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Gravitation – Class 11 Physics Chapter 7 Notes & NCERT PDF

Class 11PhysicsChapter 7NCERT book: Physics Part I - Class 11

Notes and a simple summary of Chapter 7 of the NCERT Class 11 Physics book, with the official chapter PDF, flashcards, an MCQ quiz and an AI tutor for your doubts.

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Chapter 7: Gravitation

Introduction

The fact that all objects fall towards the earth has been known forever; the motion of planets was studied by Indian and Greek astronomers for centuries. Newton's genius was to see that the force that makes an apple fall is the same force that keeps the moon in orbit. This chapter covers Kepler's laws, Newton's universal law of gravitation, the variation of g, gravitational potential energy, escape speed and satellites.

Key Concepts

7.1–7.2 Kepler's Laws

Based on Tycho Brahe's careful observations, Johannes Kepler gave three laws:

  • Law of orbits — every planet moves in an ellipse with the sun at one focus.
  • Law of areas — the line joining a planet to the sun sweeps equal areas in equal intervals of time. This follows from conservation of angular momentum, because gravity is a central force; so a planet moves fastest at perihelion and slowest at aphelion.
  • Law of periods — the square of the period is proportional to the cube of the semi-major axis: T² ∝ a³.
  • 7.3 Universal Law of Gravitation

    Every body in the universe attracts every other body with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

    F = G m₁m₂/r², directed along the line joining them. For extended bodies: a uniform spherical shell attracts an outside point mass as if all its mass were at its centre, and exerts zero force on a point mass inside it.

    7.4 The Gravitational Constant

    G was first measured by Henry Cavendish (1798) with a torsion balance. Accepted value: G = 6.67 × 10⁻¹¹ N m² kg⁻². Knowing G allowed "weighing the earth".

    7.5 Acceleration due to Gravity on the Earth's Surface

    g = G M_E/R_E². Because G and g are known, the earth's mass can be found.

    7.6 Variation of g

  • Height h above the surface (h ≪ R_E): g(h) ≈ g(1 − 2h/R_E).
  • Depth d below the surface: g(d) = g(1 − d/R_E); g is zero at the centre.
  • So g decreases both above and below the surface.

    7.7 Gravitational Potential Energy

    For a mass m at distance r from the earth's centre, V(r) = −G M_E m/r (taking zero at infinity). Near the surface, the difference reduces to m g h. The gravitational potential is the potential energy per unit mass. The potential energy of a system of particles is the sum over all pairs.

    7.8 Escape Speed

    The minimum speed that lets a body escape the earth's gravity: ½ m v² = G M_E m/R_E ⇒ v_e = √(2 G M_E/R_E) = √(2 g R_E) ≈ 11.2 km s⁻¹. For the moon it is only about 2.3 km s⁻¹, about five times smaller, which is why the moon has no atmosphere.

    7.9 Earth Satellites

    A satellite in a circular orbit of radius R_E + h: gravity supplies the centripetal force ⇒ orbital speed v = √(G M_E/(R_E + h)) and period T = 2π √((R_E + h)³/(G M_E)), which is Kepler's third law. For a satellite close to the surface, T ≈ 85 minutes. The moon is a natural satellite; artificial satellites are used for communication, weather and remote sensing.

    7.10 Energy of an Orbiting Satellite

    KE = G M_E m/[2(R_E + h)], PE = −G M_E m/(R_E + h), so total E = −G M_E m/[2(R_E + h)]. The total energy is negative, meaning the satellite is bound; if E ≥ 0 it escapes.

    Worked Examples

  • Mass of Mars (Ex 7.5). Phobos: T = 7 h 39 min = 27 540 s, orbital radius 9.4 × 10³ km. M = 4π²R³/(G T²) ≈ 6.48 × 10²³ kg. Martian year: T_M = (1.52)^{3/2} × 365 ≈ 684 days.
  • Weighing the earth (Ex 7.6). From g: M_E = g R_E²/G = 9.81 × (6.37 × 10⁶)²/(6.67 × 10⁻¹¹) ≈ 5.97 × 10²⁴ kg. From the moon's orbit (R = 3.84 × 10⁸ m, T = 27.3 days): M_E = 4π²R³/(G T²) ≈ 6.02 × 10²⁴ kg — the two agree well.
  • Kepler's constant (Ex 7.7). With T² = k R³ and k = 10⁻¹³ s² m⁻³, the moon's period is T = √(10⁻¹³ × (3.84 × 10⁸)³) s ≈ 27.3 days.
  • Escape speed. v_e = √(2 × 9.8 × 6.4 × 10⁶) ≈ 11.2 km s⁻¹.
  • Variation of g. At h = 32 km: g(h) ≈ 9.8(1 − 64/6400) ≈ 9.70 m s⁻². At depth d = R_E/2: g(d) = 9.8 × ½ = 4.9 m s⁻².
  • Formula / Key-Facts Sheet

