Orbital Period
Calculator
Results
- Orbital period (seconds)
- 31,554,148.649719
- Orbital period (days)
- 365.210053
- Orbital period (years)
- 0.99989
Astronomy results
| Orbital period (seconds) | 31,554,148.649719 |
| Orbital period (days) | 365.210053 |
| Orbital period (years) | 0.99989 |
formula-map diagram
- Orbital period (seconds)
- 31,554,148.649719
- Orbital period (days)
- 365.210053
- Orbital period (years)
- 0.99989
Astronomical relationship
Formula
T = 2π √(a³ / (G M))= 31554148.649719
Note
This result is a simplified model: it applies the displayed textbook formula to the values you entered, assuming ideal spherical bodies, circular orbits, blackbody radiation and perfect optics, and ignoring atmospheric seeing, relativistic corrections beyond those stated, cosmological models and measurement uncertainty. Use published ephemerides and catalogue data for real observations.
More in Astronomy and space
See all →Frequently asked questions
What does Kepler's third law relate in this calculator?+
It relates a planet's or satellite's orbital period to its orbital distance from the central mass: T² is proportional to a³, where a is the semi-major axis. Given the central mass and orbital radius, the calculator solves for how long one full orbit takes.
Why does a satellite farther from Earth take longer to complete an orbit?+
Both the orbital distance traveled and the required speed change with altitude: farther orbits are longer in circumference but move more slowly, and the T² ∝ a³ relationship means the period grows faster than the distance. This is why geostationary satellites, at about 35,786 km altitude, take a full 24 hours per orbit versus about 90 minutes for low Earth orbit.
Does the orbital period depend on the orbiting object's mass?+
No, for a small object orbiting a much larger body, the period depends only on the central mass and the orbital semi-major axis. A pebble and a space station at the same distance from Earth would have essentially the same orbital period.
What is a geostationary orbit, and how does the calculator relate to it?+
A geostationary orbit is the specific altitude where the orbital period exactly matches Earth's 24-hour rotation, so the satellite appears fixed over one point on the equator. You can use this calculator in reverse: find the radius that gives a 24-hour period.
Does this formula work for elliptical orbits, not just circular ones?+
Yes, Kepler's third law uses the semi-major axis a, which for an ellipse is the average of the closest and farthest distances. So enter the semi-major axis, not the current instantaneous distance, when working with an elliptical orbit.