Class 11Physics · MechanicsFull chapter

Gravitation

The whole chapter in one place — read it, then test yourself. Clear notes, formula sheet, a practice quiz, and worked NCERT solutions & PYQs.

In this chapter
Kepler's Laws of Planetary MotionNewton's Universal Law of GravitationAcceleration Due to Gravity: Variation with Altitude and DepthGravitational Potential EnergyEscape VelocityOrbital Velocity and Energy of an Orbiting Satellite

Kepler's Laws of Planetary Motion

Quick answer Kepler's three laws describe planetary orbits as ellipses with equal areas swept in equal times, and give T² ∝ r³ for all planets.

Before Newton's law of gravitation was formulated, Johannes Kepler analysed decades of planetary observations and arrived at three empirical laws that describe how planets move around the Sun.

  • Law of Orbits: Every planet revolves around the Sun in an elliptical orbit, with the Sun located at one of the two foci of the ellipse (not at the centre).
  • Law of Areas: The line joining a planet to the Sun (the radius vector) sweeps out equal areas in equal intervals of time. This means the planet moves fastest when closest to the Sun (perihelion) and slowest when farthest (aphelion). Since the gravitational force on the planet always acts along the line joining it to the Sun, it is a central force and produces zero torque about the Sun. As torque equals the rate of change of angular momentum, the planet's angular momentum about the Sun stays constant — the law of areas is therefore a direct consequence of conservation of angular momentum.
  • Law of Periods: The square of the time period T of revolution of a planet is directly proportional to the cube of the semi-major axis a of its elliptical orbit, i.e., T2 ∝ a3, with the same constant of proportionality for every planet orbiting the Sun.

Worked Example:

Given: Earth's mean orbital radius rE = 1 AU with period TE = 1 year. A planet X orbits the Sun at rX = 4 AU. Find its period of revolution.

Formula: By Kepler's third law, T2/r3 = constant, so (TX/TE)2 = (rX/rE)3.

Substitution: (TX/1)2 = (4/1)3 = 64

Result: TX = √64 = 8 years.

Law of Areas dA/dt = L/2m = constant Areal velocity is constant; L is angular momentum about the Sun, m is the planet's mass
Law of Periods T² ∝ r³ (T²/r³ = constant for all planets) r is the orbital (semi-major axis) radius
Remember
  • Law of Orbits: planetary paths are ellipses with the Sun at one focus, not perfect circles.
  • Law of Areas: equal areas are swept in equal times, so a planet moves faster near the Sun and slower far from it.
  • The law of areas follows directly from conservation of angular momentum because gravity is a central force (zero torque about the Sun).
  • Law of Periods: T² ∝ r³ holds for every planet orbiting the Sun with the same proportionality constant.
  • Newton later used the law of periods to confirm that gravitational force must vary as the inverse square of distance.

Newton's Universal Law of Gravitation

Quick answer Every pair of point masses attracts each other with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

Newton's universal law of gravitation states that any two point masses m1 and m2, separated by a distance r, attract each other with a force F given by:

F = G·m1·m2 / r2

where G is the universal gravitational constant, G = 6.674 × 10-11 N m2 kg-2. This law is universal — it applies to any two masses anywhere in the universe, not just to celestial bodies.

  • The force is always attractive and acts along the line joining the two masses (a central force).
  • It obeys Newton's third law: the force exerted by m1 on m2 is equal in magnitude and opposite in direction to the force exerted by m2 on m1.
  • For a system of many masses, the net gravitational force on any one mass is the vector sum of the forces due to all other masses individually — the principle of superposition.
  • G is independent of the medium between the masses and of the nature of the masses.

Worked Example:

Given: m1 = 50 kg, m2 = 80 kg, r = 0.5 m, G = 6.67 × 10-11 N m2 kg-2.

Formula: F = G·m1·m2 / r2

Substitution: F = (6.67 × 10-11 × 50 × 80) / (0.5)2 = (6.67 × 10-11 × 4000) / 0.25

Result: F = 1.067 × 10-6 N.

