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Physics AMU 11 Science Class 11 CUET SSC Railway

Gravitation – Complete AMU 11 Notes

📅 Published: June 2026

Gravitation connects the motion of planets with the falling of an apple. AMU asks 2-3 questions from this chapter — Newton's law, Kepler's laws, and variation of g are the hot topics. Mass vs weight always has a trick question. This expanded note covers Newton's law, Kepler's 3 laws, g and its variation, escape velocity, orbital velocity, satellites, gravitational field, gravitational PE, and 15+ PYQs. Let's go.

1) Newton's Universal Law of Gravitation

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.

F = G (m₁ m₂) / r² G = Universal gravitational constant = 6.67 × 10⁻¹¹ N m²/kg²

Vector Form

The gravitational force on m₁ due to m₂ is: F₁₂ = -G(m₁m₂/r²) r̂₂₁
where r̂₂₁ is the unit vector from m₂ to m₁. The negative sign indicates attraction.

Characteristics of Gravitational Force:

🔥 AMU PYQ: Newton's universal law of gravitation is given by:
(A) F = G m₁ m₂ / r⁴ (B) F = G m₁ m₂ / r² (C) F = G (m₁ + m₂) / r² (D) F = G (m₁ - m₂) / r²
Ans: B

🔥 AMU PYQ: The value of G was first experimentally determined by:
(A) Newton (B) Einstein (C) Cavendish (D) Kepler
Ans: C (Cavendish used a torsion balance in 1798)

🔥 AMU PYQ: Gravitational force between two bodies does NOT depend on:
(A) Product of masses (B) Distance between them (C) Medium between them (D) Universal constant G
Ans: C (gravitational force is independent of the medium)

Importance of the Universal Law of Gravitation

Newton's Universal Law of Gravitation was a revolutionary breakthrough because it unified terrestrial physics (objects falling on Earth) with celestial physics (motion of planets). Before Newton, people thought different laws governed the heavens and the Earth. Newton showed that the same force that makes an apple fall also keeps the Moon in orbit!

Cavendish Experiment — Measuring G

In 1798, Henry Cavendish used a torsion balance to measure the gravitational force between two lead spheres. From this, he calculated G ≈ 6.67 × 10⁻¹¹ N m²/kg². This is why Cavendish is credited with "weighing the Earth" — once G was known, M = gR²/G gave the mass of Earth.

Cavendish torsion balance principle:
A horizontal rod with small lead spheres is suspended by a thin wire. Large lead spheres placed near small ones create gravitational attraction. The rod rotates, twisting the wire. Angle of twist → force → G.

Why G is a Universal Constant

2) Gravitational Field and Intensity

The gravitational field at a point is the gravitational force experienced by a unit mass placed at that point. It is a vector quantity.

E = F / m₀ = GM / r² Direction: Towards the centre of the mass creating the field SI unit: N/kg (same as m/s² — note that g is the gravitational field intensity!)

So g is actually the gravitational field intensity at Earth's surface. That's why g = 9.8 N/kg = 9.8 m/s².

3) Kepler's Laws of Planetary Motion

Johannes Kepler (1571-1630) gave 3 laws based on Tycho Brahe's precise observations of planetary motion. Newton later used these laws to derive his Universal Law of Gravitation.

Law 1 — The Law of Orbits (Elliptical Orbit Law)

Every planet moves around the Sun in an elliptical orbit, with the Sun at one of the two foci.

An ellipse has two foci. The Sun is at one focus, not the centre. Eccentricity (e) = distance between foci / length of major axis Earth's orbit: e ≈ 0.0167 (nearly circular) Mercury's orbit: e ≈ 0.205 (most elliptical among planets)

Law 2 — The Law of Areas (Equal Area Law)

The line joining the planet to the Sun sweeps out equal areas in equal intervals of time.

