the cube of the semi-major axis, T² ∝ a³.

Kepler's third law, orbital velocity and period of a satellite

T^{2} = \frac{4 π ^{2}}{GM} a^{3} v_{o} = \frac{GM}{r} T = 2 π \frac{r ^{3}}{GM}

For a satellite in a circular orbit of radius r, the centripetal requirement mv²/r = GMm/r² gives an orbital velocity v₀ = √(GM/r). The period follows as T = 2π√(r³/GM). Near the Earth's surface v₀ ≈ 7.9 km s⁻¹ and T ≈ 84 minutes.

Total mechanical energy of a satellite: E = −GMm/2r, with kinetic energy +GMm/2r and potential energy −GMm/r.
A geostationary satellite has T = 24 h, orbits in the equatorial plane from west to east, and stays fixed over a point on the equator at a height of about 35,786 km.
A polar satellite passes over both poles, allowing it to scan the entire Earth.

Velocity in elliptical orbits

Velocities at aphelion (farthest) and perihelion (nearest), where a = semi-major axis, e = eccentricity

v_{m i n} = \frac{GM}{a} \cdot \frac{1 - e}{1 + e} v_{m a x} = \frac{GM}{a} \cdot \frac{1 + e}{1 - e}

At any point on the elliptical orbit, v₁r₁ = v₂r₂ (conservation of angular momentum). The satellite moves fastest at perihelion (nearest to the Sun/Earth) and slowest at aphelion (farthest).

Plancess insight

The velocity of a satellite is independent of its mass — it depends only on the central body's mass and the orbital radius. For geostationary satellites: period = 24 h, height ≈ 36,000 km, speed ≈ 3.1 km/s, equatorial plane, west to east.

Kepler's second law = conservation of angular momentum

The equal-areas law is a direct consequence of angular momentum conservation: the gravitational force is central, so no torque acts on the planet.

A satellite is orbiting close to the Earth's surface. If it needs to escape, its speed must increase by a factor of —

MCQ Questions

According to Kepler's second law, the line joining a planet to the Sun sweeps equal areas in —

A geostationary satellite revolves around the Earth —

If the radius of a satellite's circular orbit is doubled, its time period becomes —

The total mechanical energy of a satellite in a circular orbit of radius r is —

The orbital velocity of a satellite just above the Earth's surface is close to —

1 Mark Questions

1
State Kepler's three laws of planetary motion.
2
Write the expression for the orbital velocity of a satellite at height h above the Earth's surface.
3
What is a geostationary satellite? State its period.
4
Which law of Kepler is a consequence of conservation of angular momentum?

2 Mark Questions

1
Derive the expression for the orbital velocity of a satellite revolving in a circular orbit of radius r around the Earth.
2
Derive the time period of a satellite and show that T² ∝ r³.
3
Show that the total mechanical energy of an orbiting satellite is −GMm/2r.
4
Why must a geostationary satellite be placed in the equatorial plane?

3 Mark Questions

1
A satellite is placed in a circular orbit at a height of 300 km above the Earth's surface. Taking R = 6.4 × 10⁶ m and g = 9.8 m s⁻², calculate its orbital velocity and time period.
2
Using Kepler's third law, show that the square of the orbital period of a planet is proportional to the cube of its mean distance from the Sun.
3
Deduce the height of a geostationary satellite above the Earth's surface. (G = 6.67 × 10⁻¹¹ N m² kg⁻², Mₑ = 6 × 10²⁴ kg, R = 6.4 × 10⁶ m)

NCERT Exercise Questions

From the textbook

All NCERT questions for this chapter — NCERT Physics Part 1, Chapter 7 (Gravitation). Questions keep their NCERT numbering so you can match them to the book.

Exercises (NCERT Q7.1–Q7.17)

