Class 11 · Physics · Chapter 5
⚡ Gravitation
Complete Notes · Key Formulas · Important Q&A for Board Exam
5.1 Introduction
5.2 Kepler's Laws
5.3 Universal Law of Gravitation
5.4 Cavendish Balance
5.5 Acceleration due to Gravity
5.6 Variation of g
5.7 Potential Energy & Escape Velocity
5.8 Earth Satellites
📖 5.1 Introduction
★ Key Facts
- Gravitation: Force of mutual attraction between any two objects by virtue of their masses.
- It is always an attractive force with infinite range.
- Does not depend on the intervening medium.
- Gravitational force is 10⁻³⁹ times weaker than strong nuclear force — weakest fundamental force.
- Galileo demonstrated all bodies fall with the same acceleration (Leaning Tower of Pisa experiment).
- Aryabhatta (476–550 AD): Earth revolves on its own axis and moves around the Sun in circular orbit.
- Kepler (1571–1630): Established 3 laws of planetary motion from Tycho Brahe's data.
- Newton: Explained gravitation as the reason planets obey Kepler's laws.
🪐 5.2 Kepler's Laws of Planetary Motion
Law 1 — Law of Orbit
All planets move in elliptical orbits around the Sun, with the Sun at one of the foci of the ellipse.
- Perihelion (P): closest point of orbit from Sun
- Aphelion (A): farthest point of orbit from Sun
- Major axis = 2a (PA) | Semi-major axis = a
- Minor axis = 2b (MN) | Semi-minor axis = b
Law 2 — Law of Areas
The line joining a planet and the Sun sweeps equal areas in equal intervals of time.
- Planets move faster when nearer to Sun (at perihelion) and slower when farther (at aphelion).
- This law is a consequence of conservation of angular momentum.
- Valid for any central force (force always directed towards a fixed centre).
- ΔA/Δt = L/2m = constant (L = angular momentum)
Law 3 — Law of Periods
The square of the time period of revolution of a planet is proportional to the cube of the semi-major axis of its elliptical orbit.
T² ∝ r³ or T²/r³ = constant
T = time period, r = semi-major axis (length)
| Planet | Semi-major axis (×10¹⁰ m) | Period (years) | T²/r³ (×10⁻³⁴ y²m⁻³) |
|---|
| Mercury | 5.79 | 0.24 | 2.95 |
| Venus | 10.8 | 0.615 | 3.00 |
| Earth | 15.0 | 1 | 2.96 |
| Mars | 22.8 | 1.88 | 2.98 |
| Jupiter | 77.8 | 11.9 | 3.01 |
| Saturn | 143 | 29.5 | 2.98 |
⭐ Board tip
T²/r³ is nearly
constant for all planets — this confirms Kepler's third law experimentally.
⚖️ 5.3 Newton's Universal Law of Gravitation
★ Statement
Every particle of matter attracts every other particle of matter with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
F = G m₁m₂ / r²
G = Universal Gravitational Constant = 6.67 × 10⁻¹¹ N m² kg⁻²
Dimensions of G = [L³ M⁻¹ T⁻²]
★ Properties of Gravitational Force
- Always attractive — never repulsive.
- Acts along the line joining the two bodies.
- Forces form an action-reaction pair (equal magnitude, opposite directions): F⃗₁₂ = −F⃗₂₁
- Independent of intervening medium.
- Infinite range — acts even at very large distances.
- For a hollow thin spherical shell: Force on a point mass inside = zero.
- For a uniform sphere/shell and a point mass outside: Force acts as if all mass is at the centre.
⚗️ 5.4 Measurement of G — Cavendish Balance
★ Method
- Two small lead spheres (s₁, s₂) of mass m are mounted at ends of a light rod suspended by a fine vertical metallic fibre (~100 cm).
- A small mirror M is attached to the fibre to measure angle of twist.
- Two large lead spheres (L₁, L₂) of mass M are brought close — they attract the small spheres.
- A torque is generated: τ = F × L = G(mM/r²) × L
- At equilibrium: G(mM/r²)L = Kθ where K = restoring torque per unit angle, θ = angle of twist.
- Measuring θ and knowing m, M, r, L → G = 6.67 × 10⁻¹¹ N m² kg⁻²
🌍 5.5 Acceleration due to Gravity (g)
g = GM/r² (general)
g (surface) = GM/R² (r = R, radius of Earth)
g = 9.8 m/s² (at Earth's surface)
★ Key Points
- g depends only on the Earth's mass and radius — NOT on the mass of the falling object.