Areal velocity = dA/dt = constant Mathematically: dA/dt = L / 2m = constant L = angular momentum, m = mass of planet

Consequence: When a planet is closer to the Sun (perihelion), it moves faster. When farther (aphelion), it moves slower. For Earth:

🔥 AMU PYQ: Kepler's second law is based on which fundamental conservation law?
(A) Conservation of energy (B) Conservation of angular momentum (C) Conservation of linear momentum (D) Conservation of mass
Ans: B (Conservation of angular momentum)

🔥 AMU PYQ: Kepler's second law implies that a planet moves ______ when it is closer to the Sun.
(A) slower (B) faster (C) at same speed (D) stops
Ans: B

Law 3 — The Law of Periods (Harmonic Law)

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 T² = (4π² / GM) r³ Where: r = semi-major axis (average distance from Sun) T = time period, M = mass of the central body (Sun)
🟡 Senior's Trick — If distance changes:
T² ∝ r³ → T ∝ r^(3/2)
If r → 2r: T → 2^(3/2) = 2√2 ≈ 2.83 times
If r → 3r: T → 3^(3/2) = 3√3 ≈ 5.2 times
🔥 AMU PYQ: According to Kepler's third law, T² is proportional to:
(A) r (B) r² (C) r³ (D) r⁴
Ans: C (r³)

🔥 AMU PYQ: If the distance of Jupiter from the Sun is 5.2 times that of Earth, find Jupiter's time period.
Ans: T_J² / T_E² = (r_J / r_E)³ = (5.2)³ = 140.6
T_J = √140.6 × 1 year = 11.86 years ≈ 12 years

4) Acceleration Due to Gravity (g)

The acceleration produced in a body due to the gravitational pull of Earth is called acceleration due to gravity (g).

Derivation of g = GM/R²

From Newton's law: F = G M m / R² (where M = mass of Earth, m = mass of body, R = Earth's radius)
Also from Newton's 2nd law: F = mg
Equating: mg = G M m / R²
→ g = G M / R²
Value:
G = 6.67 × 10⁻¹¹ N m²/kg²
M = 6 × 10²⁴ kg (mass of Earth)
R = 6.4 × 10⁶ m (radius of Earth)
g = (6.67×10⁻¹¹ × 6×10²⁴) / (6.4×10⁶)²
g = (4.002×10¹⁴) / (4.096×10¹³)
g = 9.8 m/s²

Mass of Earth from g:

M = gR² / G = (9.8 × (6.4×10⁶)²) / 6.67×10⁻¹¹ ≈ 6 × 10²⁴ kg

5) Mass vs Weight

This is a very common AMU trick topic. Know the difference cold.

Feature Mass (m) Weight (W)
DefinitionAmount of matter in a bodyGravitational force on a body
Formula—W = mg
SI unitkg (kilogram)N (newton)
NatureScalarVector (towards Earth's centre)
Constant?Yes (same everywhere in universe)No (varies with location)
Can be zero?NoYes (free fall, space, centre of Earth)
Measuring devicePhysical balance / beam balanceSpring balance
At MoonSame as on Earth≈ 1/6 of Earth weight
At centre of EarthSameZero
Weight on Moon: gmoon = 1.62 m/s² ≈ g/6
If a person weighs 60 kg (mass), weight on Earth = 60×9.8 = 588 N
Weight on Moon = 60×1.62 = 97.2 N ≈ 1/6 of Earth weight
🔥 AMU PYQ: A body of mass 10 kg on Earth. Its weight on Earth is (g = 10 m/s²):
(A) 10 N (B) 100 N (C) 1 N (D) 1000 N
Ans: B (W = mg = 10 × 10 = 100 N)

🔥 AMU PYQ: Mass of a body on Moon is 60 kg. Its mass on Earth is:
(A) 10 kg (B) 60 kg (C) 360 kg (D) 6 kg
Ans: B (Mass is constant everywhere!)

6) Variation of g with Height, Depth, and Latitude

The value of g is not constant everywhere on or above Earth. It varies with:

6.1 Variation with Height (above Earth's surface)

g_h = GM / (R + h)²
g_h / g = R² / (R + h)²
Formula: g_h = g [R / (R + h)]²

For small h (h << R), using binomial approximation:
g_h = g (1 - 2h/R)
As h increases, g decreases.

6.2 Variation with Depth (below Earth's surface)

g_d = GM' / d² (where M' = mass of sphere of radius R-d)
Assuming uniform density:
Formula: g_d = g (1 - d/R)

At the centre of Earth (d = R): g = 0
As depth increases, g decreases linearly.

