By the end of this chapter you'll be able to…

  • 1State and apply Newton's universal law of gravitation
  • 2Explain how g varies with height, depth, and latitude
  • 3Define gravitational potential and potential energy
  • 4Derive escape velocity and orbital velocity
  • 5State and apply Kepler's three laws
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Why this chapter matters
Gravitation links the falling apple to orbiting planets. Newton's universal law, the variation of g, escape velocity, satellite motion, and Kepler's laws underpin space exploration, satellite communication, and our understanding of the cosmos.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

Gravitation

'I deduced that the forces which keep the planets in their orbs must be reciprocally as the squares of their distances from the centres about which they revolve.' — Isaac Newton

1. Chapter Overview

GRAVITATION is the UNIVERSAL force of attraction between all matter. This chapter begins with NEWTON'S LAW OF GRAVITATION, explores how g varies with altitude/depth/latitude, introduces GRAVITATIONAL POTENTIAL ENERGY, ESCAPE VELOCITY, and ORBITAL MECHANICS, and concludes with KEPLER'S LAWS of planetary motion.


2. Newton's Universal Law of Gravitation

  • Statement: Every point mass attracts every other point mass with a force DIRECTLY proportional to the product of their masses and INVERSELY proportional to the SQUARE of the distance between them.
  • Formula: F = GMm/r² (where G = 6.67 × 10⁻¹¹ N·m²/kg²)
  • G is the UNIVERSAL GRAVITATIONAL CONSTANT (same everywhere in the universe)
  • The force ALONG the line joining the two masses

Vector Form

  • F₁₂ = -Gm₁m₂/r² × r̂₁₂ (the negative sign indicates attraction)
  • The force on m₁ due to m₂ is TOWARD m₂

3. Acceleration Due to Gravity (g)

Variation with Height (Altitude)

  • g_h = g₀(1 — 2h/R) for h << R (near surface)
  • g decreases as height INCREASES

Variation with Depth

  • g_d = g₀(1 — d/R) for depth d below surface
  • At the centre of Earth (d = R): g = 0
  • g decreases LINEARLY with depth

Variation with Latitude (Rotation of Earth)

  • g_λ = g₀ — Rω²cos²λ (where λ = latitude)
  • Maximum at POLES (λ = 90°): g_pole = g₀
  • Minimum at EQUATOR (λ = 0°): g_equator = g₀ — Rω²
  • Earth's rotation causes about 0.034 m/s² reduction at equator

Key Comparison

Locationg (m/s²)Reason
Surface9.8Standard value
Pole9.83No centrifugal effect
Equator9.78Maximum centrifugal effect
Mountain top< 9.8Increased distance from centre
Deep mine< 9.8Reduced effective mass

4. Gravitational Field and Potential

  • Gravitational Field E_g = F/m = GM/r² (force per unit mass)
  • Gravitational Potential V_g = -GM/r (work per unit mass to bring from infinity)
  • Relation: E_g = -dV_g/dr
  • Gravitational POTENTIAL ENERGY: U = -GMm/r (U = 0 at infinity, NEGATIVE for bound systems)

Worked Problem

Q: Calculate the gravitational potential at a point 2R above Earth's surface. (M = 6×10²⁴ kg, R = 6400 km) A: Distance from centre = R + 2R = 3R. V = -GM/3R = -(6.67×10⁻¹¹×6×10²⁴)/(3×6.4×10⁶) = -2.08×10⁷ J/kg


5. Escape Velocity

  • Definition: The MINIMUM velocity needed for an object to escape Earth's gravitational pull (reach infinity with KE = 0)
  • Formula: v_esc = √(2GM/R) = √(2gR)
  • For Earth: v_esc = √(2×9.8×6.4×10⁶) = 11.2 km/s
  • Independent of mass of the escaping object

Comparison

Celestial BodyEscape Velocity
Earth11.2 km/s
Moon2.4 km/s
Sun618 km/s
Jupiter59.5 km/s
Black Hole> c (speed of light)

6. Satellites — Orbital Motion

Orbital Velocity

  • v₀ = √(GM/r) (for a circular orbit)
  • Near Earth's surface: v₀ ≈ √(gR) ≈ 7.92 km/s

Time Period

  • T = 2πr/v₀ = 2π√(r³/GM)
  • Geostationary Satellite: Orbits above EQUATOR at distance where T = 24 hours
    • Height ≈ 36,000 km above surface
    • Used for COMMUNICATION (appears stationary from Earth)

Energy of an Orbiting Satellite

  • KE = GMm/2r
  • PE = -GMm/r
  • Total Energy = KE + PE = -GMm/2r (NEGATIVE, meaning bound system)
  • To increase orbit radius (r ↑): Total energy increases (less negative), KE decreases, PE increases

