Kepler's Third Law (AU–Year Form) Calculator
Solve the classic harmonic law T² = a³ for either of its two unknowns — orbital period or semi-major axis — in the natural units of the solar system, with step-by-step math, a live orbital diagram showing the star at one focus, and companion metrics including orbital velocity, perihelion and aphelion.
Why the exponents are 2 and 3
The exponents follow from circular motion plus Newtonian gravity. For a circular orbit, gravity supplies the centripetal force: GMm/r² = mv²/r. Substituting v = 2πr/T and rearranging gives T² = 4π²r³/(GM). The exponents 2 and 3 are therefore not empirical fits — they are the direct consequence of an inverse-square force law and the geometry of circular motion. Kepler discovered them in 1619 by fitting Tycho Brahe's observations; Newton explained them in 1687 from first principles. This is one of the great unifications in physics.
The AU–year form is a units trick
The simplified T² = a³ works without any constants because the units have been chosen to absorb them. For the Sun, GM☉ = 4π² AU³/year². Substituting that into T² = 4π²a³/(GM) leaves exactly T² = a³. This is why the AU–year–solar-mass system is so convenient for solar system work: every constant becomes 1, and the arithmetic is just squares and cubes. For any other central mass the constant is 1/M, and for any other unit system it is 4π²/(GM) in those units.
The period does not depend on eccentricity
Kepler's third law relates T to a alone — the eccentricity of the orbit does not appear. A comet on a long elliptical orbit with a = 5 AU has exactly the same period as a body on a circular orbit at 5 AU, even though the comet spends most of its time far from the Sun and speeds through perihelion. This is a consequence of the fact that the ellipse's semi-major axis, not its shape, sets the orbital energy, and the energy sets the period. This is one of the most counter-intuitive and most useful facts in orbital mechanics.
Assumptions — and where the simple form breaks
The simplified form assumes the central mass is exactly one solar mass and that the orbiting body's mass is negligible. For a planet orbiting the Sun this is an excellent approximation — even Jupiter is only 0.1% of the Sun's mass. But it fails badly for binary stars of comparable mass, for moons around planets (where the central mass is the planet, not the Sun), and for any exoplanet whose host star is not exactly solar-mass. In those cases use the general form T² = a³/(M₁ + M₂) with both masses in solar masses. The formula also assumes Newtonian gravity — it does not account for general-relativistic precession, which matters only very close to a compact object.
How to Use This Calculator
The harmonic law links two quantities: T and a. Pick which one you want the tool to solve, then enter the other. The auxiliary inputs drive the orbital diagram and the companion metrics but do not change the primary relation.
1. Choose what to solve for
Pick T or a from the toggle. The calculator applies T = a3/2 or a = T2/3 automatically. The active field becomes read-only and shows the SOLVED badge. The exponents 3/2 and 2/3 are exact.
2. Use realistic presets
Presets cover the solar system: Mercury (0.39 AU, 0.24 yr), Venus (0.72, 0.62), Earth (1, 1), Jupiter (5.20, 11.86), Neptune (30.1, 164.8) and Pluto (39.5, 248). Use them as anchors when exploring the relation.
3. Set the orbit shape and central mass
Eccentricity drives the diagram — it changes the ellipse shape, the focus offset, and the perihelion/aphelion distances, but not the period. Central mass drives the general-form correction; with M = 1 the simplified T² = a³ is exact, and with M ≠ 1 the period scales as 1/√M.
4. Read the result and verdict
The result card shows the solved value plus T², a³ and the orbital velocity, with a colour-coded verdict classifying the orbit from inner terrestrial through Earth-like, Mars-like, main belt, Jovian, outer planets, Kuiper belt, scattered disk and extreme trans-Neptunian.
5. Check the diagram and charts
The diagram draws the orbit as an ellipse with the star at one focus, the semi-major and semi-minor axes marked, the body at its current position, and the T² = a³ identity displayed as a numerical check. The charts plot T against a on linear axes and on log–log axes with the solar system planets overlaid.