6.3 Variation with Latitude (Rotation of Earth)

geff = g - R ω² cos² φ
At equator (φ = 0°): g is minimum (due to maximum centrifugal effect) At poles (φ = 90°): g is maximum (no centrifugal effect)

Comparison Table — Variation of g:

Location Value of g Reason
Surface of Earthg = 9.8 m/s²Standard value
Height h above surfaceg_h = g [R/(R+h)]²Distance from centre increases → g decreases
Depth d below surfaceg_d = g (1 - d/R)Mass below reduces → g decreases linearly
Centre of Earth0Equal gravitational pull from all sides cancels
Equator≈ 9.78 m/s²Centrifugal effect + larger radius
Poles≈ 9.83 m/s²No centrifugal effect + smaller radius
MountainsLess than sea levelGreater distance from Earth's centre
Moon≈ 1.62 m/s²Smaller mass and radius
🔥 AMU PYQ: Acceleration due to gravity with increasing altitude:
(A) Increases (B) Decreases (C) Remains same (D) First increases then decreases
Ans: B (Decreases)

🔥 AMU PYQ: At the centre of the Earth, the value of g is:
(A) 9.8 m/s² (B) 4.9 m/s² (C) 0 (D) Infinite
Ans: C (Zero)

🔥 AMU PYQ: Where is the value of g maximum on Earth's surface?
(A) Equator (B) Tropic of Cancer (C) Poles (D) At 45° latitude
Ans: C (Poles)

7) Gravitational Potential Energy (Basic Concept)

Gravitational potential energy (U) of a body at a point is the work done in bringing it from infinity to that point against the gravitational field. It is always negative — which means the force is attractive.

U = -G M m / r At infinity: U = 0 (by convention) Near Earth's surface: U = mgh (approximate for small h)
🟡 Senior's Insight:
• U is negative because work is done by the gravitational field (attractive force).
• As a body falls towards Earth, its gravitational PE decreases (becomes more negative).
• The kinetic energy increases as PE decreases — conservation of mechanical energy!

8) Free Fall (Detailed)

When a body falls only under the influence of gravity (no air resistance), it is called free fall.

Key Points about Free Fall:

Equations for Free Fall (taking downward direction as positive):
v = u + g t h = u t + ½ g t² v² = u² + 2 g h

Weightlessness

Weightlessness is the feeling of zero weight (g = 0). This happens when:

🟡 Classic AMU Trick:
A feather and a hammer dropped on the Moon hit the ground at the same time because g is same for all bodies in vacuum. On Earth, air resistance makes the feather fall slower — but in vacuum, even on Earth, they'd hit together!

9) Escape Velocity

The minimum velocity required for a body to escape Earth's gravitational pull and never return is called escape velocity.

Derivation:

For a body to escape, its kinetic energy must be ≥ the magnitude of gravitational potential energy.
KE ≥ |PE|
½ m vₑ² ≥ G M m / R
vₑ² ≥ 2 G M / R
vₑ = √(2GM / R) = √(2gR)
For Earth:
vₑ = √(2GM/R) = √(2 × 6.67×10⁻¹¹ × 6×10²⁴ / 6.4×10⁶)
vₑ = √(1.25×10⁸) ≈ 1.12×10⁴ m/s
vₑ = 11.2 km/s
Celestial Body Escape Velocity
Earth11.2 km/s
Moon2.4 km/s
Sun618 km/s
Jupiter59.5 km/s
Mars5.0 km/s
Mercury4.3 km/s
🟡 Senior's Trick:
• Escape velocity does NOT depend on the mass of the escaping body!
• Escape velocity depends only on the mass and radius of the planet.
• If Earth's radius shrinks to half, escape velocity becomes √2 times = 15.8 km/s.
• Moon's escape velocity is only 2.4 km/s — that's why it can't hold an atmosphere!

10) Orbital Velocity and Satellites

Orbital Velocity (v₀)

The velocity required for a satellite to revolve around Earth in a circular orbit is called orbital velocity.

For a satellite in circular orbit:
Centripetal force = Gravitational force
mv₀²/r = GMm/r²
v₀ = √(GM / r) = √(gR² / r)

For a satellite close to Earth (r ≈ R):
v₀ = √(gR) = √(9.8 × 6.4×10⁶) ≈ 7.9 km/s

Relation Between Escape Velocity and Orbital Velocity:

vₑ = √2 × v₀
vₑ / v₀ = √2 ≈ 1.414
Escape velocity is 41.4% more than orbital velocity.

