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Class 11 Physics Notes Chapter 8 (Gravitation) – Physics Part-I Book

Philoid

Philoid

— 10 min read

Physics Part-I
Alright class, let's delve into Chapter 8: Gravitation from your NCERT Class 11 Physics textbook. This is a fundamental chapter, not just for your Class 11 exams but also crucial for various government competitive exams like NDA, SSC, Railways, and even parts of Civil Services Prelims. Pay close attention to the concepts and formulas.

Chapter 8: Gravitation - Detailed Notes for Government Exam Preparation

1. Introduction:

  • Gravitation is the fundamental force of attraction acting between any two objects with mass.
  • It's a universal force, meaning it applies everywhere in the universe, governing the motion of planets, stars, galaxies, and even objects on Earth.
  • It's the weakest of the four fundamental forces but has an infinite range.

2. Kepler's Laws of Planetary Motion:
(Based on observations by Tycho Brahe, formulated by Johannes Kepler)

  • First Law (Law of Orbits): Every planet revolves around the Sun in an elliptical orbit with the Sun situated at one of the foci of the ellipse.
    • Exam relevance: Shape of the orbit is elliptical, not circular. Sun is at a focus, not the center.
  • Second Law (Law of Areas): The line joining a planet and the Sun sweeps out equal areas in equal intervals of time.
    • This implies the areal velocity (dA/dt) of the planet is constant.
    • Consequence: Planets move faster when closer to the Sun (perihelion) and slower when farther away (aphelion).
    • Exam relevance: This law is a consequence of the conservation of angular momentum (L = Iω = constant) because gravitational force is a central force (acts along the line joining the bodies).
  • Third Law (Law of Periods): The square of the time period of revolution (T) of a planet around the Sun is directly proportional to the cube of the semi-major axis (a) of its elliptical orbit.
    • Mathematically: T² ∝ a³ or T²/a³ = constant.
    • For a nearly circular orbit, 'a' can be approximated by the radius 'r'. So, T² ∝ r³.
    • Exam relevance: Used to compare time periods or orbital radii of different planets or satellites.

3. Newton's Universal Law of Gravitation:

  • Statement: Every particle 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 centers.
  • Mathematical Form: F = G (m₁m₂ / r²)
    • F = Gravitational force
    • m₁, m₂ = Masses of the two objects
    • r = Distance between the centers of the two objects
    • G = Universal Gravitational Constant
  • Universal Gravitational Constant (G):
    • Value: G ≈ 6.67 × 10⁻¹¹ N m² kg⁻² (Value determined experimentally by Cavendish)
    • Dimensions: [M⁻¹ L³ T⁻²]
    • Nature: Scalar quantity. It's universal - same value everywhere in the universe.
    • Exam relevance: Know the value, units, dimensions, and the fact that it's constant.
  • Vector Form: F⃗₁₂ = - G (m₁m₂ / r²₁₂) r̂₁₂
    • F⃗₁₂ is the force on m₁ due to m₂.
    • r̂₁₂ is the unit vector pointing from m₁ to m₂.
    • The negative sign indicates that the force is attractive.
    • Obeys Newton's Third Law: F⃗₁₂ = - F⃗₂₁
  • Principle of Superposition: The net gravitational force on any particle due to a number of other particles is the vector sum of the gravitational forces exerted by all other individual particles on it.
    • F⃗_net = F⃗₁ + F⃗₂ + F⃗₃ + ...