- This explains Galileo's finding: all objects fall with the same acceleration.
- Mass of Earth: M = gR²/G = 5.97 × 10²⁴ kg
📉 5.6 Variation of g with Altitude, Depth, Latitude & Shape
(A) With Altitude (h)
gₕ = g R²/(R+h)²
For small h (h << R):
gₕ ≈ g(1 − 2h/R)
→ g
decreases with altitude.
(B) With Depth (d)
g_d = g(1 − d/R)
→ g
decreases linearly with depth.
At centre (d = R):
g = 0
→ g is maximum at Earth's surface.
(C) With Latitude (θ)
g' = g − Rω²cos²θ
At equator (θ=0°): g' = g − Rω² →
minimum
At poles (θ=90°): g' = g →
maximum
Reduction at equator: Rω² = 0.034 m/s²
(D) With Shape of Earth
Earth is an
oblate spheroid — bulged at equator.
Equatorial radius (6378 km) > Polar radius (6356 km).
∴ g at equator = 9.7804 m/s² (less)
g at poles = 9.8322 m/s² (more)
⭐ Summary — where g is max/min
Maximum g: At the poles (least radius + no rotation effect)
Minimum g: At the equator (greatest radius + maximum rotation effect)
Zero g: At the centre of the Earth
| Location | Formula | Effect on g |
|---|
| Surface | g = GM/R² | Standard value 9.8 m/s² |
| Height h above | gₕ = g(R/R+h)² | Decreases (inverse square) |
| Depth d below | g_d = g(1−d/R) | Decreases linearly |
| Latitude θ | g' = g − Rω²cos²θ | Increases from equator→pole |
| Centre of Earth | g_d = 0 (d=R) | Zero |
⚡ 5.7 Gravitational Potential Energy & Escape Velocity
★ Gravitational Potential Energy
Definition: Work done against gravitational force in bringing a mass from infinity to a point r from the centre of the Earth.
U = −GMm/r
(Negative sign: system is bound; reference at r = ∞ where U = 0)
Change in PE when lifting mass m from r_i to r_f:
ΔU = GMm (1/r_i − 1/r_f)
For small height h (<< R):
ΔU = mgh (familiar formula)
★ Gravitational Potential (V)
V = Gravitational PE per unit mass = U/m =
−GM/r
Gravitational PE = V × m
Potential difference: V₂ − V₁ = (U₂ − U₁)/m = Work done per unit mass
★ Escape Velocity
Definition: The minimum velocity with which a body must be projected vertically upward from the surface of the Earth so that it escapes the Earth's gravitational field (reaches infinity).
At surface: Total energy = ½mv_e² − GMm/R
At infinity: Total energy = 0
Energy conservation: ½mv_e² − GMm/R = 0
v_e = √(2GM/R) = √(2gR)
v_e = 11.2 km/s (for Earth)
★ Key Points — Escape Velocity
- v_e = √(2gR) — depends on mass and radius of Earth only.
- Independent of mass of the projected body.
- v_e = √2 × v_c (escape velocity = √2 × critical/orbital velocity)
🛰️ 5.8 Earth Satellites
★ Critical / Orbital Velocity (v_c)
Definition: The exact horizontal velocity given to a satellite at height h so that it revolves in a circular orbit around the Earth.
Centripetal force = Gravitational force
mv_c²/r = GMm/r²
v_c = √(GM/r) = √(GM/(R+h)) = √(gₕ(R+h))
For satellite close to surface (h ≈ 0):
v_c = √(gR) = 7.92 km/s
★ v_c Key Points
- v_c is independent of mass of satellite.
- v_c decreases with increase in height of orbit.
- Maximum v_c (7.92 km/s) is when h ≈ 0 (close to surface).
★ Possible Orbits Based on Horizontal Velocity (v_h)
| Condition | Path of Satellite |
|---|
| v_h < v_c | Ellipse — point of projection is apogee (farthest). Falls to Earth if it passes through atmosphere. |
| v_h = v_c | Stable circular orbit around Earth. |
| v_c < v_h < v_e | Ellipse — point of projection is perigee (closest). |
| v_h = v_e | Parabolic path — escapes, speed = 0 at infinity. |
| v_h > v_e | Hyperbolic path — escapes from Earth's gravity completely. |
★ Time Period of Satellite
T = 2π√(r³/GM) = 2π√((R+h)³/GM)
Also: T = 2π√((R+h)/gₕ) = 2π√(r/gₕ)
For satellite close to surface (h ≈ 0):
T_min = 2π√(R/g) ≈ 85 minutes
T² ∝ r³ — verifies Kepler's Third Law for satellites.
