Gravitation – Chapter 7 | AAI ATC Physics

✈️ AAI ATC Relevance: Satellite communication & navigation systems rely on gravitational principles

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7.1 Introduction

Gravity is the force that governs the universe — from apples falling to satellites orbiting. Key historical milestones:

Galileo (1564–1642): All bodies fall with the same acceleration (independent of mass).
Ptolemy (~2000 yrs ago): Geocentric model — all celestial objects revolve around Earth.
Copernicus (1473–1543): Heliocentric model — planets revolve around the Sun.
Tycho Brahe (1546–1601): Precise naked-eye planetary data collection.
Johannes Kepler (1571–1640): Derived three laws of planetary motion from Brahe's data.
Isaac Newton: Proposed Universal Law of Gravitation inspired by Kepler's laws.

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[ Add Image: Historical timeline of Gravitation discoveries ]

🔢 Practice MCQs — Introduction

Q1. Galileo established experimentally that all bodies, irrespective of their masses, fall towards earth with:

Q2. Who proposed the heliocentric model of the solar system in 1543?

Q3. Kepler derived his three laws of planetary motion by analysing data compiled by:

✈️ AAI ATC Relevance: Satellite orbit design is based directly on Kepler's laws

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7.2 Kepler's Laws of Planetary Motion

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Law of Orbits

All planets move in elliptical orbits with the Sun at one of the two foci. (Perihelion = closest point; Aphelion = farthest)

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Law of Areas

The line joining Sun to planet sweeps equal areas in equal time intervals. (Conservation of angular momentum)

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Law of Periods

T² ∝ a³ — square of orbital period is proportional to cube of semi-major axis.

T² = (4π²/GM_s) × R³ | T²/R³ = constant for all planets

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[ Add Image: Kepler's Elliptical Orbit with Perihelion & Aphelion ]

💡 Law of Areas & Angular Momentum

ΔA/Δt = L/(2m) = constant. This shows Kepler's 2nd law is a consequence of conservation of angular momentum, valid for any central force.

🔢 Practice MCQs — Kepler's Laws

Q4. A planet is at perihelion with speed v_P and at aphelion with speed v_A. If the perihelion distance is r_P and aphelion distance is r_A, then by conservation of angular momentum:

Q5. The time period of Earth is 1 year and its orbital radius is R. A planet has orbital radius 4R. What is its time period?

Q6. Kepler's law of equal areas is a consequence of which conservation law?

✈️ AAI ATC Relevance: Foundation of all satellite orbital mechanics

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7.3 Universal Law of Gravitation & 7.4 Gravitational Constant

Newton's Universal Law of Gravitation: Every body in the universe attracts every other body with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

F = G·m₁·m₂ / r²

G — Universal Gravitational ConstantG = 6.67 × 10⁻¹¹ N m² kg⁻². Measured by Cavendish (1798) using torsion balance.

Nature of ForceAttractive, central force. Acts along the line joining the two masses. Follows Newton's 3rd law: F₁₂ = −F₂₁.

SuperpositionNet force on a mass = vector sum of all individual gravitational forces from other masses.

Cavendish ExperimentMeasured G using a torsion wire, two small and two large lead spheres. Hence "weighed the Earth".

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[ Add Image: Cavendish Experiment Diagram ]

🔢 Practice MCQs — Universal Law & G

Q7. If the distance between two masses is doubled, the gravitational force between them becomes:

Q8. Two masses 5 kg and 10 kg are separated by 1 m. What is the gravitational force between them? (G = 6.67 × 10⁻¹¹ N m² kg⁻²)

Q9. A point mass is placed inside a hollow spherical shell of uniform density. The gravitational force on it is:

✈️ AAI ATC Relevance: Altitude variations in g affect pressure altimeter calibration

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7.5 & 7.6 Acceleration Due to Gravity

On Earth's surface: g = GM_E/R_E² ≈ 9.8 m/s²

📈 Above the Surface (height h)

g(h) = GM_E/(R_E+h)²

For h << R_E: g(h) ≈ g(1 − 2h/R_E)

g decreases as altitude increases.

📉 Below the Surface (depth d)

g(d) = g(1 − d/R_E)

g decreases linearly with depth.

g = 0 at the centre of Earth.

🔑 Key Fact for Exams

g is maximum at Earth's surface. It decreases whether you go up (above surface) or down (below surface). g = 0 at Earth's centre.

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[ Add Image: Variation of g with height and depth graph ]

🔢 Practice MCQs — Variation of g

Q10. A body weighs 63 N on Earth's surface. What is the gravitational force on it at height h = R_E/2 from the surface?

Q11. A body weighs 250 N on Earth's surface. How much will it weigh half-way to the centre of Earth?

