Circular Motion and Gravitation

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Uniform Circular Motion — Constant Speed, Changing Direction:

When a body moves in a circle of radius r with constant speed v, it undergoes uniform circular motion. Although speed is constant, velocity changes because the direction changes continuously. This means there is an acceleration — the centripetal acceleration a_c = v²/r, directed toward the centre of the circle. The force causing this acceleration (net force toward centre) is the centripetal force: F_c = mv²/r.

Key quantities:

• Angular velocity ω = 2π/T = v/r (rad/s)
• Period T = 2πr/v = 2π/ω
• Frequency f = 1/T = ω/(2π)
• Centripetal acceleration a_c = ω²r = v²/r

Newton’s Universal Law of Gravitation:

Every particle in the universe attracts every other particle with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between them: F = G(m₁m₂)/r², where G = 6.674 × 10⁻¹¹ N·m²/kg² (gravitational constant).

On Earth’s surface: weight W = mg = G(M_E × m)/R_E², so g = GM_E/R_E². With R_E ≈ 6400 km and M_E ≈ 5.97 × 10²⁴ kg, g ≈ 9.8 m/s². This is why g decreases with altitude (r increases) and with latitude (Earth’s rotation causes equatorial bulging).

⚡ ECAT Tip: The centripetal force is NOT a new type of force — it is whatever real force keeps the body moving in a circle. For a car going around a level curve: centripetal force = friction. For a satellite orbiting Earth: centripetal force = gravitational force = GMm/r². For an electron in a circular orbit in a uniform magnetic field: centripetal force = magnetic force = qvB.

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🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Gravitational Field and Potential:

The gravitational field strength g_field = F/m = GM/r² at distance r from a point mass M. This points toward the mass (attractive). For a uniform spherical mass distribution (like Earth), outside the sphere the field is the same as if all the mass were concentrated at the centre.

Gravitational potential V = -GM/r (negative, because zero is defined at infinity). This is the potential energy per unit mass. The potential difference between two points = work done per unit mass moving between them against gravity. Equipotential surfaces around a point mass are spheres.

Kepler’s Laws of Planetary Motion:

1. Law of Orbits: All planets move in elliptical orbits with the Sun at one focus (not exactly circular, but close for most planets).
2. Law of Equal Areas: A line from planet to Sun sweeps out equal areas in equal times — this means the planet moves faster when closer to the Sun (conservation of angular momentum).
3. Law of Periods: T² ∝ r³, or T² = (4π²/GM) × r³ for circular orbits. More precisely, for elliptical orbits, r is replaced by the semi-major axis a: T² = (4π²/GM) × a³.

For a satellite orbiting Earth at height h: orbital radius r = R_E + h, orbital speed v = √(GM/r), orbital period T = 2π√(r³/GM).

⚡ ECAT Tip: Geostationary satellites orbit at approximately 36,000 km above Earth’s surface. They have T = 24 hours (same as Earth’s rotation), so they appear fixed above the equator. This is useful for communications and weather satellites.

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Derivation of Orbital Speed from Gravitational Force:

For a satellite in a circular orbit of radius r around Earth (mass M_E), the centripetal force required is mv²/r. This is provided entirely by Earth’s gravitational pull: GM_E m/r² = mv²/r. Cancelling m and multiplying by r: v² = GM_E/r. Therefore orbital speed v = √(GM_E/r). This is independent of the satellite’s mass — a 1 kg satellite and a 1000 kg satellite have the same orbital speed at the same altitude.

The escape velocity from Earth’s surface: v_e = √(2GM_E/R_E) = √(2gR_E) ≈ √(2 × 9.8 × 6.4 × 10⁶) ≈ 11.2 km/s. This is √2 times the orbital velocity at Earth’s surface (v_orbital ≈ 7.9 km/s). Note: escape velocity is independent of the object’s mass and direction of launch (as long as it’s not aimed into the ground).

Gravitational Potential Energy — Full Derivation:

The work done moving a mass m from infinity to a point at distance r from Earth’s centre: W = ∫(from ∞ to r) F·dr = ∫(∞ to r) (-GMm/r²) dr = GMm [1/r - 1/∞] = GMm/r. Since we do negative work against gravity (gravity pulls inward, displacement is inward), the gravitational potential energy of the bound system is negative: U = -GMm/r. The total mechanical energy of an orbiting satellite: E = KE + PE = ½mv² - GMm/r = ½m(GM/r) - GMm/r = -½mGM/r = -½mv²_orbital.

For a circular orbit, KE = -½ PE (virial theorem for inverse-square law forces), and total energy is negative (bound orbit). For unbound trajectories (escape), E ≥ 0.

⚡ ECAT Pattern: ECAT frequently tests: (1) v_orbital = √(GM/r) and v_escape = √(2GM/r) = √2 v_orbital; (2) variation of g with altitude g’ = g(R_E/(R_E + h))²; (3) Kepler’s third law T² ∝ r³ with numerical values; (4) centripetal force problems where the tension in a string provides the centripetal force (conical pendulum). A typical ECAT question: “A satellite orbits Earth at height 500 km. Given R_E = 6400 km, g = 9.8 m/s², find the orbital speed.” r = 6900 km = 6.9 × 10⁶ m. v = √(gR_E²/r)… better: v = √(gR_E²/r) = √(9.8 × (6.4 × 10⁶)² / (6.9 × 10⁶)) = √(9.8 × 6.4² × 10⁶ / 6.9) ≈ √(5.9 × 10⁷) ≈ 7680 m/s ≈ 7.68 km/s.