| Property | Geostationary (Communication) Satellite | Polar Satellite |
|---|
| Orbit | Equatorial plane | Polar orbit (N-S direction) |
| Altitude | ~36,000 km | 500–800 km |
| Time Period | 24 hours (same as Earth) | ~85 minutes |
| Appears from Earth | Stationary | Moves across sky |
| Use | Communication, TV, telephone, radio signals (INSAT) | Weather forecasting, astronomy, solar radiation study |
| Orbits/day | 1 | ~16 times |
★ Binding Energy of an Orbiting Satellite
Definition: Minimum energy required by a satellite to escape from Earth's gravitational influence.
KE = ½mv_c² = GMm/2r
PE = −GMm/r
Total Energy (TE) = KE + PE = −GMm/2r
(Negative → satellite is bound to Earth)
Binding Energy = +GMm/2r
(Energy needed to make TE = 0, i.e., unbind the satellite)
Note: KE = −TE = BE | |PE| = 2 × KE = 2 × BE
★ Weightlessness in a Satellite
- A satellite in circular orbit is in a state of free fall — centripetal acceleration = gravitational acceleration g at that height.
- Normal reaction N = mg − ma_d = mg − mg = 0
- ∴ Apparent weight = 0 → total weightlessness
- Astronauts float inside the satellite because both satellite and astronaut have the same centripetal acceleration.
- The satellite doesn't fall on Earth because its horizontal tangential velocity keeps it in circular orbit.
Board Exam Preparation
❓ Important Questions & Answers
Chapter 5 — Gravitation | Class 11 Physics
📝 1-Mark Questions (Define / State)
Q1. State Kepler's Law of Areas. 1 Mark
Ans: The line joining a planet and the Sun sweeps equal areas in equal intervals of time. This is a consequence of conservation of angular momentum, as gravitational force is a central force.
Q2. State Kepler's Law of Periods. 1 Mark
Ans: 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 the ellipse traced by it. T² ∝ r³ or T²/r³ = constant.
Q3. State Newton's Universal Law of Gravitation. 1 Mark
Ans: Every particle of matter attracts every other particle with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between them. F = Gm₁m₂/r²
Q4. Define escape velocity. 1 Mark
Ans: The minimum velocity with which a body must be projected vertically upward from the Earth's surface so that it escapes the Earth's gravitational field and never returns. v_e = √(2GM/R) = 11.2 km/s for Earth.
Q5. Define binding energy of a satellite. 1 Mark
Ans: The minimum energy required by a satellite to escape from the Earth's gravitational influence is called its binding energy. BE = GMm/2r (for a satellite in circular orbit at radius r).
Q6. What is a geostationary satellite? 1 Mark
Ans: A satellite that revolves around Earth in the equatorial plane with the same period of rotation as the Earth (24 hours) and in the same direction (west to east), so it appears stationary from the Earth's surface. Used for communication, TV, telephone (e.g., INSAT).
Q7. What are the dimensions and SI units of G? 1 Mark
Ans: SI unit of G: N m² kg⁻². Dimensions: [L³ M⁻¹ T⁻²]. Value: G = 6.67 × 10⁻¹¹ N m² kg⁻².
📝 2-Mark Questions
Q8. Why do astronauts in an orbiting satellite feel weightless? 2 Marks
Ans: In a revolving satellite, the centripetal acceleration = gravitational acceleration (g at that height). This means the satellite is in a state of free fall. The normal reaction N = mg − mad = mg − mg = 0. Since we feel weight only through normal reaction, apparent weight = 0 → total weightlessness. Astronauts float because both they and the satellite have the same downward acceleration (free fall together).
Q9. How does acceleration due to gravity vary with (i) altitude and (ii) depth? 2 Marks
Ans:
(i) With altitude h: gₕ = gR²/(R+h)² ≈ g(1 − 2h/R) for h << R. g decreases with altitude (inversely proportional to square of distance from centre).
(ii) With depth d: g_d = g(1 − d/R). g decreases linearly with depth. At centre (d = R), g = 0. Thus g is maximum at Earth's surface.
Q10. Why is g minimum at equator and maximum at poles? 2 Marks
Ans: Two reasons:
1. Rotation of Earth: g' = g − Rω²cos²θ. At equator (θ=0°), cos²θ = 1 → maximum reduction. At poles (θ=90°), cos²θ = 0 → no reduction.
