Orbital Period Calculator
How long one orbit takes (Kepler's third law).
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How long does it take to go around? Kepler's third law links the size of an orbit to the time it takes to complete. Enter a distance from the Sun, or a custom central mass and orbit radius, to find the orbital period.
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Orbital Period — The Quick Answer
For objects orbiting the Sun, Kepler's third law simplifies beautifully: the period in years equals the square root of the distance in AU cubed. At 1 AU (Earth), that's 1 year.
Shortcut
Sun orbits: T(years) = √(a³) with a in AU | General: T = 2π√(a³/GM)
Orbital periods in the Solar System
| Planet | Distance (AU) | Period |
|---|---|---|
| Mercury | 0.39 | 0.24 years |
| Venus | 0.72 | 0.62 years |
| Earth | 1.00 | 1.00 year |
| Mars | 1.52 | 1.88 years |
| Jupiter | 5.20 | 11.86 years |
| Neptune | 30.1 | 165 years |
Kepler discovered this relationship in 1619, decades before Newton explained why it works. The further out a planet orbits, the slower it moves and the longer its year.
How It Works
The Formula
Kepler's third law
Formula
Sun orbit: T(years) = √(a³), a in AU General: T = 2π√(a³ ÷ GM)
Variables
Semi-major axis
The average orbital distance. In AU for Sun orbits, or in metres for the general formula.
Central mass
The mass being orbited, in kilograms — the Sun, a planet, or any central body.
Orbital period
The time to complete one orbit. Larger orbits take dramatically longer.
Note: This assumes a roughly circular orbit and a central mass far larger than the orbiting object, which holds well for planets, moons, and satellites.
Worked Example: Mars's Orbital Period
Kepler's law in its simplest Sun-orbit form
Take Mars's distance
Mars orbits at a ≈ 1.52 AU from the Sun.
Cube it
1.52³ = 3.51.
Take the square root
√3.51 ≈ 1.87 years — matching Mars's real 1.88-year orbit.
Reference Guide
| Planet | Distance (AU) | Period |
|---|---|---|
| Mercury | 0.39 | 0.24 years |
| Venus | 0.72 | 0.62 years |
| Earth | 1.00 | 1.00 year |
| Mars | 1.52 | 1.88 years |
| Jupiter | 5.20 | 11.86 years |
| Neptune | 30.1 | 165 years |
What Sets an Orbit's Length
Distance dominates — but the central mass matters too
For objects orbiting the Sun, the period depends only on the semi-major axis: T = √(a³) in years and AU.
Best for: Planets and comets around the Sun
The full law T = 2π√(a³/GM) brings in the mass being orbited — essential for moons, satellites, and exoplanets.
Best for: Satellites, moons, and other star systems
The simple form assumes the central mass dominates and the orbit isn't wildly elliptical. Use the general form for the rest.
Best for: Knowing when the tidy shortcut applies
Common Mistakes
Using T = √(a³) for non-Sun orbits
The tidy 'years = √(AU³)' form only works for the Sun. For a satellite around Earth, use the general T = 2π√(a³/GM) with Earth's mass.
Mixing up the units
The simple form needs distance in AU and gives years; the general form needs SI units (metres, kg, seconds). Don't feed AU into the GM formula.
Confusing radius with semi-major axis
For elliptical orbits, a is the semi-major axis (the average), not the closest or furthest distance. Using perihelion alone skews the period.
Expecting constant orbital speed
Orbiting bodies speed up near the central mass and slow far away (Kepler's second law). The period is the full-loop figure, not a fixed speed.
Research & Citations
Sources and standards referenced for the factual claims on this page
- [kepler]
Kepler, J. (1619). Harmonices Mundi — the third law of planetary motion (T² ∝ a³).
The period–distance relationship used for Sun orbits.
- [newton]
Newton, I. (1687). Philosophiæ Naturalis Principia Mathematica — generalised form T = 2π√(a³/GM).
The general two-body form used for custom central masses.
This calculator is a reference tool, not a substitute for professional advice. Verify important health, financial, legal, or safety decisions with a suitably qualified professional.
Last updated: October 8, 2026
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Frequently Asked Questions
What is Kepler's third law?
It states that the square of an orbital period is proportional to the cube of the orbit's semi-major axis. For Sun orbits in convenient units, the period in years is simply the square root of the distance in AU cubed.
How do I calculate a planet's orbital period?
For a planet orbiting the Sun, take its distance in AU, cube it, and take the square root — that's the period in years. Mars at 1.52 AU gives √(1.52³) ≈ 1.88 years, matching its real orbit.
How do I find a satellite's orbital period?
Use the general form T = 2π√(a³/GM), with M as Earth's mass and a as the orbit radius from Earth's centre. Switch to custom mode and enter Earth's mass (5.97 × 10²⁴ kg) and your orbit radius.
Why do outer planets take so much longer to orbit?
Two reasons combine: they travel a much longer path, and they move more slowly because the Sun's gravity is weaker at greater distances. That's why Neptune's year is 165 Earth years.