7. Kepler's Laws of Planetary Motion

First Law (Law of Orbits)

  • Each planet moves in an ELLIPTICAL orbit with the Sun at ONE FOCUS

Second Law (Law of Areas)

  • The line joining planet to Sun sweeps EQUAL areas in EQUAL intervals of time
  • Consequence: v_areal = L/2m = CONSTANT → Angular momentum is CONSERVED

Third Law (Law of Periods)

  • T² ∝ a³ (where a = semi-major axis)
  • T²/T₁² = a³/a₁³
  • For a circular orbit: T² = (4π²/GM)a³

Worked Problem

Q: If Earth's orbital period is 1 year at radius R, what is the period of a planet at 4R? A: Using Kepler's third law: T²/T_E² = (4R/R)³ = 64. T = 8 years.


8. Common Mistakes

  1. g is NOT the same as G: G is universal constant (6.67×10⁻¹¹), g is acceleration = GM/R²
  2. Weightlessness ≠ zero gravity: Astronauts are in FREE FALL (g ≈ 8.7 m/s² at ISS height), but they feel weightless because the spacecraft is also accelerating at the same rate
  3. g decreases with height AND depth: But for different reasons
  4. Satellite KE is half of |PE|: For a circular orbit, KE = |PE|/2 = |E|
  5. Escape velocity from Earth is 11.2 km/s, NOT 7.92 km/s: 7.92 km/s is orbital velocity near surface

9. CBSE Exam Focus

  1. Newton's law of gravitation — derivation of g (3-mark)
  2. Variation of g with height, depth, latitude (5-mark)
  3. Escape velocity derivation (3-mark)
  4. Orbital velocity, time period of satellite (5-mark)
  5. Kepler's laws — statements and derivations
  6. Energy of orbiting satellite — binding energy

10. Key Formulas

  • F = GMm/r², G = 6.67×10⁻¹¹ N·m²/kg²
  • g = GM/R² (surface), g_h = g(1 — 2h/R), g_d = g(1 — d/R)
  • v_esc = √(2GM/R) = √(2gR)
  • v₀ = √(GM/r)
  • T = 2π√(r³/GM)
  • U = -GMm/r, E_total = -GMm/2r
  • T² ∝ a³ (Kepler's third law)

11. Self-Test (5+ Q&A)

Q1: Find g at a height equal to R above Earth's surface. A: Distance = 2R. g' = GM/(2R)² = (GM/R²)/4 = g/4 = 2.45 m/s².

Q2: What is the orbital velocity of a satellite at height 300 km? (R = 6400 km, M = 6×10²⁴ kg) A: r = 6400 + 300 = 6700 km = 6.7×10⁶ m. v₀ = √(GM/r) = √(6.67×10⁻¹¹×6×10²⁴/6.7×10⁶) = √(5.97×10⁷) = 7.73 km/s.

Q3: Why is g zero at the centre of the Earth? A: At the centre, mass of Earth surrounds the point equally in all directions. The NET gravitational force is zero because forces from all sides cancel.

Q4: Calculate escape velocity on the Moon (g_m = 1.62 m/s², R_m = 1740 km). A: v_esc = √(2gR) = √(2×1.62×1.74×10⁶) = √(5.64×10⁶) = 2.37 km/s.

Q5: State Kepler's second law. Which conservation law does it correspond to? A: Kepler's second law: The line joining planet to Sun sweeps equal areas in equal times. It corresponds to the CONSERVATION OF ANGULAR MOMENTUM.


12. Conclusion

Gravitation connects TERRESTRIAL physics (g, free fall) with CELESTIAL physics (planetary motion). Newton showed that the SAME force makes an apple fall and holds the Moon in orbit. The concepts of orbital mechanics — escape velocity, geostationary satellites, Kepler's laws — have direct practical applications in space exploration and satellite technology.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Universal law of gravitation
F = G M m / r^2, G = 6.67e-11 N m^2/kg^2
g = GM/R^2 at the surface.
Variation of g
g_h = g(1 - 2h/R); g_d = g(1 - d/R)
g decreases with both height and depth; zero at Earth's centre.
Escape and orbital velocity
v_esc = sqrt(2GM/R) = sqrt(2gR); v_o = sqrt(GM/r)
Earth escape velocity is 11.2 km/s; independent of the object's mass.
Satellite energy and Kepler's third law
E = -GMm/2r; T^2 proportional to a^3
Total energy is negative for a bound orbit; T = 2 pi sqrt(r^3/GM).
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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT
Confusing g with G
G is the universal constant (6.67e-11); g = GM/R^2 is the local acceleration due to gravity.
WATCH OUT
Thinking astronauts experience zero gravity
They are in free fall; gravity is still strong at orbital height, but they feel weightless because they and the spacecraft accelerate together.
WATCH OUT
Mixing up orbital and escape velocity
Near-surface orbital velocity is about 7.9 km/s; escape velocity is sqrt(2) times larger at 11.2 km/s.
WATCH OUT
Taking satellite total energy as positive
For a bound orbit total energy E = -GMm/2r is negative; KE = |PE|/2.

