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Class 11 Physics · NCERT Chapter 7

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.

ChapterGravitationClassClass 11SubjectPhysicsBoardCBSEExamsCBSE · JEE · NEET

Written byDeep Narayan· Science & Mathematics EducatorReviewed byPushpanjali

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.

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.

F=Gm1m2r2F = G \frac{m_1 m_2}{r^2}
Gravitational force between two point masses

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.

MCQ Questions

1If the distance between two bodies is doubled, the gravitational force between them becomes —

2The gravitational force between two bodies is —

3The SI unit of the gravitational constant G is —

4Two 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

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.

g=GMR2g = \frac{GM}{R^2}
Relation between g and G at the surface

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.

gh=g(RR+h)2≈g(1−2hR)gd=g(1−dR)g_h = g\left(\frac{R}{R+h}\right)^2 \approx g\left(1 - \frac{2h}{R}\right) \qquad g_d = g\left(1 - \frac{d}{R}\right)
Variation with height h above and depth d below the surface
  • 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

1The acceleration due to gravity at a height R above the Earth's surface (where R is the Earth's radius) is —

2The value of acceleration due to gravity at the centre of the Earth is —

3The acceleration due to gravity is maximum at —

4A 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

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.

E=Fm=GMr2E = \frac{F}{m} = \frac{GM}{r^2}
Gravitational field intensity of a point mass M

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.

MCQ Questions

1The SI unit of gravitational field intensity is —

2The gravitational field intensity of a point mass M at a distance r varies as —

3Gravitational 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

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.

V=−GMrU=−GMmrV = -\frac{GM}{r} \qquad U = -\frac{GMm}{r}
Potential and potential energy of a point mass m at distance r from M

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.

U=mgh(h≪R)U = mgh \qquad (h \ll R)
Potential energy near the Earth's surface

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

1The gravitational potential at a point at a distance r from a mass M is proportional to —

2The gravitational potential energy of a system of two masses is —

3The 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?

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.

Escape Velocity

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.

ve=2GMR=2gRv_e = \sqrt{\frac{2GM}{R}} = \sqrt{2gR}
Escape velocity from a planet of mass M, radius 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 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

1The escape velocity from the Earth's surface is approximately —

2The escape velocity from a planet depends upon —

3If the radius of a planet is doubled keeping its density constant, the escape velocity —

4The 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.
T2=4π2GMa3vo=GMrT=2πr3GMT^2 = \frac{4\pi^2}{GM} a^3 \qquad v_o = \sqrt{\frac{GM}{r}} \qquad T = 2\pi\sqrt{\frac{r^3}{GM}}
Kepler's third law, orbital velocity and period of a satellite

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.

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.

MCQ Questions

1According to Kepler's second law, the line joining a planet to the Sun sweeps equal areas in —

2A geostationary satellite revolves around the Earth —

3If the radius of a satellite's circular orbit is doubled, its time period becomes —

4The total mechanical energy of a satellite in a circular orbit of radius r is —

5The 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)

Revision

Key formulas at a glance

Memorise these before attempting numericals — most exam questions hinge on one of them.

Universal law of gravitation

F=Gm1m2r2F = G \frac{m_1 m_2}{r^2}

G = 6.67 × 10⁻¹¹ N m² kg⁻².

Acceleration due to gravity

g=GMR2g = \frac{GM}{R^2}

9.8 m s⁻² on the Earth's surface.

Variation with height

gh=g(RR+h)2g_h = g\left(\frac{R}{R+h}\right)^2

≈ g(1 − 2h/R) for h ≪ R.

Variation with depth

gd=g(1−dR)g_d = g\left(1 - \frac{d}{R}\right)

zero at the centre of the Earth.

Gravitational potential

V=−GMrV = -\frac{GM}{r}

negative everywhere; zero at infinity.

Gravitational potential energy

U=−GMmrU = -\frac{GMm}{r}

near the Earth's surface U = mgh.

Escape velocity

ve=2GMR=2gRv_e = \sqrt{\frac{2GM}{R}} = \sqrt{2gR}

≈ 11.2 km s⁻¹ for the Earth.

Orbital velocity

vo=GMrv_o = \sqrt{\frac{GM}{r}}

≈ 7.9 km s⁻¹ just above the surface.

Orbital period

T=2πr3GM  ⟹  T2∝r3T = 2\pi\sqrt{\frac{r^3}{GM}} \implies T^2 \propto r^3

Kepler's third law.

Energy of an orbiting satellite

E=−GMm2rE = -\frac{GMm}{2r}

K = +GMm/2r, U = −GMm/r.

Exam tips

How this chapter is asked

Where this topic appears in CBSE, JEE Main and NEET papers.

  • F ∝ m₁m₂/r² — if distance doubles, force becomes one-fourth.
  • g = GM/R²; g is minimum at the equator, zero at the Earth's centre.
  • Escape velocity = √2 × orbital velocity for a near-surface satellite.
  • Geostationary satellite: height ≈ 35,786 km, period 24 h, equatorial, west to east.
  • Total satellite energy is negative: E = −GMm/2r, so energy must increase to move to a higher orbit.
  • Kepler's second law = conservation of angular momentum (central force, zero torque).

Solved problems

Worked examples

JEE / NEET-style numericals, solved step by step.

1

Gravitational force between the Earth and the Moon

Using 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.

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

Answer

≈ 2.0 × 10²⁰ N

2

Escape velocity from the Moon

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)

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

Answer

≈ 2.38 km s⁻¹

3

Orbital velocity of a satellite at 300 km

A 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.

  1. (a) v₀ = √(gR²/(R + h)) = √(9.8 × (6.4 × 10⁶)² ÷ 6.7 × 10⁶).
  2. gR²/(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.

Answer

v₀ ≈ 7.74 km s⁻¹; T ≈ 90.6 min

FAQ

Common questions

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

Two effects add up at the poles: the Earth is flattened (radius smaller at the poles, so g = GM/R² is larger) and the equatorial rotation reduces the effective g through the term Rω²cos²λ, which vanishes only 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 within the radius below a point contributes (shell theorem). At the centre no mass is enclosed, so the net force — and hence g — is zero.

Why is gravitational potential energy everywhere negative?

Potential energy is defined with respect to infinity, where it is taken as 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 — a sign that the system is bound.

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

It comes from cancelling the kinetic energy ½mv² against the potential energy −GMm/R; the body's mass m cancels. Since escape needs the mechanical energy to reach zero at infinity, any direction without hitting the planet works.

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

To appear fixed, its period must equal Earth's 24-hour rotation, so r = (GMT²/4π²)¹ᐟ³ ≈ 42,164 km from the centre (≈ 35,786 km above the surface). Only an equatorial orbit moving west-to-east at that radius matches the Earth's spin.

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