Orbital Period & Speed Calculator (vis-viva)

Understanding the orbital period and speed calculator — concept and principle

This calculator gives the time a satellite or planet takes to complete one orbit and its speed at any point along the orbit. It uses Kepler's third law for the period and the vis-viva equation for the speed, and works for circular and elliptical orbits around any body once you enter its gravitational parameter μ = GM. Typical uses are checking satellite orbits, homework in mechanics and quick mission sketches.

Orbit length ÷ orbital speed gives the same 92.4 minutes as T = 2π√(a³/μ).

How it is calculated

Period: T = 2π × √(a³ ÷ μ), shown in minutes.
Speed at radius r: v = √(μ × (2/r − 1/a)), in km/s.

a is the semi-major axis (half the longest diameter of the ellipse), r the current distance from the centre of the central body, both in km, and μ in km³/s² (Earth 398,600.4418). For a circular orbit, r = a and the speed reduces to √(μ/a).

The principle

Newton showed that an inverse-square gravitational force produces elliptical orbits with T² proportional to a³, which is Kepler's third law in exact form: T² = 4π²a³/μ. The period depends only on the size of the orbit, not on its shape. The vis-viva equation comes from conservation of energy: the specific orbital energy −μ/(2a) is the same everywhere on the orbit and equals the kinetic part v²/2 plus the potential part −μ/r. Solving for v gives v² = μ(2/r − 1/a). So a satellite moves fastest at its closest point and slowest at its farthest.

Worked example

Defaults: a = 6,771 km, r = 6,771 km (a circular orbit about 400 km up, similar to the International Space Station), μ = 398,600.4418 km³/s².
a³ ÷ μ = 6,771³ ÷ 398,600.4418 = 778,790.54 s².
T = 2π × √778,790.54 = 5,544.86 s = 92.4142516 min.
v = √(398,600.4418 × (2/6,771 − 1/6,771)) = √(398,600.4418 ÷ 6,771) = 7.672598648 km/s.

Each point is the calculator re-run with only “Semi-major axis (km)” changed; the gold point is the default example. Change the inputs above and this chart follows your numbers.

Semi-major axis (km)	Period (min)
1,693	11.55178145
3,386	32.67337199
5,078	60.02481717
6,771	92.4142516
8,464	129.1528429
10,156	169.775821
13,542	261.3869759

How to read the result

Low Earth orbits take about 88–100 minutes. At r = 42,164 km the period is about 1,436 minutes, one sidereal day, which is why geostationary satellites appear fixed in the sky. If r is less than a, you are on the faster, inner part of an ellipse. Escape speed at any radius is √2 times the circular speed there, about 10.85 km/s at 6,771 km.

Common mistakes

Entering altitude instead of radius from the centre.
Using μ in m³/s² with distances in km (a factor of 10⁹ error).
Entering r larger than 2a, which gives no real speed because no point of the ellipse lies that far out.

Keep in mind: The two-body model ignores atmospheric drag, Earth's oblateness and the pull of the Moon and Sun. Real orbits drift, so operational work needs full orbit propagation.

Frequently asked questions

Does a heavier satellite orbit more slowly?

No. When the satellite's mass is tiny compared with the planet's, its period and speed do not depend on its own mass.

What μ do I use for other bodies?

Use GM of the central body, for example about 1.327 × 10¹¹ km³/s² for the Sun and 4,902.8 km³/s² for the Moon.

Why is it called vis-viva?

It is Latin for "living force", an old name for what became kinetic energy.

Sources: Curtis H.D., Orbital Mechanics for Engineering Students, Elsevier · Vallado D.A., Fundamentals of Astrodynamics and Applications, Microcosm Press · NASA JPL Solar System Dynamics, astrodynamic parameters

Checked 8 Nov 2026. Standard textbook formulas in SI units, computed in your browser. For learning and estimates — real designs must follow your local codes and be checked by a qualified engineer.

Your numbers