4. Acceleration Due to Gravity (g):

  • The acceleration experienced by a body falling freely towards the Earth (or any celestial body) due to its gravitational pull.
  • Relation between 'g' and 'G': Consider an object of mass 'm' on the surface of the Earth (mass M, radius R).
    • Force F = mg (from Newton's Second Law)
    • Force F = GMm / R² (from Law of Gravitation)
    • Equating these: mg = GMm / R² => g = GM / R²
    • Exam relevance: This formula is crucial. 'g' depends on the mass (M) and radius (R) of the planet/celestial body, not on the mass (m) of the falling object.
  • Variation of 'g':
    • (a) With Altitude (Height h): g<0xE2><0x82><0x95> = GM / (R+h)² = g / (1 + h/R)²
      • For h << R (small heights): g<0xE2><0x82><0x95> ≈ g (1 - 2h/R) (using binomial approximation)
      • Conclusion: 'g' decreases with increasing altitude.
    • (b) With Depth (Depth d): g<0xE2><0x82><0x96> = GM' / (R-d)² where M' is the mass of the inner sphere of radius (R-d). Assuming uniform density (ρ), M = (4/3)πR³ρ and M' = (4/3)π(R-d)³ρ.
      • Substituting and simplifying: g<0xE2><0x82><0x96> = g (1 - d/R)
      • Conclusion: 'g' decreases with increasing depth.
      • At the center of the Earth (d=R), g<0xE2><0x82><0x96> = 0.
    • (c) Due to Rotation of Earth (Latitude λ): The effective acceleration due to gravity g' at latitude λ is given by:
      • g' = g - Rω²cos²λ
      • ω = angular velocity of Earth's rotation.
      • At Poles (λ = 90°): cosλ = 0 => g' = g (Maximum value)
      • At Equator (λ = 0°): cosλ = 1 => g' = g - Rω² (Minimum value)
      • Conclusion: 'g' is maximum at the poles and minimum at the equator due to rotation.
    • (d) Due to Shape of Earth: Earth is not a perfect sphere; it's an oblate spheroid (bulges at the equator, flattened at the poles). Equatorial radius (R<0xE2><0x82><0x91>) > Polar radius (R<0xE2><0x82><0x99>).
      • Since g ∝ 1/R², g at poles > g at the equator. This effect adds to the rotational effect.
  • Exam relevance: Understand how 'g' varies and the formulas for altitude and depth variations (especially the approximations). Know where 'g' is max/min.

5. Gravitational Potential Energy (U):

  • The energy possessed by a body due to its position in a gravitational field.
  • Defined as the work done in bringing a body from infinity to that point in the field without acceleration.
  • Formula: U = - GMm / r
    • 'r' is the distance between the centers of mass M and m.
    • The negative sign indicates that the force is attractive and the potential energy decreases as the bodies come closer.
    • Zero potential energy is conventionally taken at infinity (r = ∞).
  • Change in Potential Energy (ΔU) when moving a mass 'm' from r₁ to r₂:
    • ΔU = U₂ - U₁ = -GMm/r₂ - (-GMm/r₁) = GMm (1/r₁ - 1/r₂)
  • For lifting an object of mass 'm' by a small height 'h' near the Earth's surface (h << R):
    • r₁ = R, r₂ = R+h
    • ΔU = GMm [1/R - 1/(R+h)] ≈ GMm [h / R²] = (GM/R²) mh = mgh (This familiar formula is an approximation).
  • Exam relevance: Understand the formula, the negative sign, the reference point (infinity), and the approximate formula mgh.

6. Gravitational Potential (V):

  • Gravitational potential at a point in a gravitational field is the gravitational potential energy per unit mass at that point.
  • It's the work done in bringing a unit mass from infinity to that point without acceleration.
  • Formula: V = U / m = - GM / r
  • Unit: J/kg. Dimension: [L² T⁻²]. Nature: Scalar quantity.
  • Relation between Gravitational Field Intensity (E) and Potential (V): E = - dV/dr (Field is the negative gradient of potential).

7. Gravitational Field Intensity (E or I):

  • The gravitational force experienced by a unit mass placed at a point in the gravitational field.
  • Formula: E = F / m = GM / r² (Magnitude)
  • Vector form: E⃗ = - (GM / r²) r̂ (Points towards the mass M)
  • Unit: N/kg or m/s². Dimension: [L T⁻²]. Nature: Vector quantity.
  • Near Earth's surface, E = GM/R² = g. So, gravitational field intensity is equal to acceleration due to gravity.

8. Escape Velocity (vₑ):

  • The minimum velocity required for an object to escape the gravitational influence of a celestial body (like Earth) completely, i.e., to reach infinity.
  • Derivation: At escape, the total energy (K.E + P.E) of the object must be zero (minimum energy to reach infinity where P.E = 0 and K.E = 0).
    • (1/2)mvₑ² + (-GMm / R) = 0
    • (1/2)mvₑ² = GMm / R
    • vₑ = √(2GM / R)
  • Using g = GM/R² => GM = gR²
    • vₑ = √(2gR² / R) = √(2gR)
  • For Earth: M ≈ 6×10²⁴ kg, R ≈ 6400 km = 6.4×10⁶ m, g ≈ 9.8 m/s²
    • vₑ ≈ 11.2 km/s
  • Exam relevance: Know both formulas for vₑ. Escape velocity depends on the mass (M) and radius (R) of the planet/body, but it is independent of the mass (m), shape, and direction of projection of the escaping object.

