In 1666, Isaac Newton watched an apple fall from a tree and asked a question that had never been framed so precisely before: Is the force pulling the apple down the same force that keeps the Moon in its orbit around Earth? The answer, which Newton worked out over the following two decades, was a resounding yes — and the result was the Universal Law of Gravitation, one of the greatest intellectual achievements in human history.
Gravitation is not just the reason apples fall. It is the force that holds planets in orbit around the Sun, keeps the Moon circling Earth, determines ocean tides, causes stars to collapse into black holes, and shapes the large-scale structure of the entire universe. The same mathematical law describes a feather falling on Earth and a galaxy cluster moving through space billions of light years away.
For CBSE Class 9 Physics Chapter 10 (Gravitation) and CBSE Class 11 Physics Chapter 8 (Gravitation), this topic is mandatory and regularly appears in board examinations. This guide covers: Newton’s universal law of gravitation, the gravitational constant G, Kepler’s three laws of planetary motion, acceleration due to gravity (g), variation of g with altitude and depth, mass vs weight, gravitational potential energy, escape velocity, and five exam-ready solved problems.
- 1. Newton’s Universal Law of Gravitation
- 2. Kepler’s Three Laws of Planetary Motion
- 3. Acceleration Due to Gravity (g) — Derivation
- 4. Variation of g with Altitude, Depth & Latitude
- 5. Mass vs Weight — Difference and Comparison
- 6. Gravitational Potential Energy
- 7. Escape Velocity
- 8. Solved Numerical Problems
- 9. Frequently Asked Questions (FAQ)
1. Newton’s Universal Law of Gravitation
(1) Directly proportional to the product of their masses: F ∝ Mm
(2) Inversely proportional to the square of the distance between them: F ∝ 1/r²
Combining: F ∝ Mm/r² ⇒ F = GMm/r²
Gravitational Constant G:
G = 6.674 × 10²³ N·m²/kg² (SI) — a universal constant, same everywhere in the universe.
First measured by Henry Cavendish in 1798 using a torsion balance (the Cavendish experiment).
Dimensional formula: [G] = [M²L³T²]
Key characteristics of gravitational force:
• It is always attractive (never repulsive). There is no gravitational shielding.
• It acts along the line joining the centres of the two bodies (central force).
• It is a long-range force: it acts over infinite distances (though it weakens with r²).
• It is the weakest fundamental force in nature (compared to electromagnetic, strong and weak nuclear forces).
• It obeys Newton’s Third Law: the force Earth exerts on you = force you exert on Earth (equal and opposite).
• It is independent of the medium between the two masses.
2. Kepler’s Three Laws of Planetary Motion
Johannes Kepler (1571–1630) formulated three empirical laws describing planetary orbits from the astronomical observations of Tycho Brahe. Newton later derived all three laws mathematically from his universal law of gravitation — proving that Kepler’s laws are consequences of the inverse-square gravitational law.
Sun at one Focus
T²/r³ = constant
3. Acceleration Due to Gravity (g) — Derivation
Earth’s mass Mᴡ = 6.0 × 10²&sup4; kg | Earth’s radius Rᴡ = 6.4 × 10&sup6; m | G = 6.67 × 10²³ N·m²/kg²
Gravitational force on mass m: F = GMᴡm/Rᴡ²
By Newton’s Second Law: F = mg (weight)
Therefore: mg = GMᴡm/Rᴡ²
Cancelling m from both sides: g = GMᴡ/Rᴡ²
Substituting values: g = (6.67×10²³ × 6.0×10²&sup4;) / (6.4×10&sup6;)²
g = (4.0×10²⁵) / (4.1×10²⁵) ≈ 9.8 m/s²
Key insight: The mass m of the falling object cancels out entirely! This means all objects fall with the same acceleration g regardless of their mass (Galileo’s result confirmed by Newton). In vacuum, a feather and a hammer fall together (as demonstrated famously on the Moon by Apollo 15 astronaut David Scott).
