Audited —·Last updated 27 Jul 2026·6 citations·Tier 1·0 uses

Orbital Period Calculator

Calculate orbital period using Kepler's third law. Enter semi-major axis and central mass for results in seconds, hours, days, and years.

Orbital Period Calculator

Orbital period (seconds)
5,544.9333
Orbital period (hours)
1.5403
Orbital period (days)
0.1
Orbital period (years)
0
Mean orbital velocity
7,672.4904

Background.

The orbital period calculator applies Kepler's third law—generalized by Newton—to compute the time required for one complete revolution of a body around a central mass. The inputs are minimal: the semi-major axis of the orbit and the mass of the central body. The output is the period in seconds, with automatic conversion to hours, days, and years for readability. The tool also reports the mean orbital velocity, giving a complete kinematic summary for satellites, planets, moons, or any bound two-body system.

Johannes Kepler published his third law in 1619 in Harmonices Mundi, stating that the square of a planet's orbital period is proportional to the cube of its orbit's semi-major axis. Isaac Newton later showed in the Principia (1687) that this proportionality arises from the inverse-square law of gravitation and that the constant of proportionality involves the mass of the central body. The resulting formula, T = 2π√(a³/GM), is one of the most widely applied equations in astrodynamics. It governs everything from the 90-minute orbit of the International Space Station to the 248-year orbit of Pluto, and it underpins the design of geostationary satellites that hover over a fixed longitude by matching Earth's rotation period of 23 hours, 56 minutes, and 4 seconds.

Satellite operators use this calculator daily. A geostationary orbit requires a semi-major axis of approximately 42,164 kilometers from Earth's center, which the calculator confirms yields a period matching Earth's sidereal day. Communications satellites at this altitude—Intelsat, Inmarsat, and the GOES weather satellites—must maintain station-keeping burns to counteract perturbations from the Moon and solar radiation pressure, but the nominal orbit is set by Kepler's law alone. GPS satellites occupy semi-synchronous orbits with a 12-hour period, corresponding to a semi-major axis of about 26,560 kilometers. The calculator verifies these design parameters in seconds, replacing hand computations that were once performed with slide rules and logarithm tables.

The tool is equally valuable in planetary science and exoplanet research. When astronomers detect a planet via the transit method, they measure the interval between successive transits, which equals the orbital period. Combining this observed period with an estimated stellar mass from spectroscopic classification yields the orbital semi-major axis via the same formula rearranged as a = (GM T² / 4π²)^(1/3). This is how Kepler-186f, a potentially habitable exoplanet, was shown to orbit at 0.4 AU from its M-dwarf host star with a period of 130 days. Amateur astronomers use the calculator to predict occultations, plan observations of Jupiter's moons, and verify the orbital elements published in ephemerides.

The educational reach of the calculator extends from high school physics to graduate orbital mechanics. Students learning about gravitation can see quantitatively why Mercury's year is only 88 days while Neptune's is 165 years. They can discover that a satellite 100 kilometers above Earth's surface has a period barely shorter than the space station's, while one at lunar distance takes 27.3 days. The inverse relationship between orbital radius and velocity—v = √(GM/a)—explains why low orbits are fast and high orbits are slow, a principle that governs Hohmann transfer maneuvers and interplanetary trajectory design.

What is orbital period calculator?

Orbital period is the time required for an object to complete one full revolution along its orbit around a central body. It is usually denoted by T and measured in seconds, minutes, hours, or years depending on the scale of the system. For closed orbits, the period is finite and well-defined; for open trajectories such as parabolic or hyperbolic flybys, the period is infinite because the object does not return.

In the two-body problem under Newtonian gravity, the orbital period depends only on the semi-major axis a and the total mass of the system, not on the eccentricity of the orbit or the mass of the orbiting body (provided that body is much less massive than the central one). This is the content of Kepler's third law as generalized by Newton. A circular orbit and a highly elliptical orbit with the same semi-major axis share the same period, even though the elliptical orbit spends more time near apoapsis moving slowly and less time near periapsis moving rapidly.

The concept applies to any bound gravitational system: planets around stars, moons around planets, binary stars around each other, and artificial satellites around Earth. In general relativity, the Newtonian formula acquires small corrections that manifest as perihelion precession, but the period formula remains accurate to better than one part per million for most solar-system applications.

How to use this calculator.

  1. Enter the semi-major axis of the orbit in meters, or in astronomical units with automatic conversion.
  2. Enter the mass of the central body in kilograms.
  3. Optionally enter the mass of the orbiting body if it is not negligible compared to the central mass.
  4. The calculator computes the period using T = 2π√(a³ / G(M + m)).
  5. Results are displayed in seconds, hours, days, and years.
  6. The calculator also reports mean orbital velocity for circular orbits.
  7. Verify against known values such as ISS (92.7 min) or geostationary orbit (23.93 h).

