CBSE Class 11 Physics Chapter 8 Notes: Gravitation

Chapter Introduction

Gravitation is the study of the universal force of attraction between all masses. This chapter takes you from Kepler's observational laws to Newton's universal law of gravitation, and then applies it to real-world problems — the variation of g, escape velocity, satellite orbits, and weightlessness. It carries 8–10 marks in the board exam. The derivations of escape velocity, orbital velocity, and g variation are exam favourites. Master the formulas and you can solve most problems in under two minutes.

What this guide covers: Kepler's laws, universal law of gravitation, gravitational constant G, acceleration due to gravity and its variation, gravitational potential energy, escape velocity, orbital velocity, geostationary and polar satellites, and weightlessness — all with worked numericals.

Key Definitions

• Gravitation: The force of attraction between any two masses in the universe. It is always attractive, the weakest fundamental force, and has infinite range.
• Universal gravitational constant (G): G = 6.673 × 10^-11 N m² kg^-2. Measured by Cavendish. It is the force of attraction between two 1 kg masses placed 1 m apart.
• Acceleration due to gravity (g): The acceleration produced in a freely falling body by gravity. g = GM/R² ≈ 9.8 m s^-2 on Earth's surface. It is not a universal constant — it varies with location.
• Gravitational potential energy: U = −GMm/r. Negative because work is done by gravity when masses are brought from infinity. Zero at infinity.
• Gravitational potential: V = −GM/r. Work done per unit mass to bring a test mass from infinity to that point.
• Gravitational field intensity: g = GM/r². Force per unit mass at a distance r from a mass M.
• Escape velocity (v_e): The minimum speed needed for a body to escape a planet's gravitational field entirely (reach infinity with zero residual speed). v_e = √(2GM/R) = √(2gR).
• Orbital velocity (v_o): The speed of a satellite in a circular orbit just above the surface. v_o = √(GM/R) = √(gR).
• Geostationary satellite: A satellite that orbits in the equatorial plane with the same period as Earth's rotation (24 h), appearing stationary from Earth.
• Weightlessness: The condition where the apparent weight of a body is zero, as experienced in a freely falling frame or orbit.
• Binding energy: The minimum energy required to separate a satellite from its orbit to infinity. BE = GMm/(2r) = ½(orbital energy).

Detailed Concepts

1. Kepler's Laws of Planetary Motion

Key implications:

• The second law is a consequence of angular momentum conservation (no external torque on the planet). When the planet is closer to the Sun, it moves faster to conserve L = mvr.
• The third law can be derived from Newton's law of gravitation + circular motion: GMm/r² = mω²r → ω² = GM/r³ → (2π/T)² = GM/r³ → T² = 4π²r³/GM.
• Kepler's laws are empirical (based on Tycho Brahe's data). Newton showed they follow from his law of gravitation.
• For circular orbits, a = r (radius), and the formulas simplify.

Derivation of Kepler's third law from Newton's law (important for exams): For a planet of mass m orbiting the Sun (mass M_S) in a circular orbit of radius r: Gravitational force = centripetal force: GM_Sm/r² = mω²r GM_S/r² = (2π/T)² × r = 4π²r/T² T² = 4π²r³/(GM_S) Therefore T² ∝ r³ (since 4π²/(GM_S) is constant)

2. Universal Law of Gravitation

Newton's law: Every particle attracts every other particle with a force proportional to the product of their masses and inversely proportional to the square of the distance between them.

F = GMm / r²

Direction: Along the line joining the two masses, attractive.

Properties:

• Always attractive (unlike electromagnetic force which can be repulsive)
• Follows the inverse square law
• Acts along the line joining the centres
• Independent of the presence of other masses (principle of superposition)
• Medium between the masses does not affect it
• Obeys Newton's third law (action-reaction pair)
• Is a central force (directed along the line joining centres)

Cavendish experiment: Henry Cavendish measured G using a torsion balance in 1798. The value G = 6.673 × 10^-11 N m² kg^-2 is remarkably small, explaining why gravitational force is significant only for very large masses (planets, stars).

Principle of superposition: When multiple masses act on a body, the net gravitational force is the vector sum of individual forces from each mass.

Shell theorem (important for g variation): 1. A uniform spherical shell attracts an external particle as if the entire shell's mass were concentrated at its centre. 2. A uniform spherical shell exerts no net gravitational force on a particle inside it.

These theorems explain why we can treat planets as point masses for external calculations.