Practice problems

Try each one yourself before tapping "Show solution". Active recall > rereading.

Q1EASY· Variation of g
Find g at a height equal to R above Earth's surface.
Show solution
Distance from centre = 2R. g' = GM/(2R)^2 = g/4 = 2.45 m/s^2.
Q2MEDIUM· Orbital Velocity
Find the orbital velocity of a satellite at 300 km height (R = 6400 km, M = 6e24 kg).
Show solution
r = 6700 km = 6.7e6 m. v_o = sqrt(GM/r) = sqrt(6.67e-11 x 6e24 / 6.7e6) = sqrt(5.97e7) = 7.73 km/s.
Q3MEDIUM· Escape Velocity
Calculate the escape velocity on the Moon (g = 1.62 m/s^2, R = 1740 km).
Show solution
v_esc = sqrt(2gR) = sqrt(2 x 1.62 x 1.74e6) = sqrt(5.64e6) = 2.37 km/s.
Q4MEDIUM· Kepler
If Earth's period is 1 year at radius R, find the period of a planet at radius 4R.
Show solution
Kepler's third law: T^2 proportional to a^3. (T/T_E)^2 = (4R/R)^3 = 64, so T = 8 years.
Q5EASY· Concept
Why is g zero at the centre of the Earth?
Show solution
At the centre, Earth's mass surrounds the point symmetrically in all directions, so the gravitational forces cancel and the net force (hence g) is zero.

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

  • F = GMm/r^2; g = GM/R^2 at the surface.
  • g decreases with height (1 - 2h/R) and depth (1 - d/R); zero at centre.
  • g is maximum at the poles and minimum at the equator due to rotation.
  • Gravitational potential V = -GM/r; PE U = -GMm/r (negative for bound systems).
  • Escape velocity v_esc = sqrt(2gR) = 11.2 km/s for Earth, independent of mass.
  • Orbital velocity v_o = sqrt(GM/r); satellite total energy E = -GMm/2r.
  • Kepler: elliptical orbits; equal areas in equal times (angular momentum conservation); T^2 proportional to a^3.

CBSE marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: 6-8 marks across the chapter

Question typeMarks eachTypical countWhat it tests
Satellites / orbital mechanics3-51Orbital velocity, time period, satellite energy
Variation of g / escape velocity31Height, depth, latitude, escape velocity
Kepler's laws2-31Statements and the period-radius relation
Prep strategy
  • Memorise g variation formulas and their causes
  • Derive escape and orbital velocity from energy
  • Learn Kepler's three laws and their basis
  • Note that satellite total energy is negative

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Satellite communication

Geostationary satellites relay TV, phone, and internet signals from a fixed position in the sky.

Space missions

Escape velocity and orbital mechanics determine launch requirements and interplanetary trajectories.

GPS and navigation

Precise orbital calculations keep global positioning systems accurate.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

  1. Distinguish G and g clearly in derivations
  2. Use energy conservation to derive escape velocity
  3. State Kepler's laws precisely and link the second to angular momentum
  4. Remember satellite energy relations (KE = |PE|/2)

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

  • Derive the time period for elliptical orbits and apply Kepler's third law with the semi-major axis.
  • Analyse gravitational potential inside and outside a uniform sphere using the shell theorem.

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

CBSE Class 11 Physics examHigh
JEE Main and Advanced (Gravitation)High
NEET PhysicsHigh

Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

At the height of the International Space Station gravity is only slightly weaker than at the surface (about 8.7 m/s^2). The astronauts feel weightless because they and their spacecraft are in continuous free fall around the Earth, both accelerating toward the Earth at the same rate. With no normal reaction force between them and the craft, there is no sensation of weight -- this is apparent weightlessness, not zero gravity.

A geostationary satellite is placed in a circular orbit directly above the equator at a height of about 36,000 km, where its orbital period exactly equals Earth's rotation period of 24 hours. Because it revolves at the same angular speed as the Earth turns and in the same direction, it appears stationary relative to the ground -- ideal for communication and weather satellites.
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Last reviewed on 29 May 2026. Written and reviewed by subject-matter experts — read about our process.
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