The formula.

T = 2π√(a³ ⁄ G(M + m))
T · orbital perioda · semi-major axisG · gravitational constantM · central massm · orbiting mass

The orbital period formula is derived by combining Newton's law of gravitation with the centripetal force requirement for circular motion. For a body of mass m orbiting a central mass M at radius a with speed v, the gravitational attraction provides the necessary centripetal force: GMm/a² = mv²/a. Solving for v gives v = √(GM/a). The circumference of the orbit is 2πa, so the period T = 2πa/v = 2πa/√(GM/a) = 2π√(a³/GM).

This derivation assumes a circular orbit, but Newton proved a more general result: for any elliptical orbit, the same formula holds with a interpreted as the semi-major axis. The proof uses Kepler's second law—conservation of angular momentum—to relate the areal velocity to the period, and integrates over the full ellipse. The result is remarkable because it shows that the period is independent of eccentricity. A satellite in a highly eccentric Molniya orbit with apogee 39,000 km and perigee 500 km has the same period as a circular orbit at the average of those radii, because both share the same semi-major axis of approximately 19,750 km.

When the orbiting mass m is not negligible compared to M, both bodies orbit their common center of mass, and the effective gravitational parameter becomes G(M + m). This correction is essential for binary star systems, where the masses are comparable. For the Earth-Moon system, the correction is about 1.2% because the Moon's mass is 1/81 of Earth's. For Jupiter and the Sun, the correction is about 0.1%. For artificial satellites around Earth, the correction is negligible at less than one part per billion.

The formula also reveals the relationship between period and orbital energy. The total orbital energy is E = −GMm/(2a), which is negative for bound orbits. As the semi-major axis increases, the energy becomes less negative—closer to zero—and the period lengthens. An orbit with infinite semi-major axis has zero energy and infinite period, which is the parabolic escape trajectory. This deep connection between period, energy, and geometry is what makes Kepler's third law so powerful in celestial mechanics.

A worked example.

Example

To calculate the orbital period of the International Space Station, begin with its mean altitude of approximately 400 kilometers above Earth's surface. Adding Earth's mean radius of 6,371 kilometers gives a semi-major axis of 6,771 kilometers, or 6.771 × 10⁶ meters. Cube this value to obtain 3.104 × 10²⁰ m³. Multiply the gravitational constant G = 6.67430 × 10⁻¹¹ by Earth's mass 5.972 × 10²⁴ to get the standard gravitational parameter 3.986 × 10¹⁴ m³/s². Divide the cubed semi-major axis by this parameter: 3.104 × 10²⁰ divided by 3.986 × 10¹⁴ equals 7.787 × 10⁵ s². The square root is 882.4 seconds. Multiplying by 2π gives 5,543 seconds, which is 92.4 minutes. NASA reports the actual ISS period as about 92.68 minutes; the small discrepancy arises from orbital decay due to atmospheric drag and the station's non-circular, slightly eccentric orbit. The same method applies to any satellite or moon by substituting the appropriate semi-major axis and central mass.

semi Major Axis6,771,000
central Mass5,972,000,000,000,000,000,000,000
orbiting Mass0

Frequently asked questions.