3. Acceleration Due to Gravity (g)

On the surface: g = GM/R²

Variation of g:

Derivation of g at height h: g' = GM/(R + h)² = GM/(R²(1 + h/R)²) = g/(1 + h/R)² For h << R: (1 + h/R)^-2 ≈ 1 − 2h/R, so g' ≈ g(1 − 2h/R)

Derivation of g at depth d: At depth d, the effective mass is only the inner sphere of radius (R − d): g' = G × [M(R−d)³/R³] / (R−d)² = GM(R−d)/R³ = g(1 − d/R)

g is maximum at the poles (no centrifugal reduction + Earth's oblate shape means R is smaller at poles) and minimum at the equator (maximum centrifugal effect + larger R due to equatorial bulging).

Comparison: g vs. G:

4. Gravitational Potential Energy

U = −GMm / r

Key points:

• Negative everywhere (zero at infinity by convention)
• Work done to move mass from r₁ to r₂: W = U(r₁) − U(r₂) (positive if moving away against gravity)
• Near Earth's surface (h << R): ΔU = mgh (the familiar formula is an approximation valid for small heights)

Gravitational potential: V = −GM / r (scalar, unit: J kg^-1)

Relation between g and V: g = −dV/dr (field is negative gradient of potential)

Gravitational PE of a body at height h (exact formula): U(h) = −GMm / (R + h) At surface: U(0) = −GMm / R ΔU = U(h) − U(0) = GMm [1/R − 1/(R+h)] = GMmh / [R(R+h)] For h << R: ΔU ≈ GMmh/R² = mgh (familiar formula)

5. Escape Velocity

Derivation (from energy conservation — the most important derivation in this chapter): At the surface, total energy = KE + PE = ½mv_e² + (−GMm/R) At infinity, the body just escapes: v = 0 and PE = 0, so total energy = 0

½mv_e² − GMm/R = 0 v_e = √(2GM/R) = √(2gR)

Value for Earth: v_e = √(2 × 9.8 × 6.4 × 10⁶) ≈ 11.2 km s^-1

Dependence:

• Independent of the mass of the body (a lighter body needs the same escape speed as a heavier one)
• Independent of the direction of launch (as long as it is above the horizon)
• Depends on the mass and radius of the planet

Escape velocities for different bodies:

6. Satellite Motion

Orbital velocity (circular orbit at height h above surface): v_o = √(GM / (R + h))

For orbit near surface (h << R): v_o ≈ √(gR) ≈ 7.9 km s^-1 for Earth

Derivation: Gravitational force = centripetal force GMm/(R+h)² = mv_o²/(R+h) v_o = √(GM/(R+h))

Time period: T = 2π(R + h) / v_o = 2π√((R + h)³ / GM)

Total energy of a satellite in orbit:

• KE = ½mv_o² = GMm / (2(R+h))
• PE = −GMm / (R+h)
• Total E = KE + PE = −GMm / (2(R+h))

The total energy is negative, meaning the satellite is bound to the planet.

Relation: v_e = √2 × v_o (escape velocity is √2 times orbital velocity)

Escape vs. Orbital Velocity Comparison:

7. Geostationary and Polar Satellites

Geostationary satellite conditions (must satisfy ALL three): 1. Period = 24 h (same as Earth's rotation) 2. Orbit in the equatorial plane 3. Move from west to east (same direction as Earth's rotation)

Height of geostationary satellite (derivation): T = 2π√((R+h)³/GM) (24 × 3600)² = 4π²(R+h)³ / GM Solving: R + h ≈ 42,200 km, so h ≈ 35,786 km ≈ 36,000 km

Advantages of polar satellites: Cover the entire globe over multiple orbits; lower height means better resolution for imaging.

Disadvantages of geostationary satellites: Cannot cover polar regions; high altitude means lower signal strength and longer communication delay (~240 ms round trip).

8. Weightlessness

In a freely falling lift or orbiting satellite, both the body and the frame accelerate equally under gravity. The normal reaction (apparent weight) becomes zero. This is weightlessness.

Important: Weightlessness does NOT mean the absence of gravity. In orbit, g is still significant (≈ 8.7 m s^-2 at 400 km). The body is in continuous free fall, which is why it feels weightless.

Demonstration: If you are in a lift and the cable breaks, you float inside the lift — not because g = 0, but because both you and the lift fall with the same acceleration g. Your apparent weight (normal reaction from the floor) is zero.

Problems of weightlessness in space: bone density loss, muscle atrophy, fluid redistribution, difficulty in eating and drinking.

