Gravitation Class 11 Notes: Complete Theory, Formulas, Numericals & PYQs

Gravitation Class 11 Notes | NEET Physics Complete Guide
01

Introduction to Gravitation

These Gravitation class 11 notes cover one of the most conceptually rich and NEET-relevant chapters in Class 11 Physics. Gravitation is the universal force of attraction that acts between every two objects possessing mass. It governs why an apple falls to the ground, why the Moon orbits Earth, why planets trace elliptical paths around the Sun, and how satellites remain in orbit. Unlike contact forces, gravitation acts across empty space over unlimited distances — making it the dominant force at cosmic scales.

The chapter progresses from Kepler’s observational laws through Newton’s quantitative formulation, to modern applications in satellite technology. Every concept introduced here has direct NEET application, and the formulas are closely inter-linked — escape velocity, orbital velocity, gravitational potential energy, and g-variation all derive from Newton’s single law of gravitation.

Chapter Scope

From falling bodies on Earth to orbiting satellites and planetary motion.

NEET Weightage

2–4 questions per year. Consistent, high-scoring chapter for prepared students.

Key Concept

One universal law — Newton’s law — generates all other formulas in this chapter.

02

Kepler’s Laws of Planetary Motion

Before Newton derived his law analytically, Johannes Kepler established three empirical laws by meticulously analyzing Tycho Brahe’s observational data on planetary positions. These laws describe the geometry and timing of planetary orbits with precision.

Kepler’s Three Laws
Law 1 (Orbits): All planets move in elliptical orbits with the Sun at one focus. Law 2 (Areas): A line joining a planet to the Sun sweeps equal areas in equal times. Law 3 (Periods): T² ∝ r³ ⟹ T²/r³ = constant for all planets

Law 2 (equal areas) is a direct consequence of conservation of angular momentum — no external torque acts on the planet-Sun system, so L = mvr is conserved. When the planet is closer (perihelion), r is small, so v must be large. When farther (aphelion), v is smaller.

Law 3 is the most quantitatively useful for NEET. It allows comparison of orbital periods and radii: if Earth’s period is 1 year at 1 AU, Jupiter’s period can be calculated from its orbital radius alone.

Tip

For NEET, Kepler’s third law T² ∝ r³ is tested by giving two planets’ orbital radii and asking for the ratio of their time periods. Apply (T₁/T₂)² = (r₁/r₂)³ directly.

03

Newton’s Law of Universal Gravitation

Newton unified terrestrial and celestial mechanics with a single quantitative statement: every particle in the universe 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. The force acts along the line joining their centres.

Newton’s Law of Gravitation
F = G × m₁ × m₂ / r² F = gravitational force (N) G = 6.674 × 10⁻¹¹ N·m²/kg² (gravitational constant) m₁, m₂ = masses of the two bodies (kg) r = distance between their centres (m)

Key properties of gravitational force:

  • Always attractive — never repulsive between masses.
  • Acts along the line joining the two bodies (central force).
  • Obeys Newton’s third law: F₁₂ = −F₂₁ (equal in magnitude, opposite in direction).
  • Independent of the medium between the two bodies.
  • Follows superposition: total force on a body is the vector sum of individual gravitational forces from all other bodies.
Warning

Newton’s law applies to point masses or uniform spheres. For a uniform sphere, the entire mass can be assumed concentrated at the centre — valid only outside the sphere. Inside a uniform sphere, only the mass enclosed within the radius contributes to the force.

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04

Gravitational Constant (G)

The gravitational constant G is a fundamental physical constant that quantifies the strength of the gravitational interaction. Its value is extraordinarily small, which is why gravitational forces between everyday objects are negligible — only astronomical masses make gravity perceptible.

Value and Dimensions of G
G = 6.674 × 10⁻¹¹ N·m²·kg⁻² Dimensional formula: [M⁻¹ L³ T⁻²] G is a scalar quantity; same value throughout the universe.

G was first measured experimentally by Henry Cavendish in 1798 using a torsion balance apparatus (the Cavendish experiment). Two small lead spheres were attracted to two large lead spheres, and the tiny twist in the suspension wire allowed G to be calculated. G does not vary with temperature, pressure, medium, or location — it is a true universal constant.

Tip

Do not confuse G (universal gravitational constant, 6.674 × 10⁻¹¹) with g (acceleration due to gravity, ~9.8 m/s²). G is universal and constant; g varies with location and is derived from G, M, and R of the planet.