What is the difference between sidereal and synodic orbital periods?
The sidereal period is the time for one complete orbit relative to the distant fixed stars; it is the value computed by T = 2π√(a³/GM). The synodic period is the time between successive conjunctions or oppositions as seen from a moving observer, typically Earth. For a planet orbiting the Sun, the synodic period depends on both the planet's sidereal period and Earth's. The formula 1/S = |1/T₁ − 1/T₂| relates the two. Mars has a sidereal period of 687 days but a synodic period of 780 days because Earth is also moving. The calculator returns the sidereal period.
Why does the orbital period depend on the semi-major axis and not the eccentricity?
Kepler's third law states that T² is proportional to a³, where a is the semi-major axis. This independence from eccentricity follows from the conservation of angular momentum and the geometric properties of ellipses. A highly eccentric orbit sweeps out equal areas in equal times, spending most of its period far from the focus where it moves slowly, but compensating with rapid motion near periapsis. The integral of areal velocity over the full ellipse depends only on a, not on how flattened the ellipse is. This is why comets with 100-year periods and eccentricities of 0.99 return with the same regularity as circular orbits at the same mean distance.
Can I use this calculator for elliptical orbits?
Yes. For any bound Keplerian orbit, replace the orbital radius with the semi-major axis a. For an ellipse, a = (r_periapsis + r_apoapsis) / 2. Enter this value as the semi-major axis, and the calculator returns the correct sidereal period. The mean orbital velocity output assumes a circular orbit, so for elliptical orbits use the vis-viva equation v = √[GM(2/r − 1/a)] to find the instantaneous speed at a specific point. The period formula itself is exact for all eccentricities in the two-body Newtonian approximation.
What is a geostationary orbit and what is its period?
A geostationary orbit is a circular orbit above Earth's equator with a period exactly equal to Earth's sidereal rotation period: 23 hours, 56 minutes, and 4 seconds (86,164 seconds). Using the calculator with T = 86,164 s and M = 5.972 × 10²⁴ kg, solving for a yields approximately 42,164 km from Earth's center, or about 35,786 km above the equator. At this altitude, the satellite appears stationary relative to the ground, making it ideal for communications and weather monitoring. The orbital velocity is about 3.07 km/s.
How do atmospheric drag and solar radiation pressure affect orbital period?
Atmospheric drag removes energy from low-Earth orbits, causing the semi-major axis to shrink and the period to decrease. The ISS loses about 2 km of altitude per month and requires periodic re-boosts to maintain its orbit. Solar radiation pressure exerts a small outward force that can increase the semi-major axis over time, particularly for satellites with large surface-area-to-mass ratios such as solar sails. Both effects are perturbations that cause the actual period to drift slowly away from the Keplerian ideal computed by this calculator.
Does the mass of the orbiting satellite affect its period?
In the one-body approximation where the satellite is much less massive than the central body, the satellite's mass does not appear in the period formula. This is because the gravitational force and the inertial response both scale linearly with the satellite's mass, causing it to cancel. However, in the full two-body problem, the period depends on the sum of the masses (M + m). For artificial satellites around Earth, this correction is negligible. For the Moon orbiting Earth, the correction is about 1.2%. For binary stars of comparable mass, both masses must be included.
What is the orbital period of the Moon?
The Moon's sidereal orbital period is 27.321661 days (2,360,590 s) relative to the fixed stars. Its synodic period—the interval between successive full moons—is 29.530589 days because Earth has moved partway around its own orbit in the intervening time. Using the calculator with a = 3.844 × 10⁸ m and M = 5.972 × 10²⁴ kg yields a period of 27.32 days, confirming the consistency of the formula with observation. The small difference from the exact observed value arises from perturbations by the Sun and other planets.
Can this calculator be used for interplanetary trajectories?
For a Hohmann transfer orbit between two circular coplanar orbits, the transfer ellipse has a semi-major axis equal to the arithmetic mean of the initial and final radii. Entering this semi-major axis and the Sun's mass gives the transfer period; the one-way travel time is half this period. For example, an Earth-to-Mars Hohmann transfer has a semi-major axis of about 1.262 AU, yielding a period of 1.417 years and a transfer time of 259 days. More complex trajectories involving gravity assists require numerical integration, but the calculator provides the baseline Keplerian estimate.
Why is the orbital period squared proportional to the semi-major axis cubed?
This relationship follows from dimensional analysis combined with the inverse-square law. The only parameters available are G, M, and a. The quantity T² must be proportional to a³/(GM) because this combination has dimensions of time squared. The factor of 4π² comes from integrating the equations of motion around a closed ellipse. In a hypothetical universe where gravity followed an inverse-cube law, the relationship between period and radius would be different, and closed elliptical orbits would not exist. The T² ∝ a³ law is therefore a specific signature of the inverse-square gravitational force.
What happens if the orbital speed is exactly equal to the circular orbit speed?
If the speed equals v = √(GM/a) and the velocity vector is perpendicular to the radius vector, the orbit is a perfect circle with period T = 2π√(a³/GM). If the speed is higher but still below escape velocity, the orbit becomes an ellipse with the launch point as pericenter. If the speed equals escape velocity v = √(2GM/a), the orbit is parabolic and the period is infinite. If the speed exceeds escape velocity, the orbit is hyperbolic and the object departs on an open trajectory. The calculator assumes a bound orbit; speeds yielding parabolic or hyperbolic trajectories produce imaginary periods, signaling that the inputs describe an unbound system.

References& sources.

  1. [1]Kepler, J. (1619). Harmonices Mundi. Linz: Johann Planck. Book V, Chapter 3.
  2. [2]Newton, I. (1687). Philosophiæ Naturalis Principia Mathematica. London: Royal Society. Book I, Proposition XV.
  3. [3]Halliday, D., Resnick, R., and Walker, J. (2021). Fundamentals of Physics, 11th ed. Hoboken, NJ: Wiley. Chapter 13: Gravitation.
  4. [4]NIST (2024). CODATA Recommended Values of the Fundamental Physical Constants: 2018.
  5. [5]Bate, R.R., Mueller, D.D., and White, J.E. (1971). Fundamentals of Astrodynamics. New York: Dover. Chapter 4: Two-Body Orbital Mechanics.
  6. [6]NASA (2024). International Space Station Facts and Figures.
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