Newton's Law of Gravitation F = G·m₁·m₂ / r² G = 6.674×10⁻¹¹ N m² kg⁻², r = separation between point masses
Vector form F₁₂ = −G·m₁·m₂/r² · r̂₁₂ negative sign shows the force is attractive, directed toward the other mass
Remember
  • Gravitational force is always attractive and acts along the line joining the two masses.
  • It obeys Newton's third law — action and reaction pairs are equal and opposite.
  • G = 6.674 × 10⁻¹¹ N m² kg⁻² is a universal constant, the same everywhere and for all materials.
  • For more than two masses, forces add vectorially (principle of superposition).
  • The force falls off rapidly (inverse-square) with distance but mathematically never becomes exactly zero.

Acceleration Due to Gravity: Variation with Altitude and Depth

Quick answer g = GM/R² at Earth's surface, and it decreases both with height above the surface and with depth below it, becoming zero at the Earth's centre.

Equating the gravitational force on a mass m at the Earth's surface to mg gives the surface value of acceleration due to gravity: g = GM/R2, where M is Earth's mass and R its radius. Notice g does not depend on the falling body's own mass m.

Variation with altitude: At height h above the surface, the distance from Earth's centre becomes (R + h), so gh = GM/(R+h)2 = g·R2/(R+h)2. For h much smaller than R, this simplifies (by the binomial approximation) to gh ≈ g(1 − 2h/R), showing g decreases as we go higher.

Variation with depth: Below the surface, at depth d, only the mass enclosed within a sphere of radius (R − d) contributes to gravity (the outer shell exerts no net gravitational force inside it). Treating the Earth as a uniform sphere of density ρ, this gives gd = g(1 − d/R), so g decreases linearly with depth and becomes zero at the centre (d = R).

Worked Example:

Given: h = 64 km = 6.4 × 104 m, R = 6400 km = 6.4 × 106 m, g = 9.8 m/s2.

Formula: gh ≈ g(1 − 2h/R), valid since h ≪ R.

Substitution: 2h/R = (2 × 6.4 × 104)/(6.4 × 106) = 0.02; gh = 9.8 × (1 − 0.02) = 9.8 × 0.98

Result: gh = 9.604 m/s2 ≈ 9.6 m/s2.

Surface gravity g = GM/R²
Variation with altitude (general) g_h = g·R²/(R+h)²
Variation with altitude (h ≪ R) g_h ≈ g(1 − 2h/R)
Variation with depth g_d = g(1 − d/R) g = 0 at the centre, d = R
Remember
  • g = GM/R² at Earth's surface, depending only on the planet's mass and radius, not on the falling body's mass.
  • g decreases with altitude as the effective distance from the centre increases: g_h ≈ g(1 − 2h/R) for h ≪ R.
  • g decreases with depth because only the enclosed mass within radius (R−d) contributes: g_d = g(1 − d/R).
  • g becomes exactly zero at the Earth's centre.
  • g is maximum at the Earth's surface and decreases on going either up or down from it.

Gravitational Potential Energy

Quick answer The gravitational potential energy of a two-mass system is U = −GMm/r, taken as zero at infinite separation, and is always negative for any finite separation.

Gravitational potential energy (PE) of a system of two masses M and m separated by distance r is defined as the work done by an external agent (against the gravitational force, quasi-statically) in bringing m from infinity to that separation, with PE chosen to be zero at infinite separation. This gives:

U = −GMm/r

The negative sign shows that the system is bound: since gravity is attractive, external work must be done to pull the masses apart, so bringing them together releases energy, making U negative at all finite r. As r increases, U increases (becomes less negative), approaching zero as r → ∞.

The work done in moving a mass from r1 to r2 equals the change in potential energy:

W = GMm(1/r1 − 1/r2)

Near the Earth's surface, for small heights h (h ≪ R), this reduces to the familiar U ≈ mgh (measured relative to the surface).