4. Variation of g with Altitude, Depth and Latitude
| Factor | Formula | Effect on g | Reason |
|---|---|---|---|
| Altitude (h above surface) | gₕ = g(1 − 2h/R) [for h << R] Exact: gₕ = GM/(R+h)² |
g decreases as altitude increases | Distance from Earth’s centre increases ⇒ F = GMm/r² decreases |
| Depth (d below surface) | gₒ = g(1 − d/R) | g decreases as depth increases g = 0 at Earth’s centre |
Less mass below the object contributes to attraction; shell above exerts no net force |
| Latitude (φ) | gₓ = g − Rω²cos²φ | g is maximum at poles (φ=90°) g is minimum at equator (φ=0°) |
Centrifugal effect of Earth’s rotation reduces effective g; Earth is slightly flattened at poles (smaller R) |
| Shape of Earth (oblate spheroid) | Rᴅᵓᵈᵃᴸ < Rᵃᵗᵘᵃᵗᵓᵅ | g at poles > g at equator by ~0.5% | Earth’s radius at poles (6356 km) is less than at equator (6378 km) ⇒ smaller r at poles ⇒ larger g |
| On Moon | gᴱᵓᵓᵗ = 1.62 m/s² ≈ g/6 | g on Moon = 1/6th of g on Earth | Moon’s mass is 1/81 of Earth; Moon’s radius is 1/3.7 of Earth. g = GM/R² gives 1/6th value |
5. Mass vs Weight — Complete Comparison
▶ Mass (m)
Nature: Scalar quantity
SI Unit: Kilogram (kg)
Constancy: Mass is constant everywhere — on Earth, Moon, space, or any planet.
Measured by: Physical/beam balance (compares masses; gives same reading anywhere)
Zero mass: No object has zero mass (only photons have zero rest mass).
Formula: m = F/a (Newton’s second law)
↓ Weight (W)
Nature: Vector quantity (acts downward toward centre of Earth)
SI Unit: Newton (N)
Constancy: Weight varies with location (g varies). On Moon: W = mgᴱ = mg/6.
Measured by: Spring balance (measures force; gives different reading on Moon)
Zero weight: Weight = 0 in free fall or deep space (weightlessness).
Formula: W = mg = GMm/r²
6. Gravitational Potential Energy
This is the formula used in CBSE Class 9 and for most practical problems in Class 11.
General formula (Class 11): U = −GMm/r (negative sign: work must be done against gravity to move mass away from Earth; U = 0 at r = ∞)
The gravitational potential energy of a body at Earth’s surface: U = −GMm/R
Change in GPE when moved from r₁ to r₂: ΔU = −GMm(1/r₂ − 1/r₁) = GMm(1/r₁ − 1/r₂)
7. Escape Velocity
Derivation: For escape, the total mechanical energy must be ≥ 0 (KE must overcome PE).
At surface: KE + PE = 0 (minimum escape condition)
½mv² + (−GMm/R) = 0
½mv² = GMm/R
v² = 2GM/R = 2gR (since g = GM/R²)
vᵃ = √(2gR)
Escape velocities on different bodies:
• Earth: vᵃ = √(2 × 9.8 × 6.4×10&sup6;) = 11.2 km/s (about 40,000 km/h)
• Moon: vᵃ = 2.38 km/s (Moon has no atmosphere due to low escape velocity — gas molecules are fast enough to escape)
• Sun: vᵃ = 618 km/s
• Black hole: vᵃ ≥ c = 3×10&sup8; m/s (speed of light — even light cannot escape)
Key insight: Escape velocity does not depend on the mass or direction of the projectile. Any object (rocket, stone, atom) needs the same escape velocity from a given point on Earth’s surface. However, actual rockets use continuous thrust over the journey and don’t need to reach escape velocity in one shot.
8. Solved Numerical Problems
F = 6.67×10²³ × 25 = 166.75×10²³
Weight on Moon: Wᴱ = m × gᴱ = 50 × 1.63
gₕ = 9.8 × (6400)²/(12800)² = 9.8 × 1/4
vᵃ = √(125.44×10&sup6;) = √(1.2544×10&sup8;)
vᵃ = 1.12 × 10&sup4; m/s
(8)²/(1)² = (rᵃ/rᵉ)³
64 = (rᵃ/rᵉ)³
rᵃ/rᵉ = ³√64
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9. Frequently Asked Questions (FAQ)
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 centres.
