Circular Orbits & Kepler's Third Law — Quick Summary

Mr Toogood's Physics · Gravitational fields

AQA 3.7.2.4
v = √(GM/r)
Orbital speed
mv²/r = GMm/r²
F_c = F_grav
T² = (4π²/GM)r³
Kepler's 3rd law
T = 86400 s
Geostationary

Orbital motion as free-fall

A satellite in circular orbit with gravity providing the centripetal force towards Earth's centre

Gravity provides the centripetal force; the satellite continuously "misses" the surface as it falls.

Newton's cannon: fire a cannonball fast enough and the Earth's surface curves away beneath it exactly as fast as it falls — it never lands, and instead goes into orbit. An orbiting object is in continuous free-fall.

Apparent weightlessness: an astronaut and their station fall at the identical rate, so there's no contact force between them — it's the absence of a contact force, not the absence of gravity, that's felt as weightlessness.

Orbital speed

mv²/r = GMm/r²  →  v = √(GM/r)
  • Satellite mass m cancels — orbital speed depends only on M (source) and r, never on the orbiting object's own mass.
  • Bigger M → faster orbit needed. Bigger r → slower speed needed (v∝1/√r).
Centripetal force is "adjectival" — always another named force in disguise; here, it's gravity.
Common slip: r is measured from the planet's centre, not the surface — always add the planet's radius to a given altitude.

Kepler's third law: T² ∝ r³

Equating gravitational force with centripetal force in ω form, cancelling m, and substituting ω=2π/T:

(GM/4π²)T² = r³  →  T² ∝ r³
Log-log plot of T squared against r cubed for the planets of the Solar System, showing a straight line

A log–log plot confirms T²∝r³ across the huge range of planetary distances.

Holds for any gravitational system — not just planets around the Sun; moons, satellites, any orbiting body.

Unintuitive result: raising an orbit needs more energy overall (KE converts to GPE), yet the satellite ends up moving slower once settled into the new, higher, stable orbit.

Synchronous & geostationary orbits

A geostationary satellite positioned above the equator, remaining above a fixed point on Earth's surface

A geostationary satellite stays above one fixed point — ideal for satellite TV dishes.

  • Geostationary: T=24 h, directly above the equator — stays above a single fixed point.
  • Geosynchronous: same 24 h period, but any other inclination — returns to the same point in the sky daily, without staying fixed above one location.

Solving for r with T=86400 s gives the one unique radius: r ≈ 4.23×10⁷ m. Closer means too short a period; further out, too long.

Exam essentials

Key equations

  • v=√(GM/r)
  • GM=ω²r³
  • T²=(4π²/GM)r³

Geostationary vs. geosynchronous

  • Both: T = 24 hours.
  • Geostationary: equatorial plane only, fixed point.
  • Geosynchronous: any inclination, not fixed above one point.

Common slips

  • r is always measured from the planet's centre — add its radius to the given altitude.
  • A higher orbit means a slower speed, not faster (v∝1/√r).
  • Firing thrusters to speed up ultimately results in a lower final orbital speed once the new, higher orbit is settled into.