Orbital Period Calculator
Calculate the orbital period of a body from its semi-major axis and central mass using T = 2π√(a³/GM), with presets for the Sun, Earth, and Jupiter.
Reviewed October 7, 2026 Runs in your browser — inputs are never stored
Formula
T = 2π × √(a³ ÷ GM) · G = 6.67430 × 10⁻¹¹ m³ kg⁻¹ s⁻²
This tool provides educational conversions of astronomical distances only. It is not a navigation, mission-planning, or ephemeris service. For professional or spacecraft work, use official IAU, NASA, or observatory data products.
About the Orbital Period Calculator
The orbital period calculator works out how long an object takes to complete one orbit around a central body. Using Newton’s form of Kepler’s third law, the period depends only on the size of the orbit (its semi-major axis) and the mass of the body being orbited — not on the mass of the small orbiting object. It applies to planets around the Sun, moons around planets and satellites around Earth.
How to use it
- 1Choose the central body (Sun, Earth, Jupiter, Mars, or Saturn).
- 2Enter the semi-major axis and its unit.
- 3Press Calculate to see the period in seconds, days, and years.
Worked example
a satellite 1 AU from the Sun has a period of 1 year (365.25 days). At 42,164 km from Earth's centre (geostationary orbit), the period is about 23.93 hours — matching Earth's rotation.
How it works
The orbital period formula
From Newtonian gravity, the period of a circular or elliptical orbit is T = 2π × √(a³ ÷ GM), where a is the semi-major axis and M is the central mass. This is the general form of Kepler's third law.
Reference periods
- Earth around Sun (1 AU): 365.256 days
- Geostationary orbit (~42,164 km): 23.93 hours
- ISS (~6,778 km from Earth centre): ~92.7 minutes
Who it is for
Astronomy students, orbital mechanics learners, and satellite mission planners.
Why it helps
The orbital period is the time for one full orbit — essential for predicting satellite motion, planetary years, and Keplerian dynamics.
Common use cases
- Computing a satellite's orbital period at a given altitude
- Finding a planet's year from its orbital radius
- Teaching Kepler's third law and Newtonian gravity
- Estimating geostationary or Sun-synchronous orbit periods
Tips for better results
- For the Sun, 1 AU always gives 1 year — a quick sanity check.
- Geostationary orbit requires a period matching Earth's rotation (~86,164 s).
- Use metres or km for satellite altitudes; AU for planets.
- The semi-major axis is the average orbital radius for elliptical orbits.
Common mistakes to avoid
- Using the orbital altitude instead of the distance from the body's centre.
- Forgetting to square and cube the radius in the formula.
- Mixing central-body masses (e.g. Earth mass for a heliocentric orbit).
Frequently asked questions
A:The orbital period is the time a body takes to complete one full orbit around another body. It follows T = 2π√(a³/GM) from Newton's law of gravitation.
A:Kepler stated T² ∝ a³ for planets around the Sun. Newton showed that T²/a³ = 4π²/(GM), giving the full formula T = 2π√(a³/GM).
A:A geostationary orbit has a period equal to Earth's rotation (about 23.93 hours), which occurs at roughly 42,164 km from Earth's centre.
A:Yes. It runs in your browser on CalcWisee with no sign-up required.
Reviewed for accuracy · Last updated October 7, 2026
The astronomical unit is taken as exactly 149,597,870,700 metres per the IAU 2012 definition. Related light-time and light-year factors use the defined speed of light and the IAU Julian year where applicable.
CalcWisee Editorial Team
Calculator development and formula review
CalcWisee is an independent site founded by Ayman. Every calculator is built from a published formula or official definition, tested against hand-worked examples before it goes live, and corrected promptly when a reader reports an issue to ayman@calcwisee.com.
Expertise: Financial mathematics, Health and fitness formulas, Applied mathematics, Unit conversion standards
References & sources