2. Shape of Earth: Earth is oblate spheroid — equatorial radius (6378 km) > polar radius (6356 km). Since g = GM/R², larger R at equator → smaller g. Combined effect: g at equator = 9.7804 m/s², at poles = 9.8322 m/s².
Q11. State conditions for various orbits of a satellite depending on horizontal projection speed v_h. 2 Marks
Ans:
• v_h < v_c → Elliptical orbit, projection point is apogee (falls to Earth if it enters atmosphere).
• v_h = v_c → Stable circular orbit.
• v_c < v_h < v_e → Elliptical orbit, projection point is perigee.
• v_h = v_e → Parabolic path (escapes, speed = 0 at ∞).
• v_h > v_e → Hyperbolic path (escapes with non-zero speed at ∞).
📝 3-Mark / Derivation Questions
Q12. Derive the expression for escape velocity. 3 Marks
Ans:
Let a body of mass m be projected with escape velocity v_e from Earth's surface (mass M, radius R).
On surface: KE = ½mv_e², PE = −GMm/R → Total energy = ½mv_e² − GMm/R
At infinity: KE = 0, PE = 0 → Total energy = 0
By conservation of energy: ½mv_e² − GMm/R = 0
∴ v_e = √(2GM/R)
Since GM = gR²: v_e = √(2gR) = 11.2 km/s
Note: v_e is independent of mass of projected body.
Q13. Derive the expression for critical (orbital) velocity of a satellite at height h. 3 Marks
Ans:
Let satellite of mass m revolve in circular orbit of radius r = (R+h) at height h.
Centripetal force = Gravitational force
mv_c²/r = GMm/r²
v_c² = GM/r
v_c = √(GM/r) = √(GM/(R+h)) = √(gₕ(R+h))
For satellite close to surface (h ≈ 0, r ≈ R):
v_c = √(GM/R) = √(gR) = 7.92 km/s
v_c is independent of satellite's mass; decreases with increase in h.
Q14. Derive the expression for the time period of a satellite. 3 Marks
Ans:
Critical speed = Circumference of orbit / Time period
v_c = 2πr/T and v_c = √(GM/r)
∴ √(GM/r) = 2πr/T
T = 2πr/v_c = 2πr × √(r/GM) = 2π√(r³/GM)
T = 2π√((R+h)³/GM)
Also: T = 2π√((R+h)/gₕ)
For h ≈ 0: T_min = 2π√(R/g) ≈ 85 minutes
Since T² = 4π²r³/GM → T² ∝ r³ → verifies Kepler's Third Law.
Q15. Obtain expression for binding energy of a satellite revolving in circular orbit. 3 Marks
Ans:
For satellite of mass m at orbital radius r = (R+h):
KE = ½mv_c² = ½m(GM/r) = GMm/2r
PE = −GMm/r
Total Energy = KE + PE = GMm/2r − GMm/r = −GMm/2r
Negative sign → satellite is bound to Earth.
For satellite to escape, TE must become ≥ 0.
∴ Minimum energy to be supplied = +GMm/2r
Binding Energy = GMm/2r
Note: BE = KE = −TE and |PE| = 2 × BE.
Q16. Show that g at height h above Earth's surface = g(R/R+h)². 3 Marks
Ans:
g at surface = GM/R² ...(1)
g at height h: gₕ = GM/(R+h)² ...(2)
Dividing (2) by (1): gₕ/g = R²/(R+h)²
∴ gₕ = g [R/(R+h)]²
For small h: gₕ ≈ g(1 + h/R)⁻² ≈ g(1 − 2h/R) (using binomial approximation)
Q17. Obtain formula for g at depth d below Earth's surface. 3 Marks
Ans:
Density ρ = M/V = M/(4πR³/3) → M = (4/3)πR³ρ
g at surface = G(4/3)πRρ ...(1)
At depth d, inner sphere radius = (R−d), mass M' = (4/3)π(R−d)³ρ
g_d = GM'/(R−d)² = G(4/3)π(R−d)ρ ...(2)
Dividing (2) by (1): g_d/g = (R−d)/R
∴ g_d = g(1 − d/R)
At centre (d=R): g_d = 0.
📝 Numericals (Important Examples)
N1. At what height above Earth does g decrease by 10%? R = 6400 km. 2 Marks
Ans:
gₕ = 90% of g → gₕ/g = 0.9
Using gₕ ≈ g(1 − 2h/R):
0.9 = 1 − 2h/R → 2h/R = 0.1 → h = R/20 = 6400/20 = 320 km
N2. Gravitational force between two bodies is 1 N. Distance is doubled. Find new force. 2 Marks
Ans:
F₁ = Gm₁m₂/r² = 1 N
F₂ = Gm₁m₂/(2r)² = Gm₁m₂/4r² = F₁/4
F₂ = 1/4 = 0.25 N
Force becomes one-fourth when distance is doubled.