Important Numericals

Numerical 1: Gravitational Force

Two masses of 100 kg each are placed 1 m apart. Find the gravitational force between them. (G = 6.67 × 10^-11 N m² kg^-2)

Numerical 2: g at Altitude

Find g at a height of 320 km above Earth's surface. (R = 6400 km, g = 9.8 m s^-2)

Numerical 3: Escape Velocity

Find the escape velocity from the Moon. (M_Moon = 7.36 × 10²² kg, R_Moon = 1.74 × 10⁶ m)

Numerical 4: Orbital Velocity and Period

A satellite orbits at height 400 km. Find its orbital velocity and period. (R = 6400 km, g = 9.8 m s^-2)

Numerical 5: g at Depth

Find g at a depth of 1000 km from Earth's surface. (R = 6400 km, g = 9.8 m s^-2)

Numerical 6: Height of Geostationary Satellite

Calculate the height of a geostationary satellite. (R = 6400 km, M = 6 × 10²⁴ kg, G = 6.67 × 10^-11)

Numerical 7: Binding Energy

Find the binding energy of a 500 kg satellite in a circular orbit at height 1600 km. (R = 6400 km, g = 9.8 m s^-2)

NCERT Exercise Reference

Common Mistakes

1. Confusing g and G: g = acceleration due to gravity (varies with location); G = universal gravitational constant (same everywhere). g = GM/R² connects them. 2. Wrong formula for g at height vs. depth: g at height: g' = gR²/(R+h)² (inverse square law); g at depth: g' = g(1 − d/R) (linear). These are different formulas — do not mix them. 3. Forgetting the negative sign in gravitational PE: U = −GMm/r is negative. Zero at infinity, not at the surface. 4. Escape velocity independent of mass and direction: A feather and a rocket need the same v_e to escape (neglecting air resistance). The direction of launch does not matter (as long as it is above the horizon). 5. Geostationary orbit is not at any height: It must be at approximately 36,000 km. You cannot choose an arbitrary height for a geostationary satellite. 6. Weightlessness ≠ zero gravity: In orbit, gravity still acts (it provides the centripetal force). Weightlessness arises because both body and satellite are in free fall. 7. g is zero at the centre of Earth, not at infinity: At the centre (depth d = R), g' = g(1 − R/R) = 0. At infinity, g is also zero but by a different formula (g = GM/r² → 0 as r → ∞). 8. Total energy of a satellite is negative: KE = +GMm/(2r), PE = −GMm/r, so E = −GMm/(2r). The negative sign means the satellite is bound.

Exam Tips

• Kepler's third law (T² ∝ r³) is often asked in 1-mark questions — remember the proportionality constant for the Sun: T² = (4π²/GM_S) × a³.
• Derivation of escape velocity is a 3–5 mark question — start from total energy conservation, set energy at infinity to zero, and solve. Write each step clearly.
• g variation with height and depth: Know both formulas and be able to derive them. These are 2–3 mark questions. The derivation uses the shell theorem.
• Satellite problems: Memorise orbital velocity v_o = √(GM/r) and period T = 2π√(r³/GM). Derive them from "gravity = centripetal force."
• Geostationary satellite: Know the three conditions (period, equatorial plane, west-to-east) and the approximate height (36,000 km). The derivation of height from T = 2π√((R+h)³/GM) is a 3-mark question.
• In numericals, always convert units to SI before substituting (km → m, hours → seconds).
• Binding energy questions appear in JEE/NEET: BE = GMm/(2r) = |Total energy|.
• The v_e = √2 × v_o relation is a favourite 1-mark fact.

Quick Revision Notes

• Kepler's laws: Elliptical orbits, equal areas in equal times, T² ∝ a³.
• Newton's law: F = GMm/r². Always attractive, inverse square, acts along line of centres.
• Shell theorem: External — acts as point mass at centre; internal — zero force.
• g on surface: g = GM/R² ≈ 9.8 m s^-2.
• g at height h: g' = gR²/(R+h)² ≈ g(1 − 2h/R) for h << R.
• g at depth d: g' = g(1 − d/R). Zero at centre. Linear decrease.
• g is max at poles, min at equator (rotation + oblate shape).
• Gravitational PE: U = −GMm/r. Zero at infinity, negative everywhere else. F = −dU/dr.
• Gravitational potential: V = −GM/r. Scalar.
• Escape velocity: v_e = √(2GM/R) = √(2gR). Earth: ≈ 11.2 km/s. Independent of body mass.
• Orbital velocity: v_o = √(GM/r) = √(gR) near surface. Earth: ≈ 7.9 km/s.
• v_e = √2 × v_o — key relation.
• Satellite total energy: E = −GMm/(2r). Negative (bound). KE = −E. PE = 2E.
• Geostationary: h ≈ 36,000 km, T = 24 h, equatorial, west to east.
• Weightlessness: Free fall, not zero gravity. Apparent weight = 0.