Library / Foundations: Physics / FPH-42
Foundations: Physics
Circular orbital speed
Speed needed to stay in a circular orbit.
- Ref
- FPH-42
- Rev
- v1
- Field
- Foundations: Physics
- Result
- v [m/s]
M is the mass of the body being orbited and r the orbit radius from its centre, so r = planet radius + altitude. The satellite’s own mass cancels, because gravity and the inertia that resists it both scale with it. The result is the speed in a circular orbit; it fixes the orbit period T = 2πr/v.
Speed needed to stay in a circular orbit. Higher orbits are slower.
Calculator
- Orbital period 2πr/v92.42 min
- Speed in km/h27,620 km/h
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- At 400 km the speed is 7.67 km/s and one lap takes about 92 minutes, so a crew sees roughly 16 sunrises a day.
- At the geostationary radius of 42 164 km the speed falls to 3.07 km/s and the period to 24 hours, which is why that orbit appears to hover above one point on the equator.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (m/s) | Local step | Change in result |
|---|---|---|---|---|---|---|
| M Mass of the central body | 1e20 to 1e31 kg | 31.4 | 9,928,000 | 9,928,000 | +10% | +4.88 % |
| r Orbit radius from centre | 100,000 to 1e9 m | 631.3 | 63,130 | 62,500 | +10% | −4.65 % |
Sweep of Orbit radius from centre
| Orbit radius from centre (m) | Orbital speed (m/s) |
|---|---|
| 100,000 | 63,130 |
| 316,200 | 35,500 |
| 1,000,000 | 19,960 |
| 3,162,000 | 11,230 |
| 1e7 | 6,313 |
| 3.162e7 | 3,550 |
| 1e8 | 1,996 |
| 3.162e8 | 1,123 |
| 1e9 | 631.3 |
Sweep one variable
Variables and how each one contributes
Orbital speed
What the equation gives you. Everything else feeds it.
Mass of the central body
Mass of the body at the centre of the orbit. Default is Earth. Square root: four times the mass is needed to double the speed.
Orbit radius from centre
Distance from the centre of the central body to the satellite. Inverse square root, so higher orbits are slower. A 10 % rise in r lowers the speed by about 5 %.
Worked example with the values above
- GM/r = 6.674×10−11 × 5.972×1024 / 6,771,000 = 5.887×107 m²/s².
- Take the square root.
- v = 7,672 m/s, which is 27,620 km/h.
Assumptions and limits
- A circular orbit around a spherical body with gravity as the only force.
- The satellite is much lighter than the central body, and there is no atmospheric drag.
Common mistakes
- Using altitude instead of the radius from the centre of the planet.
- Adding the satellite’s mass. It does not appear, so a heavier satellite needs the same speed.
Statistics and uncertainty
v ∝ √(M/r), so each relative error is halved: a 2 % error in r gives a 1 % error in v. The mass of the Earth is known far more precisely than the product G M is quoted separately (3.986 × 10¹⁴ m³/s²), and for orbital work that product is the number to use.
Where it is used
- Setting the speed for a circular parking orbit.
- Finding how long a satellite is in view of a ground station.
Maths you need
- Centripetal force: mv²/r
- Newton’s law of gravitation: F = GMm/r²