N3. Calculate g on Moon's surface if mass of Moon = M/80 and radius = R/4 (g = 9.8 m/s²). 3 Marks
Ans:
g_moon/g = (M_moon/M) × (R/R_moon)²
= (1/80) × (4/1)² = (1/80) × 16 = 16/80 = 1/5
g_moon = g/5 = 9.8/5 = 1.96 m/s²
N4. Calculate period of revolution of a polar satellite orbiting close to Earth. R = 6400 km, g = 9.8 m/s². 3 Marks
Ans:
h ≈ 0, so R+h ≈ R, gₕ ≈ g
T = 2π√(R/g) = 2 × 3.14 × √(6.4×10⁶/9.8)
= 2 × 3.14 × √(653061) = 2 × 3.14 × 808
= 5075 seconds ≈ 85 minutes
Quick Revision
📐 Formula Sheet + Memory Tips
Chapter 5 — Gravitation | Class 11 Physics
📐 All Formulas at a Glance
| Topic | Formula | Key Note |
|---|
| Newton's Law | F = Gm₁m₂/r² | G = 6.67×10⁻¹¹ N m² kg⁻² |
| Kepler's 3rd Law | T² ∝ r³ or T²/r³ = const | r = semi-major axis |
| g at surface | g = GM/R² | g = 9.8 m/s² |
| g at height h | gₕ = g[R/(R+h)]² ≈ g(1−2h/R) | Decreases with altitude |
| g at depth d | g_d = g(1−d/R) | Zero at centre |
| g with latitude | g' = g − Rω²cos²θ | Max at poles, min at equator |
| Escape velocity | v_e = √(2GM/R) = √(2gR) | 11.2 km/s; indep. of mass |
| Critical velocity | v_c = √(GM/r) = √(gₕ·r) | 7.92 km/s at surface |
| v_e vs v_c | v_e = √2 × v_c | Important relation |
| Time period | T = 2π√(r³/GM) = 2π√(r/gₕ) | T_min ≈ 85 min |
| Gravitational PE | U = −GMm/r | Negative; zero at ∞ |
| Gravitational Potential | V = −GM/r = U/m | PE per unit mass |
| ΔU (small h) | ΔU = mgh | Valid only when h << R |
| KE of satellite | KE = GMm/2r | Always positive |
| PE of satellite | PE = −GMm/r | Always negative |
| Total energy | TE = −GMm/2r | Always negative (bound) |
| Binding energy | BE = +GMm/2r = −TE | BE = KE |
| Critical speed (surface) | v_c = √(gR) | 7.92 km/s |
| Cavendish torque | G·mM·L/r² = Kθ | Used to find G |
🧠 Memory Tips & Tricks
- Kepler's Laws order: "O-A-P" → Orbit (ellipse), Areas (equal), Periods (T²∝r³)
- g max/min: POlar = POsitive (max g); Equator = minimum g
- At centre of Earth: g = 0 (complete symmetry, all forces cancel)
- Escape vs Critical: v_e = √2 × v_c — escape is √2 times orbital speed
- TE is always negative for an orbiting satellite = bound system
- BE = KE for orbiting satellite — easy memory trick
- |PE| = 2 × KE = 2 × BE for circular orbit — golden relation
- Weightlessness: Not because gravity is zero — it's because satellite is in free fall (gravity provides centripetal force)
- Geostationary altitude ≈ 36,000 km above equator; T = 24 hours
- Polar satellite period ≈ 85 minutes; altitude 500–800 km
- G value: 6.67 × 10⁻¹¹ — remember: "6.67, negative 11"
- Earth's mass: M = gR²/G ≈ 6 × 10²⁴ kg
⚡ MCQ Answers — Exercises
- i) g is maximum at → (C) the pole of the Earth
- ii) Weight at centre of Earth → (B) zero (g = 0 at centre, W = mg = 0)
- iii) Gravitational potential is minimum at → (A) the centre of the Earth (V = −GM/r, r smallest → V most negative → minimum)
- iv) If BE = 3×10⁹ J, KE = ? → (D) 3×10⁹ J (since KE = BE for circular orbit)
Visions Learning · Class 11 Physics · Chapter 5: Gravitation · Complete Notes & Q&A