Kepler's Third Law
Orbital period from semi-major axis
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Working out orbital period from semi-major axis is easier with Kepler's Third Law, which answers as you type. You supply Semi-major Axis (AU) and Central Mass (Solar) to Kepler's Third Law; the answer follows at once. Kepler's Third Law puts the headline number first. Anyone comparing options in Science can use Kepler's Third Law to compare scenarios. Kepler's Third Law is privacy-first: Science inputs stay offline. Kepler's Third Law sits in the Science section next to free online kepler's third law calculator and kepler's third law calculator orbital period. Kepler's Third Law goes from question to answer fast. Kepler's Third Law answers the next Science question you meet.
What the defaults give
A semi-major axis of 1.524 AU around a 1 solar-mass star gives a period of 1.881 years, or 687 days.
That is Mars. The same law gives every planet in the solar system its period from one number.
The formula
T² = a³ ÷ M, with T in years, a in AU and M in solar masses
In these units the constant of proportionality is exactly 1, which is the whole reason the astronomical unit and the solar mass exist as a pair. In SI the equation carries 4π² ÷ GM, and using AU and solar masses absorbs it.
Worked examples
a = 1.524, M = 1 → T = √(3.540) = 1.881 years (687 days)
a = 0.762 → 0.665 years (243 days), which is Venus
a = 3.048 → 5.321 years (1,944 days)
a = 1.524, M = 2 → 1.330 years (486 days)
a = 1.524, M = 0.5 → 2.661 years (972 days)
The mass term is easy to forget. A planet at Mars distance from a star twice the Sun would orbit in 1.33 years, not 1.88, because a heavier star pulls harder and the orbit can be tighter in time for the same radius.
What the law actually says
The square of the orbital period is proportional to the cube of the semi-major axis. Distance grows much faster than time: move a planet four times further out and its year becomes eight times longer, because 4³ is 64 and the square root is 8. That is why the outer planets are so slow and why a mission to Neptune takes decades while a mission to Venus takes months.
The law holds for any two-body orbit, not just planets: moons around a planet, a satellite around the Earth, a binary star pair. The mass in the equation is the total mass of the two bodies, which is why the Moon orbits the Earth-Moon barycentre rather than the Earth centre.
Limits
It assumes an elliptical orbit dominated by one central mass. A planet perturbed by a larger neighbour, an orbit with a third body of comparable mass, or a relativistic regime near a compact object needs more than this. For the solar system the deviation from pure Kepler is small but measurable — Mercury precesses by 43 arcseconds per century, which Newtonian gravity does not explain.
Eccentricity does not change the period for a given semi-major axis. A comet and a circular orbit with the same a have the same year, even though the comet spends most of its time far from the Sun.
Frequently Asked Questions
What is the orbital period at 1.524 AU?
1.881 years, or 687 days, which is the Martian year. T is the square root of 1.524 cubed.
Why is the mass term in the formula?
A heavier central body pulls harder, so the orbit completes faster at the same distance. Doubling the mass cuts the period by a factor of √2.
Does eccentricity change the period?
No, for a fixed semi-major axis. A comet with the same a as a circular orbit has the same orbital period, though its speed varies a lot.
Can I use it for satellites around Earth?
Yes, in the right units. Convert to solar masses and AU, or use T² = 4π²a³ ÷ GM directly with SI values.
How far out is a planet with a 5.3-year period?
About 3.05 AU, because a is the cube root of T squared. That is roughly the asteroid belt distance.
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