05

Acceleration Due to Gravity (g)

When an object is in free fall near Earth’s surface, it accelerates at a rate determined by Earth’s gravitational pull. This acceleration, called g, is derived by equating Newton’s gravitational force to Newton’s second law of motion.

Derivation of g
Gravitational force: F = GMm/R² Newton's 2nd law: F = mg Therefore: g = GM/R² g_Earth ≈ 9.8 m/s² (standard value at surface) M = mass of Earth = 6 × 10²⁴ kg R = radius of Earth = 6.4 × 10⁶ m

Note that g is independent of the mass of the falling body — all objects regardless of mass fall with the same acceleration in a vacuum. This is the equivalence principle, foundational to Einstein’s General Theory of Relativity.

Warning

The formula g = GM/R² gives the surface value of g. At altitude h or depth d, the formula changes. Never use the surface formula when the problem specifies a location above or below Earth’s surface.

06

Variation of g — Altitude, Depth, and Latitude

The value of g is not constant — it varies with position on or above Earth. Understanding these variations is critical for Gravitation class 11 notes and appears frequently in NEET numericals.

Variation with Altitude (h above surface)
g_h = g(1 − 2h/R) for h << R (approximate) g_h = GM/(R + h)² (exact formula) g decreases with increasing altitude.
Variation with Depth (d below surface)
g_d = g(1 − d/R) At the centre of Earth (d = R): g_d = 0 g decreases linearly with depth inside Earth.
Variation with Latitude (φ)
g_φ = g − Rω²cos²φ g is maximum at poles (φ = 90°, cos φ = 0) g is minimum at equator (φ = 0°, cos φ = 1) ω = angular velocity of Earth's rotation
Factor Effect on g Rate of Change
Altitude (h above surface) Decreases Inversely with (R + h)²
Depth (d below surface) Decreases linearly g_d = g(1 − d/R)
Latitude (towards poles) Increases g_pole > g_equator by ~0.05 m/s²
Rotation of Earth Reduces effective g Effect maximum at equator
Warning

A classic NEET trap: g at height h equals g at depth d does NOT mean h = d. The altitude formula has a squared term (exact) or a factor of 2 (approximate), while the depth formula is linear. For the approximate case: g(1 − 2h/R) = g(1 − d/R) gives d = 2h.

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07

Gravitational Potential Energy

Gravitational potential energy (U) is the work done by an external agent in moving a mass from infinity to a given point in the gravitational field, without any change in kinetic energy. Because gravity is attractive and does positive work as the mass approaches, the external agent does negative work — hence U is always negative for gravitational systems.

Gravitational Potential Energy
U = −GMm/r U → 0 as r → ∞ (reference point at infinity) Near Earth's surface: U = mgh (for small h, approximate) Change in PE: ΔU = mgh (valid only near surface where g is constant)
Tip

The negative sign in U = −GMm/r is not optional — it encodes the bound nature of the system. A satellite in orbit has total energy E = −GMm/(2r), which is negative, confirming it is gravitationally bound. A positive total energy would mean the object has escaped the gravitational field.

08

Gravitational Potential and Field Intensity

Gravitational potential (V) and gravitational field intensity (E or g-field) are field quantities — they describe the gravitational environment at a point in space, independent of what test mass is placed there.

Gravitational Potential (V)
V = U/m = −GM/r V is work done per unit mass to bring a test mass from ∞ to point P. Unit: J/kg Dimension: [L²T⁻²] V is always negative (attractive field).
Gravitational Field Intensity (I or g-field)
I = F/m = GM/r² (directed toward the source mass) Relation: I = −dV/dr Unit: N/kg or m/s² Dimension: [LT⁻²]

Inside a uniform spherical shell, the gravitational field is zero at every point — the contributions from all parts of the shell cancel out. This is the shell theorem. Outside the shell, the field is identical to that of a point mass at the centre.

09

Escape Velocity

Escape velocity is the minimum initial velocity an object must be given at the surface of a planet so that it can overcome the planet’s gravitational attraction and escape to infinity — without any further propulsion once launched.

Derivation of Escape Velocity
At launch: KE + PE = 0 (total energy = 0 at escape threshold) ½mv_e² − GMm/R = 0 v_e = √(2GM/R) = √(2gR) v_e (Earth) = √(2 × 9.8 × 6.4 × 10⁶) ≈ 11.2 km/s

Key characteristics of escape velocity:

  • Independent of the mass and direction of the escaping object.
  • Depends only on the mass (M) and radius (R) of the planet.
  • v_e = √2 × orbital velocity at the surface.
  • On Moon: v_e ≈ 2.4 km/s (Moon’s low escape velocity is why it has no atmosphere).
  • Black holes: escape velocity exceeds the speed of light — even light cannot escape.
Warning

Escape velocity is NOT the velocity needed to reach a specific altitude — it is the velocity to escape to infinity. It is derived by setting total mechanical energy to zero (KE + PE = 0). If total energy remains negative after launch, the object is still gravitationally bound.

