Topic 1 of 6 · Form 6 · Upper 6
Circular motion and gravitation
For a satellite, gravitational force provides the centripetal force: GMm/r² = mv²/r, so v = √(GM/r) and T² = (4π²/GM)r³.
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Circular motion
- Angular velocity ω = Δθ/Δt = 2π/T; v = rω.
- Centripetal acceleration a = v²/r = rω²; centripetal force F = mv²/r, directed towards the centre.
Worked example: A 0.50 kg stone on a 0.80 m string moves in a horizontal circle at 4.0 m s⁻¹. Find the tension (ignore gravity).
F = mv²/r = 0.50 × 16 ÷ 0.80 = 10 N
Gravitational fields
- Newton's law: F = GMm/r² (G = 6.67 × 10⁻¹¹ N m² kg⁻²).
- Field strength g = F/m = GM/r².
- Gravitational potential φ = −GM/r (zero at infinity); potential energy = −GMm/r.
Orbits
For a satellite, gravitational force provides the centripetal force: GMm/r² = mv²/r, so v = √(GM/r) and T² = (4π²/GM)r³.
A geostationary satellite orbits above the equator with a period of 24 hours, moving west to east – used for television and weather (about 3.6 × 10⁷ m above the surface).
Exam tip: In orbit questions, r is measured from the centre of the planet, not from its surface.
Check yourself
- Convert 120 revolutions per minute to angular velocity in rad s⁻¹.
- Why is gravitational potential negative?
- State two features of a geostationary orbit.
Show answers
- 2 rev/s × 2π = 4π ≈ 12.6 rad s⁻¹
- It is defined as zero at infinity, and work must be done to move a mass away from the planet towards infinity.
- Any two: period 24 hours, above the equator, moves west to east, fixed position relative to the ground.
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