6. Switch to Advanced for depth
Advanced mode adds the log–log chart with the planets, a sensitivity table showing the exact 3/2 and 2/3 scaling, and companion metrics including orbital velocity, perihelion, aphelion, orbital circumference, the general-form correction, and the sidereal-to-synodic comparison.
What This Tool Offers
Built around the classic harmonic law T² = a³ in the natural units of the solar system, and the physics that makes it work.
Two-way solving
Solve for the orbital period in years or for the semi-major axis in AU. The calculator applies T = a3/2 or a = T2/3 exactly — no rounding, no approximations.
Live orbital diagram
The orbit is drawn as an ellipse with the star at one focus, the semi-major and semi-minor axes marked, the focus offset shown, and the orbiting body placed at its current position along the orbit.
Orbital-zone verdict
Each result is classified by semi-major axis from inner terrestrial through Earth-like, Mars-like, main belt, Jovian, outer planets, Kuiper belt, scattered disk and extreme trans-Neptunian, with a coloured badge.
Perihelion and aphelion
The orbit's closest and farthest points follow directly from the semi-major axis and eccentricity, so the full shape of the orbit is characterised by just two numbers.
Dual comparison charts
T vs. a on linear axes showing the 3/2 power curve, plus an advanced log–log chart where the law becomes a straight line with slope 3/2 — with the actual solar system planets plotted as reference points.
Reference tables
A lookup grid across semi-major axes, a table of solar system bodies with their a, T and the Kepler check T²/a³, and a full variable reference table with unit notes.
Reference Tables & Conversions
An orbit explorer, the harmonic law lookup grid, and the solar system check table.
Orbit explorer
Harmonic Law Lookup: a → T → T²
Computed from T² = a³. T² and a³ always match exactly in these units — that is the whole content of the law.
| a (AU) | T (years) | T (days) | T² (yr²) | a³ (AU³) | v (km/s) |
|---|
The Solar System as a Kepler Test
Every planet and dwarf planet satisfies T² = a³ to the precision of the measurements. The ratio T²/a³ is the empirical test of the law.
| Body | a (AU) | T (years) | T² | a³ | T² / a³ |
|---|
Variable Reference
| Symbol | Meaning | Typical range | Units |
|---|---|---|---|
| T | Orbital period | 1e-6 – 1e8 | years |
| a | Semi-major axis | 1e-5 – 1e6 | AU |
| e | Eccentricity | 0 – 0.95 | dimensionless |
| q | Perihelion = a(1−e) | — | AU |
| Q | Aphelion = a(1+e) | — | AU |
| b | Semi-minor axis = a√(1−e²) | — | AU |
| M | Central mass | 1e-8 – 1e10 | M☉ |
| v | Mean orbital velocity | 0.001 – 200 | km/s |
What Does T² = a³ Measure?
Kepler's third law says that the square of a planet's orbital period is proportional to the cube of its mean distance from the Sun. In the natural units of the solar system — years for the period and astronomical units for the distance — the constant of proportionality is exactly 1, and the law reads T² = a³. It is one of the most elegant relations in all of physics: a single equation linking the time a planet takes to go around once with the size of its orbit, valid for every body in the solar system from Mercury to the most distant trans-Neptunian object.