10

Orbital Velocity and Satellites

An artificial satellite orbits Earth because the gravitational force provides the necessary centripetal force for circular motion. The velocity at which this balance is achieved is the orbital velocity.

Orbital Velocity Derivation
Gravitational force = Centripetal force GMm/(R + h)² = mv_o²/(R + h) v_o = √[GM/(R + h)] Near surface (h << R): v_o = √(gR) ≈ 7.9 km/s
Time Period of Satellite
T = 2π(R + h)/v_o = 2π√[(R + h)³/GM] Near surface: T ≈ 2π√(R/g) ≈ 84 minutes Note: T² ∝ (R + h)³ — this is Kepler's third law applied to satellites.
Tip

As orbital radius increases, orbital velocity decreases but time period increases. A satellite at greater height moves slower but takes longer to complete one orbit. These are inverse relationships — frequently tested in NEET conceptual questions.

11

Geostationary Satellites

A geostationary satellite appears stationary when viewed from Earth because it revolves in the same direction as Earth’s rotation and with exactly the same angular velocity. It is used for communication, weather forecasting, and television broadcasting.

Conditions for Geostationary Orbit
Time period: T = 24 hours (same as Earth's rotation) Orbital radius: r ≈ 42,400 km from Earth's centre Height above surface: h ≈ 36,000 km (≈ 6R) Orbit: equatorial plane, west to east direction Orbital velocity: v_o ≈ 3.1 km/s
  • The orbit must be in the equatorial plane — tilted orbits would appear to oscillate north-south over 24 hours.
  • Direction must be west-to-east (prograde) — same as Earth’s rotation.
  • Only one unique altitude satisfies T = 24 h. This makes geostationary orbit slots a finite resource.
  • GPS satellites are NOT geostationary — they operate at ~20,200 km altitude in medium Earth orbit (MEO) with periods of ~12 hours.
12

Energy of an Orbiting Satellite

For a satellite in circular orbit at radius r = R + h, the energy components are inter-related through the virial theorem for gravitational systems.

Satellite Energy in Circular Orbit
Kinetic energy: KE = +GMm/(2r) (always positive) Potential energy: PE = −GMm/r (always negative) Total energy: E = −GMm/(2r) (always negative) Relation: KE = −E = −PE/2 |PE| = 2 × KE always holds for circular orbits.

The negative total energy confirms the satellite is gravitationally bound. To move a satellite to a higher orbit, energy must be added (total energy increases, becomes less negative). When a satellite loses energy due to atmospheric drag, it spirals inward — into a lower orbit where its speed is paradoxically higher.

Warning

When a satellite moves to a higher orbit, its total energy increases (less negative) but its kinetic energy decreases. This seems counterintuitive — adding energy makes the satellite move slower. The potential energy gain exceeds the KE loss. This is one of NEET’s favourite conceptual traps in the Gravitation chapter.

13

Weightlessness

Weightlessness is experienced when a body is in free fall — when the only force acting on it is gravity and there is no supporting normal reaction. Astronauts in an orbiting satellite feel weightless because both the satellite and the astronauts are in the same free-fall state.

Apparent Weight in Various Situations
Lift accelerating up: W_app = m(g + a) [heavier] Lift accelerating down: W_app = m(g − a) [lighter] Free fall (a = g): W_app = m(g − g) = 0 [weightless] Orbiting satellite: Centripetal acc = g at that altitude → W_app = 0

Gravity still acts on an orbiting astronaut — in fact, gravity is providing the centripetal force keeping them in orbit. Weightlessness does not mean absence of gravity; it means absence of the contact (normal) force. This distinction is critical for NEET conceptual questions.

