๐ช Keplerโs Third Law Calculator
Calculate a solar-orbiting bodyโs orbital period (years) from its semi-major axis (astronomical units, AU), or the reverse, using Keplerโs third law T(years)=a(AU)^1.5. This is an approximation for a Sun-centered two-body system, calibrated to Earthโs orbit (1 AU, 1 year).
How to use
- Choose "axis to period" or "period to axis" mode.
- Enter the semi-major axis (AU) or orbital period (years).
- The corresponding value is calculated automatically.
How the calculation works
Keplerโs third law says that the square of a planetโs orbital period is proportional to the cube of its semi-major axis (half the long axis of the elliptical orbit). Tยฒ = aยณ (T in years, a in astronomical units, AU) Period T = a^1.5 Semi-major axis a = T^(2/3) An astronomical unit is the mean EarthโSun distance, about 149.6 million km. Earth has a = 1 AU and T = 1 year, so in these units the constant is 1 and the formula becomes very simple. It works for planets, asteroids and comets orbiting the Sun. It is an approximation that ignores the orbiting bodyโs own mass and the pull of other planets.
Worked example
Mars (semi-major axis 1.523679 AU) 1.523679^1.5 โ 1.8808 years (about 687 days) Jupiter (5.2044 AU) 5.2044^1.5 โ 11.87 years Halleyโs Comet (about 17.8 AU) 17.8^1.5 โ 75.1 years Semi-major axis for a 2-year orbit 2^(2/3) โ 1.5874 AU
Things to be aware of
- The semi-major axis is the average of the closest and farthest distances from the Sun. Even for long, thin comet orbits, this value gives the period.
- For Jupiter the formula gives 11.87 years against an actual 11.86 years. Jupiter has about a thousandth of the Sunโs mass, which causes the small difference.
- Orbits around other stars or planets need a correction for the central mass: Tยฒ = aยณ รท M, with M in solar masses.
FAQ
Why does this formula work in these units?
Because Earthโs semi-major axis is defined as 1 AU and its orbital period as 1 year, Keplerโs third law simplifies to T=a^1.5 in this unit system.
Is this accurate for other planets?
It closely matches observed values for the major planets (e.g. Mars, at 1.524 AU, gives about 1.881 years, matching the actual measured period). Itโs still an approximation that ignores gravitational perturbations from other planets.
Can this be used for bodies orbiting other stars?
No, this simplified formula is calibrated to the Sunโs mass, so it only applies to bodies within the solar system (planets, asteroids, etc. orbiting the Sun).