Definition
Orbital period (T) = the time taken for an object to complete one full orbit
- Units: seconds (s) in SI, but in practice often expressed in minutes, hours, days or years depending on the orbit
- Examples:
- International Space Station: T ≈ 90 minutes
- The Moon around Earth: T ≈ 27.3 days
- Earth around the Sun: T = 1 year
- Halley's Comet around the Sun: T ≈ 76 years
Orbital speed equation
- In one complete orbit, the body sweeps out a path whose length is the circumference of the orbit (treating it as a circle):
distance per orbit = 2πr
- where r is the average orbital radius (measured from the centre of the body being orbited to the orbiting body, not from the surface)
- Combine this with speed = distance / time to get the orbital speed equation:
v = 2πr / T
- Where:
- v = orbital speed (m/s)
- r = orbital radius (m)
- T = orbital period (s)
Watch out for the radius
- r is from centre to centre, not from the planet's surface
- For a satellite in low Earth orbit at a height h above the surface, the orbital radius is:
r = R_Earth + h
- where R_Earth ≈ 6400 km. Forgetting to add R_Earth is the single most common mistake in this calculation
Example — A weather satellite circles the Earth at an altitude of 800 km. Its full circuit round the planet takes 100 minutes. Taking the Earth's radius as 6400 km, find the satellite's orbital speed in m/s.
- Step 1 — List the known quantities in SI units
- h = 800 km = 800 000 m
- R_Earth = 6400 km = 6 400 000 m
- T = 100 minutes = 100 × 60 = 6000 s
- Step 2 — Calculate the orbital radius from the centre of the Earth
- r = R_Earth + h = 6 400 000 + 800 000 = 7 200 000 m
- Step 3 — Apply v = 2πr / T
- v = (2π × 7 200 000) / 6000
- v = 45 239 000 / 6000
- v ≈ 7540 m/s (to 3 s.f.)
- The satellite travels at roughly 7.5 kilometres every second, about 25 times the speed of sound
Example — The Moon orbits the Earth at an average radius of 3.84 × 10⁸ m, with a period of 27.3 days. Calculate its orbital speed.
- Step 1 — Convert the period to seconds
- T = 27.3 × 24 × 60 × 60 = 2 358 720 s
- Step 2 — Apply v = 2πr / T
- v = (2π × 3.84 × 10⁸) / 2 358 720
- v = 2.413 × 10⁹ / 2 358 720
- v ≈ 1020 m/s (to 3 s.f.)
- The Moon trundles around Earth at about 1 km/s, slow by satellite standards, because it is much further out
Why faster-orbiting bodies are closer in
- Looking at v = 2πr / T together with the fact that gravity falls off with distance, you can see the pattern:
- Close to the central body, gravity is strong, needs high speed to balance, period is short
- Far from the central body, gravity is weak, only slow speed needed to balance, period is long
- That is why Mercury has a year of 88 days and Neptune has a year of 165 years; and why low satellites zip around the Earth in 90 minutes while geostationary satellites take 24 hours
- Quantitatively, Kepler's third law says T² ∝ r³, but the qualitative rule "closer = faster" is enough for IGCSE