Gravitation

Chapter 8: Gravitation

Gravitation is the force of attraction that exists between every pair of objects in the universe. Newton's law of gravitation unified terrestrial and celestial mechanics — showing that the same force that makes an apple fall also keeps planets in orbit.

Newton's Law of Universal Gravitation

"Every particle of matter 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 them."

F = G M m / r²

Where:
G = 6.674 × 10^-11 N m² kg^-2 (Universal Gravitational Constant)
M, m = masses of the two objects
r = distance between their centres
F is always attractive and acts along the line joining the two objects

The gravitational force obeys Newton's third law: the force on M due to m is equal and opposite to the force on m due to M.

Acceleration Due to Gravity (g)

On Earth's surface (mass M_E, radius R_E):
g = G M_E / R_E²

With M_E = 6 × 10²⁴ kg, R_E = 6.4 × 10⁶ m: g ≈ 9.8 m/s²

Variation of g:
With altitude h: g_h = g (R_E / (R_E + h))² ≈ g(1 - 2h/R_E) for h << R_E (g decreases with altitude)
With depth d: g_d = g(1 - d/R_E) (g decreases with depth; at centre, g = 0)
With latitude: g is slightly less at equator (centrifugal effect + Earth's oblate shape) and greatest at poles.

Gravitational Field

The gravitational field at a point is the gravitational force per unit mass at that point:
E = F/m = GM/r² (directed toward the source mass)

Gravitational Potential Energy

U = -G M m / r (negative because gravitation is attractive; zero at infinity)

The negative sign indicates a bound state — energy must be supplied to escape.

Escape Velocity

The minimum speed needed to escape a planet's gravitational pull:
v_e = √(2GM/R) = √(2gR)

For Earth: v_e = √(2 × 9.8 × 6.4 × 10⁶) ≈ 11.2 km/s

Orbital Velocity

For a satellite in circular orbit at height h:
v_o = √(GM / (R + h))

For a near-Earth orbit (h << R): v_o = √(gR) ≈ 7.9 km/s

Note: v_e = √(2) × v_o (escape velocity is √(2) times orbital velocity)

Time period of satellite: T = 2 pi (R+h) / v_o = 2 pi √((R+h)³ / GM)

Geostationary orbit: T = 24 hours, h ≈ 36,000 km above equator. Satellite appears stationary.

Kepler's Laws of Planetary Motion

1.Law of Orbits: Each planet moves in an ellipse with the Sun at one focus.
2.Law of Areas: A line joining a planet to the Sun sweeps equal areas in equal time intervals. (Angular momentum conservation)
3.Law of Periods: T² is proportional to a³, where a is the semi-major axis of the ellipse. T² = (4 pi² / GM_sun) × a³

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Example 1

Find the gravitational force between Earth (6×10²⁴ kg) and Moon (7.4×10²² kg), separated by 3.8×10⁸ m.
F = G M m / r² = (6.67×10^-11)(6×10²⁴)(7.4×10²²)/(3.8×10⁸)²
= (6.67 × 6 × 7.4 × 10³⁵) / (14.44 × 10¹⁶)
≈ 2.0 × 10²⁰ N

Example 2

A satellite orbits Earth at height 400 km (R_E = 6400 km, g = 9.8 m/s²). Find orbital speed.
v_o = √(G M_E / (R+h)) = √(9.8 × (6.4×10⁶)² / (6.8×10⁶)) ≈ 7670 m/s ≈ 7.67 km/s

Example 3

Find escape velocity from Earth (g = 9.8 m/s², R = 6.4×10⁶ m).
v_e = √(2gR) = √(2 × 9.8 × 6.4×10⁶) = √(125.44×10⁶) ≈ 11,200 m/s = 11.2 km/s

Example 4

At what height above Earth is g half of its surface value?
g_h = g/2 = g R²/(R+h)² → (R+h)² = 2R² → R+h = R √(2) → h = R(√(2) - 1) ≈ 0.414 R ≈ 2650 km

Example 5

Kepler's third law — if Earth takes 1 year and Mars orbits at 1.52 times Earth's distance, find Mars's orbital period.
T² proportional to a³ → T_M²/T_E² = (a_M/a_E)³ = (1.52)³ = 3.51
T_M = √(3.51) ≈ 1.87 years

Example 6

Why does a satellite in orbit experience weightlessness?
Both the satellite and the astronaut inside are in free fall toward Earth at the same rate (same g). The normal contact force between them is zero — hence apparent weightlessness.

Example 7

Show that for a near-Earth orbit, v_e = √(2) × v_o.
v_o = √(gR), v_e = √(2gR) = √(2) × √(gR) = √(2) × v_o. Proved.

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Common mistakes

Taking g as constant everywhere — g varies with altitude, depth, and latitude.
Confusing gravitational potential energy (U = -GMm/r) with potential (V = -GM/r). U = mV.
Thinking objects in satellites are "beyond gravity" — at 400 km altitude, g ≈ 8.7 m/s² (still strong). Weightlessness is due to free fall, not absence of gravity.
Confusing orbital velocity with escape velocity — escape velocity = √(2) × orbital velocity.

Summary

Newton's law of gravitation describes the force between any two masses as F = GMm/r². On Earth's surface g = GM_E/R_E² and varies with height, depth, and latitude. Escape velocity = √(2gR) ≈ 11.2 km/s. Satellites in circular orbits satisfy T² proportional to r³ (Kepler's third law). Geostationary satellites have T = 24 h. Orbital weightlessness results from free fall, not absence of gravity.