Worked Example:

Given: A satellite of mass m = 1000 kg orbits at height h = 400 km. Earth's mass M = 6 × 1024 kg, radius R = 6.4 × 106 m, so orbital radius r = R + h = 6.8 × 106 m.

Formula: U = −GMm/r

Substitution: GMm = 6.67 × 10-11 × 6 × 1024 × 1000 = 4.002 × 1017; U = −(4.002 × 1017) / (6.8 × 106)

Result: U ≈ −5.89 × 1010 J.

Gravitational potential energy U = −GMm/r U → 0 as r → ∞
Work done moving between two separations W = GMm(1/r₁ − 1/r₂)
PE near the surface (h ≪ R) U ≈ −GMm/R + mgh
Remember
  • Gravitational PE is defined to be zero at infinite separation, so it is negative at every finite separation.
  • U = −GMm/r; as r increases, U increases (becomes less negative), approaching zero at infinity.
  • Work done against gravity to move a mass from r₁ to r₂: W = GMm(1/r₁ − 1/r₂).
  • For small heights near the surface (h ≪ R), U reduces to the familiar mgh.
  • The sign of the total mechanical energy (KE + PE), not force alone, determines whether an orbit is bound (E < 0) or unbound (E ≥ 0).

Escape Velocity

Quick answer Escape velocity is the minimum speed needed to launch a body from a planet's surface so that it never falls back, given by v_e = √(2GM/R).

Escape velocity (ve) is the minimum speed with which a body must be projected from a planet's surface so that it just escapes the planet's gravitational pull and reaches infinity with (at minimum) zero residual speed.

Using conservation of mechanical energy: at the surface the body has kinetic energy (1/2)mve2 and potential energy −GMm/R; at infinity, in the minimum-escape case, both KE and PE are zero. Setting total initial energy equal to total final energy (zero):

(1/2)mve2 − GMm/R = 0

Solving for ve:

ve = √(2GM/R) = √(2gR)

Escape velocity is independent of the mass of the escaping body and of the direction in which it is projected (as long as it clears any obstruction). For Earth, ve ≈ 11.2 km/s.

Worked Example:

Given: M = 6 × 1024 kg, R = 6.4 × 106 m, G = 6.67 × 10-11 N m2 kg-2.

Formula: ve = √(2GM/R)

Substitution: 2GM = 2 × 6.67 × 10-11 × 6 × 1024 = 8.004 × 1014; 2GM/R = (8.004 × 1014)/(6.4 × 106) = 1.2506 × 108 m2/s2

Result: ve = √(1.2506 × 108) ≈ 1.118 × 104 m/s ≈ 11.2 km/s.

Escape velocity v_e = √(2GM/R) = √(2gR) independent of the escaping body's mass
Remember
  • Escape velocity is the minimum launch speed for a body to permanently leave a planet's gravitational field.
  • v_e = √(2GM/R) = √(2gR); it does not depend on the mass or direction of the escaping object.
  • Earth's escape velocity is approximately 11.2 km/s.
  • A planet or moon retains its atmosphere better when its escape velocity is large compared with the thermal speeds of gas molecules.
  • For a circular orbit at the same radius, escape velocity is √2 times the orbital velocity.

Orbital Velocity and Energy of an Orbiting Satellite

Quick answer A satellite in circular orbit moves at v₀ = √(GM/r) with total mechanical energy E = −GMm/2r, always negative, showing it is gravitationally bound.

For a satellite of mass m moving in a circular orbit of radius r around a planet of mass M, gravity supplies exactly the centripetal force needed to maintain the orbit:

GMm/r2 = mv02/r

Solving for the orbital speed:

v0 = √(GM/r)

The time period is the orbit's circumference divided by orbital speed:

T = 2πr/v0 = 2π√(r3/GM)

which is consistent with Kepler's third law, T2 ∝ r3.

The satellite's total mechanical energy is the sum of kinetic and potential energy:

E = (1/2)mv02 − GMm/r = (1/2)(GMm/r) − GMm/r = −GMm/2r

This is always negative, showing the satellite is gravitationally bound to the planet. The magnitude GMm/2r is called the binding energy — the minimum energy that must be supplied to remove the satellite to infinity.