  • F = G m₁m₂/r²; G = 6.67 × 10⁻¹¹ N m² kg⁻²
  • g = G M_E/R_E²; g(h) ≈ g(1 − 2h/R_E); g(d) = g(1 − d/R_E)
  • V(r) = −G M m/r
  • v_e = √(2 G M/R) = √(2gR) ≈ 11.2 km s⁻¹
  • v(orbit) = √(G M/r); T = 2π √(r³/(G M)); T² ∝ r³
  • E(satellite) = −G M m/(2r); KE = −E; PE = 2E
  • Exam Focus

    Frequent questions: statements and proofs of Kepler's second law (from angular momentum conservation), derivations of g(h) and g(d), escape speed, orbital velocity and period, energy of a satellite, and numericals like the mass of Mars and "weighing the earth".

  • Outline — "Show that Kepler's second law follows from conservation of angular momentum": gravity is central ⇒ τ = 0 ⇒ L constant; area swept dA/dt = L/(2m) = constant.
  • Outline — "Derive the escape speed": KE at surface + PE at surface = 0 at infinity ⇒ ½mv² = GMm/R.
  • Outline — "Why is the total energy of a satellite negative?": KE = GMm/2r, PE = −GMm/r ⇒ E = −GMm/2r; negative means bound orbit.
  • Key Terms

  • Gravitation (गुरुत्वाकर्षण)
  • Universal gravitational constant (सार्वत्रिक गुरुत्वाकर्षण नियतांक)
  • Acceleration due to gravity (गुरुत्वीय त्वरण)
  • Ellipse (दीर्घवृत्त)
  • Kepler's laws (केप्लर के नियम)
  • Law of areas (क्षेत्रफलों का नियम)
  • Gravitational potential energy (गुरुत्वीय स्थितिज ऊर्जा)
  • Gravitational potential (गुरुत्वीय विभव)
  • Escape speed (पलायन चाल)
  • Satellite (उपग्रह)
  • Orbital speed (कक्षीय चाल)
  • Time period (आवर्तकाल)
  • Central force (केंद्रीय बल)
  • Binding energy (बंधन ऊर्जा)
  • Common Mistakes

  • Using g(h) ≈ g(1 − h/R) for height — the factor is 2h/R; the (1 − d/R) form is for depth.
  • Thinking g is maximum at the centre of the earth; it is zero there.
  • Forgetting that gravitational PE is negative with zero at infinity.
  • Confusing escape speed (√2gR) with orbital speed near the surface (√gR).
  • Assuming heavier satellites need a different orbital speed; v = √(GM/r) is independent of the satellite's mass.
  • Thinking a shell exerts some force on a mass inside it; it exerts none.
  • 💡 Key Learning Points

    • ✓State Kepler's three laws and relate the law of areas to conservation of angular momentum
    • ✓Apply Newton's law of gravitation and know the value and measurement of G
    • ✓Derive and use the variation of g with height and depth
    • ✓Derive escape speed and the orbital speed, period and energy of a satellite
    • ✓Use orbital data to find the masses of planets (Mars, Earth)

    👨‍🏫 Teaching Tips

    • →Draw an ellipse with two pins and a loop of string to show the sun at one focus
    • →Use planet period and distance data to verify T² ∝ a³ on a graph
    • →Discuss Cavendish's torsion balance and why G is so hard to measure
    • →Compare escape speeds of the earth and moon to explain why the moon has no atmosphere
    • →Let students compute the earth's mass two ways (Ex 7.6) and compare
    • →Explain geostationary satellites and their use in communication
    • →Clear the misconception that astronauts are weightless because there is no gravity

    📋 Assessment Questions

    1. State Kepler's laws and derive the law of areas
    2. Derive the expression for g at a depth d and at the centre of the earth
    3. Derive the escape speed from the earth's surface
    4. Find the mass of Mars from the period and orbit of Phobos
    5. Show that the total energy of an orbiting satellite is negative and explain its meaning

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