Kepler's Third Law Calculator (Orbital Period)
Compute the orbital period of any body from its semi-major axis and the mass it orbits using Kepler's third law.
Yksiköt
AU
M☉
Time to complete one full orbit
- 1
Semi-major axis in metres
1 AU × 149 597 870 700 m/AU = 149 597 870 700 m - 2
a³ ÷ (G × M)
25 221 604 944 996,945 s²Squaring its square root and multiplying by 2π gives the orbital period. - 3
Period P = 2π × √(a³ ÷ G·M)
2π × √(25 221 604 944 996,945 s²) = 31 554 857,83 s - 4
Period in years
31 554 857,83 s ÷ 31 557 600 s/yr = 0,9999
Miten tämä laskin toimii?
Kepler's third law gives an orbital period from the semi-major axis a and central mass M: P = 2π·√(a³/(G·M)). Enter a and M in SI or astronomical units and the calculator returns the period in seconds, days, and years. In solar units this simplifies to P(years) = √(a(AU)³ / M(M☉)).
Kaava
How this is calculated
Kepler's third law relates the orbital period P to the orbit's semi-major axis a and the mass M of the central body. In SI units the period is P = 2π·√(a³ / (G·M)), where G = 6.674×10⁻¹¹ m³·kg⁻¹·s⁻² is the gravitational constant, a is in meters, and M is in kilograms. The result is a period in seconds, which is also converted to days and years.
The Units selector lets you enter values either in SI (meters and kilograms) or in astronomical units (AU and solar masses). When astronomical units are chosen, the inputs are converted internally using 1 AU = 1.4960×10¹¹ m and 1 solar mass = 1.989×10³⁰ kg before applying the same formula. This reproduces the familiar astronomical shortcut P(years) = √(a(AU)³ / M(M☉)), which for a body orbiting the Sun (M = 1 M☉) reduces to P² = a³.
The model assumes a two-body system in which the orbiting body's mass is negligible compared with the central mass, so M is taken as the central mass alone. For comparable masses, replace M with the total mass of the system. Inputs must be positive; a non-positive semi-major axis or mass returns no result because the orbit and gravitational force would be undefined.
Usein kysytyt kysymykset
With the Sun's mass (1 solar mass) and a semi-major axis of 1 AU, Kepler's third law gives P = √(1³ / 1) = 1 year by definition — Earth's orbit is the reference for these astronomical units.
For a small body orbiting a much larger one (a planet around a star, a satellite around a planet) use the central mass. When the two masses are comparable, use the sum of both masses for the most accurate period.
Yes. Kepler's third law uses the semi-major axis, not the radius, so it applies to elliptical as well as circular orbits — the period depends only on a and the central mass, not on the eccentricity.
Tunnetaan myös nimellä
TG we-Calculate Editorial Team. (2026). Kepler's Third Law Calculator (Orbital Period) [Online calculator]. TG we-Calculate. https://we-calculate.com/fi/calculator/keplers-third-law-calculator
TG we-Calculate Editorial Team. "Kepler's Third Law Calculator (Orbital Period)." TG we-Calculate. 2026. https://we-calculate.com/fi/calculator/keplers-third-law-calculator.
TG we-Calculate Editorial Team, "Kepler's Third Law Calculator (Orbital Period)," TG we-Calculate, 2026. [Online]. Available: https://we-calculate.com/fi/calculator/keplers-third-law-calculator
@misc{wecalculate_keplers_third_law_calculator, title = {Kepler's Third Law Calculator (Orbital Period)}, author = {{TG we-Calculate Editorial Team}}, howpublished = {\url{https://we-calculate.com/fi/calculator/keplers-third-law-calculator}}, year = {2026}, note = {TG we-Calculate} }
Oliko tästä laskimesta sinulle apua?