Time Period of a Satellite:

T = 2πr / v₀ = 2π √(r³ / GM) This is Kepler's 3rd law: T² ∝ r³

Types of Satellites:

Type Properties
Natural satellitesMoons of planets. Example: Moon (natural satellite of Earth).
Artificial satellitesMan-made. Used for communication, weather, navigation, research.

Geostationary (Geosynchronous) Satellites:

Polar Satellites:

🔥 AMU PYQ: The orbital speed of a satellite very close to Earth's surface is:
(A) √(gR) (B) √(2gR) (C) √(GM/R²) (D) √(2GM/R)
Ans: A (√(gR) ≈ 7.9 km/s)

🔥 AMU PYQ: The time period of a geostationary satellite is:
(A) 12 hours (B) 24 hours (C) 48 hours (D) 1 hour
Ans: B

🔥 AMU PYQ: The ratio of escape velocity to orbital velocity for a satellite near Earth's surface is:
(A) √2 : 1 (B) 1 : √2 (C) 2 : 1 (D) 1 : 2
Ans: A (vₑ = √2 × v₀)

11) Thrust and Pressure (Brief)

Thrust is the force acting perpendicularly to a surface. Pressure is thrust per unit area.

Pressure (P) = Force (F) / Area (A) SI unit: Pascal (Pa) = N/m²

In the context of gravitation, the weight of a body (thrust) acting on a surface produces pressure. This is why:

12) Thrust and Pressure (Brief)

Thrust is the force acting perpendicularly to a surface. Pressure is thrust per unit area.

Pressure (P) = Force (F) / Area (A) SI unit: Pascal (Pa) = N/m²

In the context of gravitation, the weight of a body (thrust) acting on a surface produces pressure. This is why:

🔥 AMU PYQ: Pressure is defined as:
(A) Force × Area (B) Force / Area (C) Area / Force (D) Mass × Acceleration
Ans: B (P = F/A)

13) Applications of Gravitation in Daily Life

🟡 Senior's Insight: Gravitation is the weakest but most far-reaching force in nature. It holds galaxies together, keeps our feet on the ground, and controls the motion of everything in the universe — from an apple to entire galaxies.

14) Why the Moon has No Atmosphere

The Moon's escape velocity is only 2.4 km/s, which is very small compared to Earth's 11.2 km/s. The average speed of gas molecules at normal temperatures is around 1-2 km/s. Many gas molecules on the Moon can easily exceed 2.4 km/s and escape into space.

Key point: For a planet or moon to retain an atmosphere, its escape velocity must be at least 6 times the root-mean-square speed of gas molecules. Since the Moon's escape velocity is too low, it couldn't hold onto an atmosphere.

🔥 AMU PYQ: Why does the Moon have no atmosphere?
Ans: Because its escape velocity (2.4 km/s) is too low to retain gas molecules, which can easily escape into space.

15) Common Mistakes Students Make in Gravitation

⚠️ Mistake 1: Confusing G and g.
Fix: G (universal constant, 6.67×10⁻¹¹) is the same everywhere. g (9.8 m/s² on Earth surface) varies with location.

⚠️ Mistake 2: Thinking mass changes with location.
Fix: Mass is constant everywhere. Weight changes because g changes.

⚠️ Mistake 3: Applying g = GM/R² at height without adjusting R.
Fix: At height h, distance from centre = R + h, so g_h = GM/(R+h)²

⚠️ Mistake 4: Forgetting that Kepler's 3rd law applies to any satellite, not just planets.
Fix: T² ∝ r³ works for any body orbiting a central mass (including Earth satellites).

⚠️ Mistake 5: Thinking escape velocity depends on the mass of the escaping object.
Fix: vₑ = √(2GM/R) — no m term! It's independent of the mass of the escaping body.

⚠️ Mistake 6: Confusing orbital velocity with escape velocity.
Fix: v₀ = √(gR) ≈ 7.9 km/s (orbital), vₑ = √(2gR) ≈ 11.2 km/s (escape). vₑ = √2 × v₀

⚠️ Mistake 7: Thinking g is zero inside Earth only at one point.
Fix: g decreases linearly with depth from surface to centre, where g = 0.