9. Satellites:

  • Objects revolving around a planet in a stable orbit.
  • Orbital Velocity (v₀): The velocity required to put a satellite into a circular orbit of radius 'r' (r = R+h, where R is planet's radius, h is height above surface).
    • The necessary centripetal force (mv₀²/r) is provided by the gravitational force (GMm/r²).
    • mv₀²/r = GMm/r²
    • v₀ = √(GM / r) = √(GM / (R+h))
  • For a satellite orbiting close to the Earth's surface (h << R, so r ≈ R):
    • v₀ ≈ √(GM / R) = √(gR)
    • v₀ ≈ √(9.8 × 6.4×10⁶) ≈ 7.92 km/s ≈ 8 km/s.
  • Relation between vₑ and v₀ (near surface):
    • vₑ = √(2gR), v₀ = √(gR) => vₑ = √2 v₀
  • Time Period of a Satellite (T): Time taken for one complete revolution.
    • T = Distance / Velocity = 2πr / v₀ = 2πr / √(GM/r)
    • T = 2π √(r³ / GM) or T² = (4π² / GM) r³
    • This confirms Kepler's Third Law (T² ∝ r³).
    • Substituting GM = gR² (only valid if r ≈ R): T = 2π √(R / g) ≈ 84.6 minutes (for satellite near Earth's surface).
  • Energy of an Orbiting Satellite:
    • Kinetic Energy (K.E): K = (1/2)mv₀² = (1/2)m (GM/r) = GMm / 2r
    • Potential Energy (P.E): U = - GMm / r
    • Total Mechanical Energy (E): E = K.E + P.E = (GMm / 2r) + (-GMm / r) = - GMm / 2r
    • Observations:
      • E = - K.E (Total energy is negative of K.E)
      • E = U / 2 (Total energy is half the P.E)
      • The negative total energy indicates a bound system (satellite cannot escape). To make the satellite escape (E=0), energy equal to +GMm/2r (binding energy) must be supplied.
  • Geostationary Satellites:
    • Revolve in an equatorial orbit.
    • Time period = 24 hours (same as Earth's rotation).
    • Revolve from West to East (same direction as Earth).
    • Appear stationary from Earth.
    • Height above Earth's surface ≈ 36,000 km.
    • Uses: Communication, broadcasting, weather forecasting.
  • Polar Satellites:
    • Revolve in polar orbits (passing over North and South poles).
    • Orbit at lower altitudes (≈ 500 - 800 km).
    • Time period ≈ 100 minutes.
    • Scan the entire Earth's surface over time.
    • Uses: Remote sensing, environmental monitoring, espionage, weather data.

10. Weightlessness:

  • Weight of a body (W = mg) is the force with which a planet attracts it. It's also the reading on a spring balance supporting the body.
  • A body is in a state of weightlessness when its apparent weight is zero.
  • This occurs during free fall. A satellite orbiting the Earth is constantly in a state of free fall towards the Earth (though it never hits it due to its tangential velocity).
  • Inside an orbiting satellite, both the satellite and the astronaut inside have the same acceleration (g' = GM/r²). There is no relative acceleration between them, and the astronaut doesn't press against the floor of the satellite. Hence, the apparent weight is zero.
  • Weightlessness is also experienced at the center of the Earth (g=0) and at null points in space where gravitational fields cancel out.

Key Points Summary for Exams:

  • Memorize formulas for F, g, g<0xE2><0x82><0x95>, g<0xE2><0x82><0x96>, U, V, vₑ, v₀, T, Energies of satellite.
  • Understand the dependencies: How g varies, what vₑ and v₀ depend on.
  • Kepler's Laws: Especially the implications of the 2nd (angular momentum) and 3rd (T² ∝ r³) laws.
  • Energy of satellite: K.E is positive, P.E and Total Energy are negative for a bound orbit. E = -K.E = U/2.
  • Geostationary vs Polar satellites: Key characteristics and uses.
  • Weightlessness = Condition of free fall.