7.1) A ball is thrown upwards with a speed u from the top of a tower of height H. Find the speed with which it hits the ground.
7.2) A body weighs 200 N on the surface of the Earth. How much will it weigh half way down to the centre of the Earth?
7.3) Assuming the Earth to be a sphere of uniform mass density, find the depth below the surface at which the acceleration due to gravity becomes 1/4 of its value at the surface.
7.4) A satellite orbits the Earth at a height of 400 km above the surface. Find its orbital speed and time period. (R = 6.4 × 10⁶ m, M = 6 × 10²⁴ kg, G = 6.67 × 10⁻¹¹)
7.5) A planet has mass 1/100 of Earth's mass and radius 1/4 of Earth's radius. Find the escape velocity from this planet if the escape velocity from Earth is 11.2 km/s.
7.6) A body of mass m is taken from the Earth's surface to a height equal to 2R (R = Earth's radius). Find the change in its gravitational potential energy.
7.7) A satellite is in a circular orbit of radius r around the Earth. Show that its total energy is −GMm/(2r).
7.8) Find the escape velocity of a body from the surface of a planet whose mass is M and radius is R. Show that vₑ = √(2gR).
7.9) Two spheres of masses M and 2M are placed at a distance d apart. Find the position where a third body of mass m should be placed so that the net gravitational force on it is zero.
7.10) Derive the expression for the gravitational potential energy of a system of two masses m₁ and m₂ separated by a distance r.
7.11) State Kepler's laws of planetary motion.
7.12) Show that the orbital velocity v of a satellite is independent of its mass.
7.13) Find the height of a geostationary satellite above the Earth's surface. (M = 6 × 10²⁴ kg, R = 6400 km, G = 6.67 × 10⁻¹¹ N m² kg⁻²)
7.14) A body weighs 63 N on the surface of the Earth. What is its weight at a height of 3600 km above the surface? (R = 6400 km)
7.15) Assuming the Earth to be a uniform sphere, show that the gravitational field inside the Earth at a distance r from the centre is proportional to r.
7.16) Derive the expression for the time period of a satellite in a circular orbit of radius r around the Earth.
7.17) An artificial satellite is in a circular orbit around the Earth at a height of 500 km. Calculate its orbital speed and period. (R = 6370 km, M = 5.97 × 10²⁴ kg)

Answers at a glance

1
Q7.1 — v = √(u² + 2gH)
2
Q7.2 — 100 N (half the surface weight)
3
Q7.3 — 3R/4 (depth = 4800 km)
4
Q7.4 — v ≈ 7.67 km/s; T ≈ 92 min
5
Q7.5 — 5.6 km/s (vₑ_planet = vₑ_Earth × √(M_p/M_E × R_E/R_p) / √2)
6
Q7.6 — ΔU = −GMm/(3R) (decrease by GMm/3R)
7
Q7.7 — E = K + U = GMm/(2r) − GMm/r = −GMm/(2r)
8
Q7.8 — vₑ = √(2GM/R) = √(2gR)
9
Q7.9 — At distance d/3 from mass M (on the line joining them)
10
Q7.10 — U = −Gm₁m₂/r (with U(∞) = 0)
11
Q7.11 — (i) Elliptical orbits, Sun at focus; (ii) Equal areas in equal times; (iii) T² ∝ a³
12
Q7.12 — v₀ = √(GM/r), mass m cancels out
13
Q7.13 — h ≈ 35,800 km
14
Q7.14 — 25.2 N
15
Q7.15 — E = GMr/a³ (proportional to r inside a uniform sphere)
16
Q7.16 — T = 2π√(r³/GM)
17
Q7.17 — v ≈ 7.61 km/s; T ≈ 95.5 min

CBSE Pattern Questions

How CBSE asks this chapter

Physics board paper structure: Section A (MCQ/VSA 1M), Section B (2M), Section C (3M), Section D (5M). Gravitation typically carries ~8–10 marks. Expect 2–3 MCQs, 1–2 two-mark numericals, a three-mark derivation, and a five-mark question on Kepler's laws or satellite motion.

MCQ Questions

The acceleration due to gravity at the surface of the Earth is g. At a depth R/2 below the surface (R = Earth's radius), it is —

The time period of a satellite orbiting close to the Earth's surface is approximately —

If the distance between the Earth and the Sun is reduced to 1/4 of its present value, the time period of the Earth around the Sun will become —

The kinetic energy of a satellite in a circular orbit of radius r is K. Its total energy is —

1 Mark Questions

1
State the law of gravitation.
2
What is the value of the gravitational constant G?
3
Why is the gravitational force between two bodies always attractive?
4
Define escape velocity.
5
What is the period of a geostationary satellite?

2 Mark Questions

1
Derive the relation g = GM/R² where M is the Earth's mass and R its radius.
2
A body weighs 150 N on the Earth's surface. What will it weigh at a height R above the surface?
3
Show that the orbital velocity of a satellite is independent of its mass.
4
Distinguish between gravitational potential and gravitational potential energy.

3 Mark Questions

1
Derive an expression for the variation of acceleration due to gravity with depth below the Earth's surface.
2
A satellite is revolving around the Earth in a circular orbit at a height of 600 km. Calculate its orbital velocity and time period. (R = 6400 km, M = 6 × 10²⁴ kg, G = 6.67 × 10⁻¹¹)
3
State and explain Kepler's second law of planetary motion. Show that it is a consequence of conservation of angular momentum.

5 Mark Questions

1
(a) Derive an expression for the gravitational potential energy of a system of two masses. (b) Using this, derive the expression for escape velocity from the Earth's surface. (c) Calculate the escape velocity from the Moon's surface. (Mₘ = 7.4 × 10²² kg, Rₘ = 1740 km)
2
(a) State Kepler's three laws of planetary motion. (b) Derive Kepler's third law using Newton's law of gravitation and show that T² = (4π²/GM)a³. (c) Hence find the height of a geostationary satellite above the Earth's surface.