Worked Example:

Given: A satellite of mass m = 1000 kg orbits at height h = 400 km above Earth's surface. R = 6.4 × 106 m, so r = 6.8 × 106 m; GM = 4.0 × 1014 N m2/kg.

Formula: v0 = √(GM/r); E = −GMm/2r

Substitution: v0 = √((4.0 × 1014)/(6.8 × 106)) = √(5.88 × 107) ≈ 7670 m/s; E = −(4.0 × 1014 × 1000)/(2 × 6.8 × 106) = −(4.0 × 1017)/(1.36 × 107)

Result: v0 ≈ 7.67 km/s, and E ≈ −2.94 × 1010 J (negative, confirming a bound orbit).

Orbital velocity v₀ = √(GM/r)
Time period of satellite T = 2π√(r³/GM)
Total energy of satellite E = −GMm/2r always negative — bound orbit
Binding energy E_binding = +GMm/2r
Remember
  • Orbital velocity v₀ = √(GM/r); it decreases as the orbital radius r increases.
  • Time period T = 2π√(r³/GM), matching Kepler's third law (T² ∝ r³).
  • Total mechanical energy of an orbiting satellite E = −GMm/2r is always negative — the satellite is gravitationally bound.
  • Kinetic energy = +GMm/2r equals the magnitude of the total energy; binding energy also equals GMm/2r.
  • A geostationary satellite has a period equal to Earth's rotation period (about 24 h) and orbits at a fixed height (~36,000 km) above the equator.

The formula sheet

Every formula in this chapter, in one place — screenshot it before your exam.

dA/dt = L/2m = constant
Law of Areas
T² ∝ r³ (T²/r³ = constant for all planets)
Law of Periods
F = G·m₁·m₂ / r²
Newton's Law of Gravitation
F₁₂ = −G·m₁·m₂/r² · r̂₁₂
Vector form
g = GM/R²
Surface gravity
g_h = g·R²/(R+h)²
Variation with altitude (general)
g_h ≈ g(1 − 2h/R)
Variation with altitude (h ≪ R)
g_d = g(1 − d/R)
Variation with depth
U = −GMm/r
Gravitational potential energy
W = GMm(1/r₁ − 1/r₂)
Work done moving between two separations
U ≈ −GMm/R + mgh
PE near the surface (h ≪ R)
v_e = √(2GM/R) = √(2gR)
Escape velocity
v₀ = √(GM/r)
Orbital velocity
T = 2π√(r³/GM)
Time period of satellite
E = −GMm/2r
Total energy of satellite
E_binding = +GMm/2r
Binding energy

Test yourself

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0 correct · 0/12 answered
Q1 Kepler's Laws easy

Kepler's law of areas (equal areas swept in equal times) is a direct consequence of the conservation of:

Q2 Kepler's Third Law medium

A planet revolves around the Sun at a mean distance 4 times that of Earth's orbital radius. If Earth's period is 1 year, the planet's period of revolution is:

Q3 Universal Law of Gravitation easy

If the distance between two point masses is doubled while their masses remain unchanged, the gravitational force between them becomes:

Q4 Universal Law of Gravitation medium

Two point masses of 10 kg and 20 kg are separated by 2 m. The gravitational force between them (G = 6.67×10⁻¹¹ N m² kg⁻²) is nearly:

Q5 Variation of g with altitude easy

The acceleration due to gravity decreases with height above the Earth's surface mainly because:

Q6 Variation of g with depth medium

At what depth below the Earth's surface does the acceleration due to gravity reduce to half its surface value? (R = Earth's radius)

Q7 Variation of g with depth easy

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

Q8 Gravitational Potential Energy medium

The negative sign in the expression U = −GMm/r for gravitational potential energy indicates that:

Q9 Escape Velocity hard

A hypothetical planet has twice the mass of Earth but the same radius as Earth. Compared to Earth's escape velocity of 11.2 km/s, the escape velocity from this planet is approximately:

Q10 Orbital Velocity medium

For a satellite orbiting very close to a planet's surface, the relation between orbital velocity v₀ and escape velocity v_e (at the same radius) is:

Q11 Energy of Orbiting Satellite easy

The total mechanical energy of a satellite moving in a stable circular orbit around the Earth is:

Q12 Orbital Velocity hard

A satellite orbits at a radius 4 times the Earth's radius R. If v₀ is the orbital velocity for an orbit close to Earth's surface (radius R), the orbital velocity at radius 4R is:

NCERT solutions & previous-year questions

Step-by-step model answers — tap a question to reveal the full solution.

NCERT questions 6

1 State Kepler's three laws of planetary motion.Kepler's Laws

Law of Orbits: Every planet revolves around the Sun in an elliptical orbit with the Sun at one of the foci of the ellipse.

Law of Areas: The radius vector drawn from the Sun to a planet sweeps out equal areas in equal intervals of time, so the areal velocity of the planet is constant. This is a consequence of conservation of angular momentum, since the gravitational force on the planet acts along the line joining it to the Sun (a central force) and hence exerts no torque about the Sun.

Law of Periods: The square of the time period of revolution of a planet around the Sun is proportional to the cube of the semi-major axis of its elliptical orbit, i.e., T2 ∝ a3.

2 The Earth revolves around the Sun in a nearly circular orbit of radius 1.5×10^11 m, with a time period of 3.15×10^7 s (1 year). Using Newton's law of gravitation, estimate the mass of the Sun. (G = 6.67×10⁻¹¹ N m² kg⁻²)Kepler's Third Law / Gravitation

Given: r = 1.5×1011 m, T = 3.15×107 s, G = 6.67×10-11 N m2 kg-2

Formula: Equating gravitational force to centripetal force: GMsm/r2 = mr(2π/T)2, giving Ms = 4π2r3/(GT2).

Substitution: r3 = (1.5×1011)3 = 3.375×1033 m3; T2 = (3.15×107)2 = 9.92×1014 s2

Ms = (4π2 × 3.375×1033) / (6.67×10-11 × 9.92×1014)

Result: Ms ≈ 2.0×1030 kg, close to the accepted mass of the Sun.

3 The weight of a body on the surface of the Earth is 72 N. What will be its weight at a height equal to half the radius of the Earth?Variation of g with altitude

Given: Surface weight W = mg = 72 N; height h = R/2

Formula: gh = gR2/(R+h)2, so Wh = W × R2/(R+h)2

Substitution: R + h = 3R/2, so R2/(R+h)2 = R2/(9R2/4) = 4/9; Wh = 72 × 4/9

Result: Wh = 32 N

4 Calculate the escape velocity of a body from the surface of the Earth. (Mass of Earth M = 6×10^24 kg, radius R = 6.4×10^6 m, G = 6.67×10⁻¹¹ N m² kg⁻²)Escape Velocity

Given: M = 6×1024 kg, R = 6.4×106 m, G = 6.67×10-11 N m2 kg-2

Formula: ve = √(2GM/R)

Substitution: 2GM = 2 × 6.67×10-11 × 6×1024 = 8.004×1014; 2GM/R = 8.004×1014/6.4×106 = 1.2506×108 m2/s2

Result: ve = √(1.2506×108) ≈ 1.118×104 m/s ≈ 11.2 km/s

5 Derive the expression for the orbital velocity of a satellite revolving close to the Earth's surface and calculate its value. (g = 9.8 m/s², R = 6.4×10^6 m)Orbital Velocity

Derivation: For a satellite of mass m in a circular orbit of radius R close to the surface, gravity supplies the centripetal force: mg = mv02/R, so v0 = √(gR).