Multiple Choice Questions (MCQs)

  1. The time period of a satellite orbiting close to the surface of the Earth is approximately:
    a) 24 hours
    b) 100 minutes
    c) 84.6 minutes
    d) 365 days

  2. If the radius of the Earth were to shrink by 1% (its mass remaining the same), the acceleration due to gravity 'g' on the surface of the Earth would:
    a) Decrease by 1%
    b) Increase by 1%
    c) Decrease by 2%
    d) Increase by 2%

  3. Kepler's second law regarding the constancy of areal velocity of a planet is a consequence of the law of conservation of:
    a) Energy
    b) Linear momentum
    c) Angular momentum
    d) Mass

  4. The escape velocity from the surface of the Earth is vₑ. The escape velocity from the surface of another planet having twice the radius and the same mean density as the Earth is:
    a) 2 vₑ
    b) vₑ
    c) vₑ / 2
    d) 4 vₑ

  5. A satellite is orbiting the Earth in a circular orbit. Which of the following quantities remains constant?
    a) Velocity
    b) Angular momentum
    c) Potential energy
    d) Acceleration

  6. Where will the true weight of a body be zero?
    a) At the equator
    b) At the poles
    c) In a geostationary satellite
    d) At the center of the Earth

  7. The ratio of the kinetic energy to the total energy of an electron revolving in a Bohr orbit (similar concept applies to satellite energy) is:
    a) 1 : 1
    b) 1 : -1
    c) 1 : 2
    d) 1 : -2

  8. If the distance between the Earth and the Sun were half its present value, the number of days in a year would have been approximately:
    a) 64.5
    b) 129
    c) 182.5
    d) 730

  9. A body weighs 72 N on the surface of the Earth. What is the gravitational force on it at a height equal to half the radius of the Earth?
    a) 32 N
    b) 28 N
    c) 16 N
    d) 72 N

  10. Which of the following statements about the universal gravitational constant 'G' is correct?
    a) It has the dimensions [M L T⁻²]
    b) Its value depends on the medium between the masses.
    c) Its value is the same throughout the universe.
    d) Its value was first determined by Newton.

Answer Key for MCQs:

  1. c) 84.6 minutes
  2. d) Increase by 2% (Hint: g = GM/R². Use percentage error or differentiation: dg/g = -2 dR/R. If R decreases by 1%, dR/R = -0.01, so dg/g = -2(-0.01) = +0.02 or +2%)
  3. c) Angular momentum
  4. a) 2 vₑ (Hint: vₑ = √(2GM/R). M = Volume × Density = (4/3)πR³ρ. So vₑ = √[2G(4/3)πR³ρ / R] = R√(8πGρ/3). vₑ ∝ R. If R doubles, vₑ doubles.)
  5. b) Angular momentum (Magnitude of velocity, PE, KE, Total E are constant, but velocity and acceleration vectors change direction)
  6. d) At the center of the Earth (g=0, so W=mg=0. In a satellite, apparent weight is zero, but true weight mg is not zero)
  7. b) 1 : -1 (Hint: K.E = GMm/2r, Total E = -GMm/2r. Ratio K.E / Total E = (GMm/2r) / (-GMm/2r) = -1 or 1:-1)
  8. b) 129 (Hint: T² ∝ r³. T₂²/T₁² = (r₂/r₁)³. T₂² / (365)² = (1/2)³. T₂ = 365 / √8 ≈ 365 / 2.828 ≈ 129 days)
  9. a) 32 N (Hint: Weight W = mg. g<0xE2><0x82><0x95> = g / (1 + h/R)². Here h = R/2. So g<0xE2><0x82><0x95> = g / (1 + (R/2)/R)² = g / (1 + 1/2)² = g / (3/2)² = g / (9/4) = 4g/9. New weight W' = m g<0xE2><0x82><0x95> = m(4g/9) = (4/9)mg = (4/9)W = (4/9) * 72 N = 32 N)
  10. c) Its value is the same throughout the universe.

Study these notes thoroughly. Focus on understanding the concepts behind the formulas. Practice numerical problems based on these formulas, especially those involving ratios and variations. Good luck with your preparation!

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