13) Important Formulae — At a Glance

Quantity Formula
Newton's law of gravitationF = G m₁ m₂ / r²
Gravitational constantG = 6.67 × 10⁻¹¹ N m²/kg²
Acceleration due to gravityg = GM / R² = 9.8 m/s²
Mass of EarthM = gR² / G ≈ 6 × 10²⁴ kg
WeightW = mg
g at height hg_h = g [R/(R+h)]² ≈ g(1 - 2h/R) for h << R
g at depth dg_d = g (1 - d/R)
Kepler's 1st lawElliptical orbits, Sun at one focus
Kepler's 2nd lawAreal velocity = constant (angular momentum conserved)
Kepler's 3rd lawT² ∝ r³, T² = (4π²/GM) r³
Gravitational fieldE = GM/r² = g
Gravitational PEU = -GMm/r
Escape velocityvₑ = √(2GM/R) = √(2gR) = 11.2 km/s
Orbital velocityv₀ = √(GM/r) = √(gR²/r)
Near Earth orbital velocityv₀ = √(gR) ≈ 7.9 km/s
vₑ : v₀vₑ = √2 × v₀
Satellite time periodT = 2π √(r³/GM)
Geostationary satellite height≈ 36,000 km, T = 24 h
PressureP = F/A, SI unit: Pa (N/m²)

14) PYQ Battle — 16 AMU Past Questions on Gravitation

Q1. Which law describes the orbits of planets around the Sun?
(A) Newton's law (B) Faraday's law (C) Kepler's law (D) Kirchhoff's law
Ans: C
Q2. The force of attraction between two masses does NOT depend on:
(A) Product of masses (B) Distance between them (C) Medium between them (D) G
Ans: C (gravitational force is independent of the medium)
Q3. If the distance between two masses is halved, the gravitational force becomes:
(A) Half (B) Double (C) Four times (D) One-fourth
Ans: C (F ∝ 1/r², so halving r → 4× force)
Q4. A body weighs 63 N on Earth's surface. What is the gravitational force on it at height = R/2 above the surface?
Ans: g_h = g [R/(R+R/2)]² = g(R/(3R/2))² = g(2/3)² = (4/9)g
W' = (4/9)×63 = 28 N
Q5. A body weighs 250 N on Earth's surface. Its weight halfway down to the centre of Earth is:
Ans: g_d = g(1 - d/R) = g(1 - (R/2)/R) = g/2
W' = 250/2 = 125 N
Q6. The escape velocity from Earth's surface is about:
(A) 7.9 km/s (B) 11.2 km/s (C) 3×10⁸ m/s (D) 5 km/s
Ans: B
Q7. A planet has double the mass of Earth but same density. The escape velocity compared to Earth is:
(A) Same (B) Double (C) Half (D) Four times
Ans: A (if density same, vₑ ∝ R)
Q8. Correct expression for escape speed from Earth:
(A) √(gR) (B) √(2gR) (C) gR (D) R/√g
Ans: B
Q9. The value of g is minimum at:
(A) Poles (B) Equator (C) Centre of Earth (D) Mountains
Ans: C (g = 0 at centre)
Q10. The time period of a geostationary satellite is:
(A) 12 hours (B) 24 hours (C) 48 hours (D) 1 hour
Ans: B
Q11. Kepler's second law is a consequence of conservation of:
(A) Energy (B) Linear momentum (C) Angular momentum (D) Mass
Ans: C
Q12. If Earth's mass doubles and radius doubles, g becomes:
(A) 2g (B) g/2 (C) 4g (D) Same
Ans: D (g ∝ M/R² → 2M/(2R)² = 2M/4R² = M/2R² — wait, that's g/2. Let me recalculate: g₂ = G(2M)/(2R)² = 2GM/4R² = GM/2R² = g/2. So B)
Q13. The orbital velocity of a satellite close to Earth is:
(A) 7.9 km/s (B) 11.2 km/s (C) 3×10⁸ m/s (D) 9.8 m/s
Ans: A
Q14. A satellite orbits Earth at height equal to Earth's radius. Its orbital velocity is:
Ans: v₀ = √(GM/(R+R)) = √(GM/2R) = √(gR/2) = √(9.8×6.4×10⁶/2) = √(31.36×10⁶) ≈ 5.6 km/s
Q15. A body of mass m is taken from Earth's surface to height equal to R. The change in gravitational potential energy is:
Ans: ΔU = U₂ - U₁ = (-GMm/2R) - (-GMm/R) = GMm(-1/2R + 1/R) = GMm/2R
Q16. The gravitational force between two objects is 100 N. If the distance is tripled, the new force is:
(A) 100 N (B) 33.3 N (C) 11.1 N (D) 900 N
Ans: C (F ∝ 1/r², triple r → 1/9 × 100 = 11.1 N)

15) Practice Exercise — 15 Problems

Try these. Cover the answers — honest practice is the key.