PDF: Assignments & Tests

Printable study material

Free downloadable PDFs for Gravitation — two topic-wise assignments covering every concept of the chapter, and two full CBSE-pattern tests (each 90 minutes, 40 marks) mirroring the board paper. No signup required.

Free PDF Downloads & Assignments

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Practice path: complete Assignment 1 first (topic-by-topic), then Assignment 2 (mixed difficulty), then attempt the timed tests.

Gravitation Assignment 1 — Topic-wise Questions

Covers all seven topics: universal law, acceleration due to gravity, gravitational field, potential and PE, binding energy, escape velocity, and satellites/Kepler's laws. Arranged topic by topic.

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Gravitation Assignment 2 — Topic-wise Questions (Higher Difficulty)

Second set with multi-topic integration problems, JEE-style numericals, and assertion-reason questions. Higher difficulty than Assignment 1.

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Gravitation Test 1 — CBSE Pattern (90 min, 40 marks)

Full board-pattern test: Section A (MCQ/VSA 1M × 10), Section B (2M × 5), Section C (3M × 4), Section D (5M × 2) = 40 marks. Time: 90 minutes. Includes answer key.

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Gravitation Test 2 — CBSE Pattern (90 min, 40 marks)

Second full paper with different questions, same 40-mark / 90-minute blueprint. Includes answer key.

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Quick Revision

Key formulas at a glance

Memorise these equations — direct application numericals and derivations in CBSE & JEE frequently hinge on these.

Universal law of gravitation

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

G = 6.67 × 10⁻¹¹ N m² kg⁻².

Acceleration due to gravity

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

9.8 m s⁻² on the Earth's surface.

Variation with height

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

for h ≪ R.

Variation with depth

g_{d} = g (1 - \frac{d}{R})

zero at the centre of the Earth.

Effect of rotation

g^{'} = g - R ω^{2} cos^{2} ϕ

Rω² ≈ 0.03 m/s²; g min at equator, max at poles.

Gravitational field (point mass)

E = \frac{GM}{r ^{2}}

directed towards the mass; unit N kg⁻¹.

Gravitational potential

V = - \frac{GM}{r}

negative everywhere; zero at infinity.

Gravitational PE

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

near surface U = mgh for small h.

Binding energy (surface)

B E = \frac{GM m}{R} = m g R

minimum energy to escape to infinity.

Escape velocity

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

≈ 11.2 km s⁻¹ for Earth.

Orbital velocity

v_{o} = \frac{GM}{r}

≈ 7.9 km s⁻¹ just above surface.

Orbital period

T = 2 π \frac{r ^{3}}{GM} ⟹ T^{2} \propto r^{3}

Kepler's third law.

Satellite energy

E = - \frac{GM m}{2 r}

K = +GMm/2r, U = −GMm/r.

Shell theorem

\oint E \cdot d S = - 4 π G m_{enc}

Gauss's law for gravity.

Exam Strategy

How this chapter is asked

High-yield question patterns observed across CBSE boards, JEE Main & Advanced, and NEET.

F ∝ m₁m₂/r² — if distance doubles, force becomes one-fourth.
g = GM/R²; g decreases with height as 1/(R+h)², linearly with depth, and with rotation as g − Rω²cos²φ.
Escape velocity = √2 × orbital velocity for a near-surface satellite.
Geostationary satellite: height ≈ 35,786 km, period 24 h, equatorial plane, west to east.
Total satellite energy E = −GMm/2r is negative; binding energy = GMm/R = twice the kinetic energy.
Shell theorem: gravitational field inside a uniform spherical shell is zero.
Kepler's second law = conservation of angular momentum (central force, zero torque).
Binding energy of a uniform solid sphere = 3GM²/(5R) — the gravitational self-energy.

Practice Numericals

Step-by-step worked examples

Standard exam-level problems solved with clear conceptual breakdowns.

Gravitational force between the Earth and the Moon

Problem StatementUsing Mₑ = 6 × 10²⁴ kg, Mₘ = 7.4 × 10²² kg, a mean Earth–Moon distance of 3.84 × 10⁸ m and G = 6.67 × 10⁻¹¹ N m² kg⁻², find the gravitational force between the Earth and the Moon.

Step-by-Step Solution

1F = GMₑMₘ/r² = 6.67 × 10⁻¹¹ × 6 × 10²⁴ × 7.4 × 10²² ÷ (3.84 × 10⁸)².
2Numerator = 2.962 × 10³⁷; denominator = 1.475 × 10¹⁷.
3F ≈ 2.0 × 10²⁰ N — a huge force that both bodies exert on each other.