14

Numerical Framework

NEET-level numericals in Gravitation class 11 notes fall into predictable solution patterns. Here is the strategic approach for each type:

Problem Type Key Formula Watch Out For
Gravitational force between two bodies F = Gm₁m₂/r² Use centre-to-centre distance, not surface distance
Value of g at altitude h g_h = g(1 − 2h/R) approx Use exact formula for large h
Value of g at depth d g_d = g(1 − d/R) g = 0 at centre; linear not inverse-square
Escape velocity v_e = √(2gR) Direction does not matter; no propulsion after launch
Orbital velocity at height h v_o = √[GM/(R+h)] Use R + h, not just h
Time period of satellite T = 2π√[(R+h)³/GM] T² ∝ r³ — Kepler’s third law applies
Satellite energy E = −GMm/(2r) Total energy is negative; KE = |E|, PE = 2E
Critical Relationship: v_e vs v_o
v_e = √2 × v_o (at the same radius) v_e (surface) ≈ 11.2 km/s v_o (near surface) ≈ 7.9 km/s 11.2 / 7.9 ≈ √2 ✓
15

Conceptual Questions

  • The gravitational force between two bodies does not depend on the medium between them — unlike electrostatic force, which is affected by the permittivity of the medium.
  • If Earth’s rotation suddenly stopped, g at the equator would increase (the centrifugal reduction of ~0.034 m/s² would be removed), but g at poles would remain unchanged.
  • A tunnel through Earth’s centre would allow a body to oscillate in simple harmonic motion — g inside varies linearly, providing restoring force proportional to displacement.
  • On the Moon, time period of a simple pendulum is longer (T = 2π√(L/g), and g_moon = g_earth/6), but an astronaut’s mass remains the same as on Earth.
  • Two satellites at different altitudes cannot be in the same orbit — each altitude has a unique orbital velocity and time period.
  • The binding energy of a satellite (energy needed to free it from orbit) = GMm/(2r) = |Total energy|.
16

PYQ Trends

Topic Frequency Type
Very HighOrbital velocity and satellite energy 2–3 questions/year Numerical + Conceptual
HighEscape velocity and v_e vs v_o relation 1–2 questions/year Conceptual + Numerical
HighVariation of g (altitude, depth, latitude) 1–2 questions/year Numerical
ModerateKepler’s third law (T² ∝ r³) 1 question/year Numerical
ModerateGravitational potential energy and binding energy 1 question/year Numerical
LowWeightlessness and geostationary satellites Occasional Conceptual
Tip

NEET consistently tests the energy of orbiting satellites — specifically the relationship KE = −E and PE = 2E. Master the signs and magnitudes of all three energy components (KE, PE, total) and their behaviour as orbital radius changes. This topic appears in nearly every NEET paper.

17

Chapter Summary

Gravitation class 11 notes build a coherent framework from Kepler’s empirical laws through Newton’s universal law to modern satellite applications. Every major formula in the chapter — g, escape velocity, orbital velocity, satellite energy — is a direct consequence of F = GMm/r².

Quick Revision — Must-Know Points

  • Kepler’s third law: T² ∝ r³; apply as (T₁/T₂)² = (r₁/r₂)³ for planet comparisons.
  • Newton’s law: F = Gm₁m₂/r². Always centre-to-centre distance. Always attractive.
  • G = 6.674 × 10⁻¹¹ N·m²/kg². Dimensions: [M⁻¹L³T⁻²]. Universal constant.
  • g = GM/R² at Earth’s surface ≈ 9.8 m/s². Independent of mass of falling body.
  • g decreases with altitude: g_h ≈ g(1 − 2h/R). Decreases with depth: g_d = g(1 − d/R).
  • g is maximum at poles and minimum at equator due to Earth’s rotation and oblate shape.
  • Gravitational PE: U = −GMm/r. Always negative. Near surface: ΔU = mgh.
  • Escape velocity: v_e = √(2gR) ≈ 11.2 km/s. Independent of mass and direction.
  • Orbital velocity: v_o = √[GM/(R+h)] ≈ 7.9 km/s near surface. v_e = √2 × v_o.
  • Satellite energy: KE = GMm/2r, PE = −GMm/r, Total E = −GMm/2r. E = −KE = PE/2.
  • Geostationary: T = 24 h, height ≈ 36,000 km, equatorial, west-to-east orbit.
  • Weightlessness = free fall state. Gravity acts; normal force = 0.
18