Q12. At what depth is the value of g equal to half of its value on the surface? (R_E = 6400 km)

✈️ AAI ATC Relevance: Satellite launch energy calculations use gravitational PE

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7.7 Gravitational Potential Energy

Gravitational potential energy (GPE) of a particle of mass m at distance r from Earth's centre:

V(r) = −GM_Em / r (V = 0 as r → ∞)

For two masses m₁ and m₂ separated by r:

V = −Gm₁m₂ / r

GPE is always negative (attractive force).
Work done in moving mass from r₁ to r₂: W₁₂ = −GM_Em(1/r₂ − 1/r₁)
Near Earth's surface approximation: V(h) ≈ mgh (valid only for h << R_E)

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[ Add Image: Gravitational PE vs Distance graph ]

🔢 Practice MCQs — Gravitational Potential Energy

Q13. The gravitational potential energy at infinity is taken as:

Q14. A system of 4 equal masses m at vertices of a square of side l has how many mass pairs at distance l?

Q15. A 400 kg satellite moves from circular orbit of radius 2R_E to 4R_E. The change in total energy is: (g = 9.81 m/s², R_E = 6.37 × 10⁶ m)

✈️ AAI ATC Relevance: Rocket launch missions require knowledge of escape velocity

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7.8 Escape Speed

The minimum speed required for an object to escape Earth's gravitational pull (reach infinity with zero speed):

v_e = √(2GM_E/R_E) = √(2gR_E) ≈ 11.2 km/s

Escape Speed – Earth11.2 km/s. Depends on mass & radius of planet, NOT on the mass or direction of the projected body.

Escape Speed – Moon≈ 2.3 km/s (5× less than Earth). This is why Moon has no atmosphere — gas molecules escape easily.

Total EnergyAt escape, total mechanical energy = 0. ½mv² = GMm/R → v = √(2GM/R)

Not Escape VelocityEscape speed is a scalar — independent of direction of projection.

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[ Add Image: Escape Speed concept — projectile reaching infinity ]

🔢 Practice MCQs — Escape Speed

Q16. The escape speed from Earth's surface is 11.2 km/s. A body is projected with 3 times this speed. What is the speed of the body far away from Earth?

Q17. The escape speed on a planet of mass 4M and radius 2R (compared to Earth with mass M, radius R) is:

Q18. The Moon has no atmosphere primarily because:

✈️ AAI ATC Relevance: Communication satellites, GPS, weather satellites — all ATC-critical systems

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7.9 & 7.10 Earth Satellites & Orbital Energy

A satellite in circular orbit at height h above Earth's surface:

Orbital Speed: v = √(GM_E/(R_E+h))

Time Period: T = 2π(R_E+h)^(3/2) / √(GM_E) | T² = k(R_E+h)³

Energy of Orbiting Satellite:

KE = +GM_Em / 2(R_E+h) | PE = −GM_Em / (R_E+h) | Total E = −GM_Em / 2(R_E+h)

Total Energy = NegativeSatellite is bound. KE = −(Total E); PE = 2×(Total E).

Geostationary OrbitT = 24 h, height ≈ 36,000 km. Used for communication satellites (INSAT series).

Close Orbit (h≈0)v ≈ √(gR_E) ≈ 7.9 km/s; T₀ ≈ 85 minutes.

WeightlessnessAstronaut experiences weightlessness NOT because g = 0, but because both astronaut and satellite are in free fall.

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[ Add Image: Satellite orbit diagram with energy labels ]

🔢 Practice MCQs — Earth Satellites

Q19. For a satellite in circular orbit, the kinetic energy (KE) is related to its total mechanical energy (E) as:

Q20. A satellite orbits Earth at height 400 km. Mass of satellite = 200 kg, M_E = 6×10²⁴ kg, R_E = 6.4×10⁶ m, G = 6.67×10⁻¹¹. Energy required to rocket it out of Earth's gravitational influence is approximately:

Q21. The time period of a satellite very close to Earth's surface (h ≈ 0) is approximately:

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Chapter 7 — Quick Formula Summary

Quantity	Formula	Value / Unit
Universal Gravitational Constant	G	6.67 × 10⁻¹¹ N m² kg⁻²
Newton's Law of Gravitation	F = Gm₁m₂/r²	N (Newton)
g on Earth's surface	g = GM_E/R_E²	≈ 9.8 m/s²
g at height h (h << R_E)	g(h) ≈ g(1 − 2h/R_E)	m/s²
g at depth d	g(d) = g(1 − d/R_E)	m/s²
Gravitational PE	V = −Gm₁m₂/r	J (Joule)
Escape Speed	v_e = √(2GM_E/R_E)	11.2 km/s
Orbital Speed	v = √(GM_E/(R_E+h))	m/s
Orbital Time Period	T = 2π(R_E+h)^(3/2)/√(GM_E)	s
Satellite Total Energy	E = −GM_Em/2(R_E+h)	J (negative)
Kepler's 3rd Law	T² ∝ a³ (T²/a³ = const)	—

Chapter 7 — Gravitation

Law of Orbits

Law of Areas

Law of Periods

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