Circular Motion, Oscillations and Gravitation
Scope of the domain
This domain covers motion in a circle, simple harmonic motion (SHM), and gravitational fields and orbits. The exam asks you to relate angular speed, linear speed and centripetal force, to use SHM equations and period formulas, and to calculate gravitational field strength and orbital quantities.
Core rules
- Angular speed ω = 2π/T = 2πf; linear speed v = ωr; centripetal acceleration a = v²/r = ω²r, directed towards the centre.
- The resultant force towards the centre has magnitude mv²/r. It is provided by tension, friction, gravity, or a normal reaction, not an additional force.
- SHM: acceleration is proportional to displacement and opposite in direction, a = −ω²x. Maximum speed v₀ = ωA at equilibrium; maximum acceleration ω²A at the extremes.
- SHM periods: mass–spring T = 2π√(m/k); simple pendulum T = 2π√(l/g) for small angles.
- Newton's law of gravitation F = GMm/r²; field strength g = GM/r².
- Orbital speed v = √(GM/r); Kepler's third law T² = (4π²/GM)r³.
- A geostationary satellite orbits above the equator with a 24-hour period, so it stays above one point.
Worked example
A satellite orbits at radius 7.0 × 10⁶ m from Earth's centre. Take GM = 3.99 × 10¹⁴ m³ s⁻². Orbital speed v = √(3.99 × 10¹⁴ ÷ 7.0 × 10⁶) = √(5.70 × 10⁷) = 7.55 × 10³ m/s. Period T = 2πr/v = 2π × 7.0 × 10⁶ ÷ 7.55 × 10³ = 5.83 × 10³ s, about 97 minutes. Note that r is measured from Earth's centre, not from the surface.
Traps that cost marks
- Using altitude instead of orbital radius and forgetting to add Earth's radius.
- Adding an extra outward force to a circular-motion diagram; centripetal force is the resultant, not a separate force.
- Writing v = ωr but using the wrong radius, such as the diameter.
- Forgetting the minus sign in a = −ω²x, which shows acceleration points towards equilibrium.
- Applying the pendulum formula at large angles, where the SHM approximation fails.
- Treating g as constant with altitude; it falls as 1/r².