Case 1
- Semi-major axis
- 149600000000m
- Mass of the central body
- 198900010²⁴ kg
31,554,187.748s
Open with these valuesOrbital Period Calculator
31,554,187.748s
Result: 31,554,187.748 sKepler's third law gives the time for one full orbit: T = 2π × √(a³ / (G × M)). Earth at 1.496e11 m from the Sun takes 31554188 seconds, which is 365.21 days. Enter the semi-major axis in metres and the central mass in units of 10²⁴ kg — the Sun is 1989000, Earth 5.972.
Worked examples
Case 2
- Semi-major axis
- 384400000m
- Mass of the central body
- 5.97210²⁴ kg
2,371,877.064s
Open with these valuesCase 3
- Semi-major axis
- 42200000m
- Mass of the central body
- 5.97210²⁴ kg
86,275.162s
Open with these valuesHow it's calculated
T = 2π × √(a³ ÷ (G × M))
- StepEnter the semi-major axis in metres — for a circle, the orbital radius.
- StepEnter the central mass in units of 10²⁴ kg: the Sun is 1989000.
- ResultRead the period in seconds; divide by 86400 to get days.
What this number means
The 3/2 power, not a straight line
Kepler's third law puts the semi-major axis under a 3/2 power, so an orbit twice as wide takes about 2.83 times as long rather than twice. Jupiter orbits 5.2 times farther out than Earth and needs 4331.27 days against Earth's 365.21.
Metres in, units of 10²⁴ kg, seconds out
Enter the Sun as 1989000 and the Earth as 5.972, exactly as planetary tables list the masses. The semi-major axis stays in plain metres, and the period comes out in seconds. Divide by 86400 for days: Earth's 31554187.747615 s are 365.21 days.
The orbiting body's own mass drops out
As long as the orbiting body is much lighter than the central one, the period depends only on the axis and the central mass. A small satellite and a large one at the same altitude come round in the same time.
G is measured; the day is a definition
The calculation uses G = 6.6743e-11 m³·kg⁻¹·s⁻², the CODATA value with a relative uncertainty of 2.2e-5, which makes it the least precisely known quantity in the formula. The 86400 seconds per day behind the reference column are fixed by definition and carry no uncertainty at all.
Commonly misread
Doubling the semi-major axis doubles the orbital period.
It multiplies the period by about 2.83, because the axis enters to the 3/2 power. Four times the axis means eight times the period.
Entering the Sun's mass as 1989000000000000000000000000000 kg.
The mass box already counts in units of 10²⁴ kg, so the Sun is 1989000 and the Earth 5.972. The multiplication by 10²⁴ happens inside the calculation.
Reading the output of 31554188 as days.
The period is in seconds: 31554187.747615 s divided by 86400 gives 365.21 days. Jupiter's 374221737.152048 s become 4331.27 days.
Measuring the semi-major axis from the surface of the central body.
It runs from the centre of the central body, like every distance in this law. The geostationary row therefore uses 42200000 m, not a height above the ground.
Reference table
| Semi-major axis, central mass | Orbit | Orbital period (s) |
|---|---|---|
| 6780000, 5.972 | Low Earth orbit, 0.064 days | 5555.992461 |
| 42200000, 5.972 | Geostationary, 0.999 days | 86275.162000 |
| 384400000, 5.972 | The Moon, 27.45 days | 2371877.064000 |
| 149600000000, 1989000 | Earth around the Sun, 365.21 days | 31554187.747615 |
| 778000000000, 1989000 | Jupiter around the Sun, 4331.27 days | 374221737.152048 |
Questions
How do I calculate the orbital period?
Use Kepler's third law: T = 2π × √(a³ / (G × M)), with a in metres, M in kilograms and G = 6.6743e-11. Earth at 1.496e11 m around the Sun at 1.989e30 kg gives about 3.155e7 s, or 365.2 days. The mass box here counts in units of 10²⁴ kg, so the Sun is entered as 1989000.
What is the orbital period?
It is the time a body takes to complete one full revolution around the object it orbits. It depends only on the size of the orbit and the mass of the central body, not on the orbiting body's own mass when that mass is small in comparison.
What is the semi-major axis?
The semi-major axis is half the longest diameter of an elliptical orbit and stands for the orbit's average size. For a circular orbit it is simply the orbital radius, measured from the centre of the central body. Enter it in metres: the Earth–Sun distance is about 1.496e11 m.
Does the mass of the orbiting object matter?
For the standard form of Kepler's third law, no. As long as the orbiting body is much lighter than the central mass, its own mass cancels out and the period depends only on the axis and the central mass. A small satellite and a large one at the same altitude share the same period.
Why do larger orbits take longer?
Because the period grows with the 3/2 power of the semi-major axis. Doubling the orbital distance multiplies the period by about 2.83, so an orbit twice as wide takes nearly three times as long. A larger central mass works the other way, pulling harder and shortening the period.
How do I read the answer in days?
Divide the seconds by 86400, the number of seconds in a day. Earth's 31554188 s become 365.21 days, and Jupiter's 374221737 s become 4331.27 days. The reference table below already shows both.
Sources and last check
Information, not professional advice.