1. If the mass of a planet is 4× Earth's mass and radius is 2× Earth's radius, find g on that planet.
Answerg' = G(4M)/(2R)² = 4GM/4R² = GM/R² = g. So g' = 9.8 m/s² (same as Earth)
2. Calculate the gravitational force between two 1 kg masses placed 1 m apart.
AnswerF = 6.67×10⁻¹¹ × 1 × 1 / 1² = 6.67×10⁻¹¹ N (extremely small!)
3. The mass of Earth is 6×10²⁴ kg and radius 6.4×10⁶ m. Find escape velocity.
Answervₑ = √(2GM/R) = √(2×6.67×10⁻¹¹×6×10²⁴/6.4×10⁶) = √(1.25×10⁸) = 1.12×10⁴ m/s = 11.2 km/s
4. At what height above Earth's surface does g become half of its surface value?
Answerg/2 = g[R/(R+h)]² → 1/2 = R²/(R+h)² → (R+h)/R = √2 → h = (√2 - 1)R = 0.414R ≈ 2650 km
5. The time period of a satellite orbiting at height 2R from Earth's surface is how many times that of a satellite close to Earth?
AnswerT/T₀ = (r/r₀)^(3/2) = ((R+2R)/R)^(3/2) = (3)^(3/2) = 3√3 ≈ 5.2 times
6. A planet has escape velocity 22.4 km/s. What is its value of gR?
Answervₑ = √(2gR) → gR = vₑ²/2 = (22.4×10³)²/2 = 501.76×10⁶/2 = 2.5088×10⁸ m²/s²
7. Find the gravitational force between Earth (6×10²⁴ kg) and Moon (7.4×10²² kg) if distance = 3.84×10⁸ m.
AnswerF = 6.67×10⁻¹¹ × 6×10²⁴ × 7.4×10²² / (3.84×10⁸)² ≈ 2×10²⁰ N
8. A body weighs 60 N at Earth's surface. What is its weight at a depth of R/4?
Answerg_d = g(1 - d/R) = g(1 - 1/4) = 3g/4. W' = 60×3/4 = 45 N
9. What is the value of g at a height of 100 km above Earth's surface? (R = 6400 km)
Answerg_h = g(1 - 2h/R) = 9.8(1 - 200/6400) = 9.8(1 - 0.03125) = 9.8×0.96875 ≈ 9.49 m/s²
10. Two bodies of masses 5 kg and 10 kg are placed 2 m apart. Find gravitational force.
AnswerF = 6.67×10⁻¹¹ × 5 × 10 / 4 = 6.67×10⁻¹¹ × 50/4 = 8.34×10⁻¹⁰ N
11. What is the time period of a satellite orbiting at Earth's surface?
AnswerT = 2π√(R³/GM) = 2π√(R/g) = 2π√(6.4×10⁶/9.8) = 2π×808 = 5076 s ≈ 84.6 min
12. A pendulum clock calibrated at sea level will run faster or slower on a mountain?
AnswerSlower (because g is less on the mountain, so T = 2π√(L/g) increases, clock ticks slower)
13. The mass of Jupiter is 318× Earth mass and radius is 11.2× Earth radius. Find g on Jupiter.
Answerg_J = G(318M)/(11.2R)² = 318/125.44 × g = 2.53 × 9.8 ≈ 24.8 m/s²
14. A body is projected with velocity 15 km/s from Earth. Will it escape?
AnswerEscape velocity = 11.2 km/s. Since 15 > 11.2, YES, it will escape.
15. If Earth stops rotating, what happens to your weight at the equator?
AnswerYour weight will increase slightly (because the centrifugal effect that reduces apparent weight at equator will be gone).

16) Quick Revision — One-liners

✅ Gravitation done. Remember: Kepler → motion of planets, Newton → why they move that way. Gravitation is one of the most beautiful chapters — it explains both a falling apple and orbiting galaxies. Next up: Work, Power, Energy!
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