Mathematical form: F = GMm/r²
where F = gravitational force (N), G = 6.674×10²³ N·m²/kg² (universal gravitational constant), M and m = masses of the two bodies (kg), r = distance between their centres (m).
Characteristics:
(1) Always attractive — never repulsive.
(2) Acts along the line joining the centres of the two bodies.
(3) Obeys Newton’s Third Law: force on M by m = −force on m by M.
(4) Follows inverse-square law: doubling distance reduces force by 4 times.
(5) Independent of the medium between the masses.
Derivation: Let M = mass of Earth, R = radius of Earth, m = mass of falling object.
Gravitational force: F = GMm/R² …(1)
By Newton’s Second Law: F = mg …(2)
From (1) and (2): mg = GMm/R²
Cancelling m: g = GM/R²
Value: G = 6.67×10²³ N·m²/kg², Mᴡ = 6.0×10²&sup4; kg, Rᴡ = 6.4×10&sup6; m
g = (6.67×10²³ × 6.0×10²&sup4;) / (6.4×10&sup6;)² = 9.8 m/s² ≈ 10 m/s²
Key point: The mass m cancels out, proving all objects fall with the same acceleration regardless of their mass (Galileo’s conclusion). In vacuum, a feather and a cannon ball fall identically.
Mass (m): Amount of matter. Scalar. Unit: kg. Constant everywhere. Measured by beam balance.
Weight (W): Gravitational force on body = mg. Vector (downward). Unit: Newton. Varies with location. Measured by spring balance.
On the Moon: gᴱ = 1.63 m/s² ≈ gᴡ/6 (Moon’s mass is 1/81 of Earth’s; Moon’s radius is 1/3.7 of Earth’s).
Mass on Moon = Same as on Earth (mass doesn’t change).
Weight on Moon = mgᴱ = m × (gᴡ/6) = Wᴡ/6 (only 1/6th of weight on Earth).
Example: A 60 kg person:
• Mass = 60 kg everywhere (Earth, Moon, space)
• Weight on Earth = 60×9.8 = 588 N
• Weight on Moon = 60×1.63 = 97.8 N (≈ 98 N, one-sixth of Earth weight)
• Weight in deep space = 0 N (weightlessness)
First Law (Law of Orbits): Every planet revolves around the Sun in an elliptical orbit with the Sun at one focus of the ellipse. The closest point is perihelion; the farthest is aphelion.
Second Law (Law of Areas): The line joining the Sun to a planet sweeps out equal areas in equal intervals of time. This means planets move faster near perihelion and slower near aphelion. The law is a consequence of conservation of angular momentum (no torque from the central gravitational force).
Third Law (Law of Periods): The square of the orbital period (T) of a planet is directly proportional to the cube of its mean orbital radius (r): T² ∝ r³, or T²/r³ = 4π²/(GM) = constant (same for all planets orbiting the Sun). Newton derived all three laws from his universal law of gravitation.
Why g decreases with altitude: g = GM/r². As altitude h increases, the distance from Earth’s centre r = (R+h) increases, so g decreases as 1/(R+h)². At height h: gₕ = GM/(R+h)² = g[R/(R+h)]². For small h: gₕ ≈ g(1 − 2h/R).
At h = R (one Earth radius above surface): gₕ = g/4 = 2.45 m/s².
At the centre of Earth: g = 0 at Earth’s centre. Using the formula gₒ = g(1 − d/R), at d = R (centre): gₒ = g(1 − 1) = 0. The reason: at the centre, mass is equally distributed in all directions. Gravitational forces from all sides cancel exactly, giving zero net gravitational force. An object placed at Earth’s centre would be in a state of weightlessness (W = 0), though its mass is unchanged.
Summary of g variations:
• g maximum at poles (smaller R, no centrifugal effect) = 9.83 m/s²
• g minimum at equator = 9.78 m/s²
• g decreases going up (altitude) or going down (depth)
• g = 0 at Earth’s centre
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