Common Mistakes

  • Confusing G and g: G is the universal constant (6.674 × 10⁻¹¹); g is the local acceleration (9.8 m/s² at surface). They are related by g = GM/R² but are entirely different quantities.
  • Using surface g formula at altitude or depth: Once the object is not at Earth’s surface, the formulas change. Use g_h or g_d as applicable — never plug altitude directly into g = GM/R².
  • Forgetting the negative sign in U = −GMm/r: The negative sign is essential. It indicates a bound system. Dropping it completely changes the physics of escape velocity and satellite energy problems.
  • Using wrong distance in F = Gm₁m₂/r²: r is the centre-to-centre distance, not the surface-to-surface distance. For two spheres of radii R₁ and R₂ separated by gap d, use r = R₁ + R₂ + d.
  • Escape velocity has direction dependency: False. v_e is independent of the launch direction — a body launched at v_e in any direction (not just vertical) will escape, provided it doesn’t hit Earth first.
  • Confusing orbital velocity decrease with energy loss: A satellite at a higher orbit moves slower but has higher total energy (less negative). The increase in PE more than compensates for the KE decrease.
  • Thinking weightlessness means no gravity: Gravity is fully acting on an orbiting astronaut. Weightlessness is the absence of contact force (normal reaction), not absence of gravitational force.
  • Applying depth formula at surface: At the surface, d = 0, so g_d = g — this is consistent. At the centre, d = R, so g_d = 0. The formula is linear, not inverse-square inside Earth.

Frequently Asked Questions — Gravitation Class 11

What is the difference between gravitational potential and gravitational potential energy? +
Gravitational potential energy (U) is the energy of a specific mass m at a point in the gravitational field: U = −GMm/r. It depends on the test mass m. Gravitational potential (V), on the other hand, is a property of the field itself at that point — it is the potential energy per unit mass: V = U/m = −GM/r. Potential is independent of the test mass and describes the field environment. This distinction is analogous to the difference between electric potential energy and electric potential in electrostatics.
Why does g decrease both with altitude and with depth? +
Above the surface, g decreases because the distance from Earth’s centre increases — the inverse-square law applies (g ∝ 1/r²). Below the surface, only the mass enclosed within the sphere of radius (R − d) contributes to the gravitational force (shell theorem). This enclosed mass decreases as d increases, and decreases faster than the distance decreases, resulting in a net linear decrease in g. At the exact centre, all forces cancel and g = 0.
How is escape velocity derived, and why is it independent of mass? +
Escape velocity is derived by setting the total mechanical energy equal to zero — the minimum condition for escape to infinity. Starting with: ½mv_e² − GMm/R = 0, solving for v_e gives √(2GM/R). The mass m of the escaping object cancels out completely in this equation, which is why escape velocity is independent of the escaping object’s mass. It only depends on the mass M and radius R of the planet being escaped from.
What happens to the orbital speed of a satellite when it moves to a higher orbit? +
Orbital speed decreases as the satellite moves to a higher orbit. From v_o = √[GM/(R+h)], as h increases, (R+h) increases, so v_o decreases. However, the time period T increases (T ∝ r^(3/2)). This seems paradoxical because energy must be added (via rocket thrust) to move to a higher orbit, yet the satellite ends up moving slower. The explanation is that the large increase in gravitational potential energy more than accounts for the decrease in kinetic energy — total energy increases (becomes less negative).
Why do astronauts experience weightlessness in a satellite even though gravity is acting on them? +
Weightlessness is the sensation caused by the absence of a contact (normal) force, not the absence of gravity. In an orbiting satellite, both the satellite and the astronaut are in the same free-fall state — they are both falling toward Earth at the same rate (the centripetal acceleration equals the local g at that altitude). Because both fall together, the satellite floor never needs to push up against the astronaut. The normal reaction is zero, creating the sensation of weightlessness. Gravity is fully present and is in fact providing the centripetal force for the orbit.
What is the significance of the negative total energy of a satellite? +
The negative total energy (E = −GMm/2r) signifies that the satellite is gravitationally bound to Earth — it cannot escape without additional energy input. The magnitude of the total energy equals the binding energy: the minimum energy that must be supplied to free the satellite from its orbit and send it to infinity. If total energy were zero, the satellite would be at the escape threshold. If positive, the object would have already escaped. This sign convention is universal for all bound gravitational systems, including planetary orbits.

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Table of Contents — Physics Class 11

Table of Contents

Physics — Class 11

01Units and MeasurementsGo to page
02Motion in a Straight LineGo to page
03Motion in a PlaneGo to page
04Laws of MotionGo to page
05Work, Energy and PowerGo to page
06System of Particles and Rotational MotionGo to page
07GravitationGo to page
08Mechanical Properties of SolidsGo to page
09Mechanical Properties of FluidsGo to page
10Thermal Properties of MatterGo to page
11ThermodynamicsGo to page
12Kinetic TheoryGo to page
13OscillationsGo to page
14WavesGo to page

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