ZIMSEC A Level · Physics · Form 6
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.
Questions about this topic (0)
No questions about this topic yet.
Sign in or create a free account to join in.