Orbital Period Calculator
Calculate orbital period using Kepler's laws. Find orbital velocity, angular velocity, and perihelion/aphelion distances.
About this calculator
This calculator applies Kepler's Third Law, T² = (4π²/GM) × a³, to find how long a body takes to complete one orbit around a central mass. Working in astronomical units (AU), solar masses, and years lets the gravitational constant collapse to the tidy 4π² instead of the full SI value, which is why the formula looks so clean here — it's the same physics as G = 6.674×10⁻¹¹ m³/(kg·s²), just rescaled to a solar-system-friendly unit system. Once the period is known, the tool converts it to days and derives the average orbital velocity by dividing the orbit's circumference (2πa in meters) by the period in seconds, plus the angular velocity in radians per day. If you supply an eccentricity, it also reports perihelion and aphelion — the closest and farthest points in an elliptical orbit, computed as a(1−e) and a(1+e).
A key assumption worth knowing: this is the idealized two-body problem, meaning it treats the orbiting body as a point mass and ignores perturbations from other planets, drag, or relativistic corrections — accurate enough for planets and most exoplanets, but not for precision spacecraft trajectory work. The orbital velocity reported is an average over the full ellipse, not the instantaneous speed, which actually varies throughout the orbit — fastest at perihelion, slowest at aphelion. A common mixup is entering eccentricity as a percentage instead of a decimal fraction between 0 and 1; Earth's is 0.017, not 1.7.
Inputs
Results
Orbital Period
1 years
Orbital Period
365.25 days
Orbital Velocity
29,785.68 m/s
How to Use This Calculator
- Enter the Semi-Major Axis (AU) of the orbit — Earth's is 1 AU, Mars is 1.524 AU.
- Enter the Central Mass (solar masses) of the star or body being orbited — the Sun is 1 M☉.
- Set Orbital Eccentricity (0–0.99): 0 is circular, Earth's is 0.017, Pluto's is 0.25.
- Read Orbital Period (years and days), Mean Orbital Velocity (m/s), and Perihelion/Aphelion distances (AU).
- Use the orbital period to plan observation timing or to verify radial velocity measurements.
How the result changes with Semi-Major Axis
| Semi-Major Axis | Orbital Period | Orbital Period | Orbital Velocity |
|---|---|---|---|
| 0.5 | 0.354 years | 129.14 days | 42,123.31 m/s |
| 0.75 | 0.65 years | 237.24 days | 34,393.54 m/s |
| 1.5 | 1.837 years | 671.01 days | 24,319.9 m/s |
| 2.5 | 3.953 years | 1,443.78 days | 18,838.12 m/s |
What each input means
- Semi-Major Axis
- Average distance from central body (1 AU = Earth-Sun distance)
- Central Mass
- Mass of central body in solar masses
- Orbital Eccentricity
- Orbit eccentricity (0 = circular, <1 = elliptical)
How this is calculated
Formula
T² = (4π² / GM) × a³ (Kepler's Third Law)Worked example, using the default values
- Identify Input ParametersSemi-Major Axis = 1, Central Mass = 1, Orbital Eccentricity = 0 = 3 input(s) provided
- Calculate Orbital PeriodOrbital Period1 = 1
- Calculate Orbital PeriodOrbital Period365.25 = 365.25
- Calculate Orbital VelocityOrbital Velocity29785.68 = 29785.68
- Calculate Angular VelocityAngular Velocity0.017202 = 0.017202
- Calculate Perihelion DistancePerihelion Distance1 = 1
Engine last updated . Checked against 4 independently-derived tests — how we verify calculators. Built by Paul Gunder, a software engineer, not a licensed financial, medical, or legal professional.
Frequently Asked Questions
Why does a heavier central mass shorten the orbital period?
Kepler's Third Law puts the central mass in the denominator of T² = (4π²/GM) × a³, so a larger M directly shrinks T for the same semi-major axis. Physically, more mass means stronger gravitational pull, which forces the orbiting body to move faster to avoid falling in — a faster orbit at the same distance means less time to complete one lap. Doubling the central mass while holding the semi-major axis fixed shortens the period by a factor of √2, not by half.
Why does the calculator report only one orbital velocity when real orbital speed varies?
The velocity shown here is the mean orbital velocity — the orbit's circumference (2πa) divided by the period — not the instantaneous speed at any specific point. Real orbital speed actually changes continuously across an elliptical orbit, moving fastest at perihelion and slowest at aphelion, a consequence of angular momentum conservation (Kepler's Second Law). Only for a perfectly circular orbit (eccentricity 0) does the instantaneous speed match this average value everywhere.
How are perihelion and aphelion computed from eccentricity?
The calculator multiplies the semi-major axis by (1 − e) for perihelion and (1 + e) for aphelion, where e is the orbital eccentricity you enter. At e = 0 both values collapse to the semi-major axis itself, since a circular orbit has no near or far point. Entering a value above 0.99 isn't accepted because eccentricities approaching 1 describe parabolic, not elliptical, trajectories where this closed-orbit math no longer applies.
Why does the Kepler curve on the chart use fixed reference points instead of my input?
The chart plots Kepler's Third Law across six fixed semi-major-axis values roughly matching Mercury through Jupiter, so you can see how period scales with distance around your chosen central mass — it isn't meant to trace your specific planet's position. Each point still uses your entered central mass in the calculation, so changing that mass reshapes the whole reference curve, not just your single result.
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