Given: g = 9.8 m/s2, R = 6.4×106 m

Substitution: v0 = √(9.8 × 6.4×106) = √(6.272×107)

Result: v0 ≈ 7.92×103 m/s ≈ 7.92 km/s

6 A satellite orbits the Earth at a height of 3600 km above the surface. Taking the Earth's radius as 6400 km and GM = 4.0×10^14 N m²/kg, find the orbital period of the satellite.Time Period of a Satellite

Given: h = 3600 km, R = 6400 km ⟹ r = R + h = 10000 km = 1×107 m; GM = 4.0×1014 N m2/kg

Formula: v0 = √(GM/r); T = 2πr/v0

Substitution: v0 = √(4.0×1014/1×107) = √(4.0×107) ≈ 6325 m/s; T = 2π×1×107/6325 ≈ 6.2832×107/6325

Result: T ≈ 9933 s ≈ 2.76 hours

Previous-year board questions 4

Q1 Derive an expression for the orbital velocity of a satellite revolving in a circular orbit close to the Earth's surface. Hence obtain an expression for its time period in terms of the Earth's radius R and g. CBSE 2023 3 marks

Consider a satellite of mass m revolving in a circular orbit of radius R (close to the Earth's surface) with orbital speed v0. Gravity provides the necessary centripetal force:

GMm/R2 = mv02/R

Since g = GM/R2, the left side equals mg, so:

mg = mv02/R ⟹ v0 = √(gR)

The time period equals the circumference divided by speed:

T = 2πR/v0 = 2πR/√(gR) = 2π√(R/g)

Thus, v0 = √(gR) and T = 2π√(R/g).

Q2 Define escape velocity. Derive an expression for the escape velocity of a body from the surface of the Earth. CBSE 2022 2 marks

Escape velocity is the minimum speed with which a body must be projected from the surface of a planet so that it just escapes the planet's gravitational field and reaches infinity with zero residual speed.

Derivation: By conservation of energy, at the surface the body has kinetic energy (1/2)mve2 and potential energy −GMm/R. At infinity, in the minimum-escape case, both are zero. Equating total energies:

(1/2)mve2 − GMm/R = 0

ve2 = 2GM/R ⟹ ve = √(2GM/R) = √(2gR)

Note that ve is independent of the mass of the escaping body.

Q3 (a) State Kepler's law of areas and show that it follows from the conservation of angular momentum. (b) A satellite revolves around the Earth in a circular orbit at a height of 400 km above the Earth's surface. Calculate its orbital speed. (Take Earth's radius R = 6400 km, GM = 4.0×10^14 N m²/kg) CBSE 2020 5 marks

(a) Kepler's law of areas states that the line joining a planet to the Sun sweeps out equal areas in equal intervals of time, i.e., the areal velocity dA/dt is constant.

The gravitational force on a planet acts along the line joining it to the Sun, so its torque about the Sun is zero. Since torque = dL/dt, and the torque is zero, the angular momentum L of the planet about the Sun stays constant. The area swept in time dt is dA = (1/2)|r × v|dt, so dA/dt = L/2m, which is constant since both L and m are constant. This proves the law of areas is equivalent to conservation of angular momentum.

(b) Given: h = 400 km, R = 6400 km ⟹ r = R + h = 6800 km = 6.8×106 m; GM = 4.0×1014 N m2/kg

Formula: v0 = √(GM/r)

Substitution: v0 = √(4.0×1014/6.8×106) = √(5.88×107)

Result: v0 ≈ 7.67×103 m/s ≈ 7.67 km/s

Q4 The mass of the Earth is 6×10^24 kg and its radius is 6.4×10^6 m. Calculate the energy required to move a body of mass 100 kg from the surface of the Earth to infinity. (G = 6.67×10⁻¹¹ N m² kg⁻²) CBSE 2019 3 marks

Given: M = 6×1024 kg, R = 6.4×106 m, m = 100 kg, G = 6.67×10-11 N m2 kg-2

Formula: Since PE at infinity is zero, the energy required equals the magnitude of the gravitational potential energy at the surface: E = GMm/R

Substitution: GMm = 6.67×10-11 × 6×1024 × 100 = 4.002×1016; E = 4.002×1016/6.4×106

Result: E ≈ 6.25×109 J

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