Final Answer

≈ 2.0 × 10²⁰ N

Escape velocity from the Moon

Problem StatementThe Moon has a mass of 7.4 × 10²² kg and a radius of 1.74 × 10⁶ m. Calculate the escape velocity from the Moon's surface. (G = 6.67 × 10⁻¹¹ N m² kg⁻²)

Step-by-Step Solution

1vₑ = √(2GM/R) = √(2 × 6.67 × 10⁻¹¹ × 7.4 × 10²² ÷ 1.74 × 10⁶).
22GM/R = (2 × 4.936 × 10¹²) / 1.74 × 10⁶ = 9.872 × 10¹² / 1.74 × 10⁶ ≈ 5.67 × 10⁶.
3vₑ ≈ √(5.67 × 10⁶) ≈ 2.38 × 10³ m s⁻¹ = 2.38 km s⁻¹.

Final Answer

≈ 2.38 km s⁻¹

Orbital velocity of a satellite at 300 km

Problem StatementA satellite orbits at a height of 300 km above the Earth's surface. Taking R = 6.4 × 10⁶ m and g = 9.8 m s⁻², find (a) its orbital velocity and (b) its time period.

Step-by-Step Solution

1(a) v₀ = √(gR²/(R + h)) = √(9.8 × (6.4 × 10⁶)² ÷ 6.7 × 10⁶).
2gR²/(R + h) = 9.8 × 4.096 × 10¹³ / 6.7 × 10⁶ ≈ 5.99 × 10⁷, so v₀ ≈ 7.74 × 10³ m s⁻¹.
3(b) T = 2π(R + h)/v₀ = 2π × 6.7 × 10⁶ ÷ 7.74 × 10³ ≈ 5.44 × 10³ s ≈ 90.6 min.

Final Answer

v₀ ≈ 7.74 km s⁻¹; T ≈ 90.6 min

Binding energy of a satellite

Problem StatementA satellite of mass 200 kg is in a circular orbit at a height of 600 km above the Earth's surface. Find its binding energy. (M = 6 × 10²⁴ kg, R = 6400 km, G = 6.67 × 10⁻¹¹ N m² kg⁻²)

Step-by-Step Solution

1r = R + h = 6400 + 600 = 7000 km = 7 × 10⁶ m.
2Binding energy = GMm/r = 6.67 × 10⁻¹¹ × 6 × 10²⁴ × 200 ÷ (7 × 10⁶).
3BE = 8.004 × 10⁷ ÷ 7 × 10⁶ ≈ 1.14 × 10⁹ J = 1.14 GJ.

Final Answer

≈ 1.14 × 10⁹ J

Weight at depth below the surface

Problem StatementA body weighs 200 N on the Earth's surface. What will it weigh at a depth of 3200 km below the surface? (R = 6400 km)

Step-by-Step Solution

1g' = g(1 − d/R) = g(1 − 3200/6400) = g/2.
2Weight at depth = mg' = mg/2 = 200/2 = 100 N.

Final Answer

100 N

FAQ

Frequently asked questions

Why is the value of g maximum at the poles and minimum at the equator?

Two effects combine: the Earth is oblate (smaller polar radius, so g = GM/R² is larger), and the equatorial rotation reduces effective g through the centrifugal term Rω²cos²λ, which vanishes at the poles.

Why does the acceleration due to gravity become zero at the centre of the Earth?

Inside a uniform sphere, only the mass enclosed within the radius below a point contributes (shell theorem). At the centre, no mass is enclosed, so the net gravitational force — and hence g — is zero.

Why is gravitational potential energy everywhere negative?

Potential energy is defined with respect to infinity (where it is zero). Because gravity is attractive, work must be done against the field to pull a mass away, so the energy stored at a finite distance is negative — indicating a bound system.

Why is escape velocity independent of the mass and direction of the projectile?

Setting kinetic energy ½mv² equal to the binding energy GMm/R gives v = √(2GM/R); the body's mass m cancels. Since escape requires mechanical energy to reach zero at infinity, any direction that avoids hitting the planet works.

Why is a geostationary satellite placed only over the equator at about 35,786 km?

To appear fixed, its period must match Earth's 24-hour rotation, giving r ≈ 42,164 km from the centre. Only an equatorial orbit at that radius, moving west to east, matches the Earth's spin.

What is the shell theorem and why is it important?

Newton's shell theorem states: (1) a uniform spherical shell attracts an external particle as if all its mass were at the centre, and (2) it exerts no net gravitational force on an internal particle. This lets us treat planets and stars as point masses for external calculations.

How are binding energy and escape velocity related?

Binding energy is the minimum energy to disassemble a system to infinity. Setting it equal to kinetic energy ½mv² gives v = √(2GM/R), which is exactly the escape velocity formula. Both measure how tightly gravity holds a body.

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