Kepler's Third Law Formula
The defining relationship in AU–year units is:
Rearranged to solve for the two unknowns:
a = T2/3 = (T²)1/3
The general form with a central mass M (in solar masses) and a negligible orbiting mass is:
T = a3/2 / √M
a = ( T² · M )1/3
And the full Newtonian form with both masses:
The companion orbital quantities follow from a and e:
Aphelion: Q = a ( 1 + e )
Semi-minor axis: b = a √( 1 − e² )
Mean orbital speed: v ≈ 29.78 / √a km/s (for M = 1)
Kepler Variables
| Symbol | Meaning | Notes |
|---|---|---|
| T | Orbital period | Time for one complete orbit |
| a | Semi-major axis | Half the longest diameter of the orbit |
| e | Eccentricity | Shape of the ellipse; does not affect T |
| M | Central mass | Sets the constant of proportionality |
| q, Q | Perihelion, aphelion | Closest and farthest orbital points |
| v | Mean orbital velocity | Circular approximation, 29.78/√a km/s |
How Each Variable Affects the Result
The period scales as the 3/2 power of the semi-major axis. Doubling the distance multiplies the period by 21.5 ≈ 2.83. Tripling the distance multiplies the period by 31.5 ≈ 5.20. Multiplying by 10 gives 101.5 ≈ 31.6. This is a steep, but not explosive, scaling — it is much gentler than the cube would suggest (10³ = 1000) and much steeper than a linear relation (10¹ = 10). The exponent 3/2 is the signature of an inverse-square force law acting on a circular orbit, and it is why the outer planets take so long to complete a single circuit: Neptune at 30 AU is 30 times farther than Earth, but its period is 301.5 ≈ 165 years, not 30 years.
The inverse relation a = T2/3 is even gentler. To double the period you must move the body outward by 22/3 ≈ 1.59, or about 59%. To increase the period tenfold you need a2/3 such that a3/2 = 10, which gives a = 102/3 ≈ 4.64 AU. This is why all the inner planets are crowded into a small range of semi-major axes — they need those small values to have periods of months rather than decades.
Kepler Calculation Examples
T = 13/2 = 1.000 year
T = 5.2031.5 = 11.86 years
T = 30.071.5 = 164.8 years
a = 2482/3 = 39.48 AU
a = 762/3 = 17.83 AU
With e = 0.967: q = a(1−e) = 0.59 AU, Q = a(1+e) = 35.1 AU
T = a3/2/√M = 1 / √0.5 = 1.414 years
M = a³/T² = 21.54³/100² = 9992/10000 = 0.9992 M☉
Frequently Asked Questions
Because the units are defined so that the constant equals 1. The general form is T² = 4π²a³/(GM), and for the Sun GM☉ = 4π² AU³/year². Substituting that gives T² = 4π²a³/(4π²) = a³, with the 4π² cancelling exactly. This is not a coincidence — the AU is defined as Earth's mean orbital distance, and the year is defined as Earth's orbital period, so by construction Earth must satisfy T = a = 1 and therefore the constant must be 1 for the Sun. For any other central mass the constant becomes 1/M, and for any other unit system it becomes the appropriate value of 4π²/(GM) in those units.
No. This is one of the most surprising and most useful facts in orbital mechanics. The period of an orbit depends only on its semi-major axis, not on how elongated it is. A comet on a highly eccentric orbit with a = 5 AU has exactly the same period as a body on a circular orbit at 5 AU — about 11.2 years. The comet speeds through perihelion and crawls near aphelion, but the two effects balance exactly so that the round trip takes the same time. The physical reason is that the semi-major axis determines the orbital energy, and the energy determines the period. Eccentricity changes the shape of the path and the speed at each point along it, but not the total time for a complete circuit.
Yes, provided you correct for the host star's mass. Most exoplanet host stars are not exactly solar-mass — they range from about 0.1 to 2 solar masses, and the period scales as 1/√M. A planet at 1 AU around a 0.5 solar-mass star has a period of 1.414 years, not 1 year. Astronomers routinely compute periods and semi-major axes in the AU–year–solar-mass system precisely because the mass is the only correction needed. In practice, when you read an exoplanet paper and see "a = 0.05 AU, P = 4.2 days" for a planet around a solar-type star, that is Kepler's third law in action, with the mass correction absorbed into the numbers.
The sidereal period is the time for a body to complete one orbit relative to the fixed stars — this is the T in Kepler's third law. The synodic period is the time between two successive identical configurations as seen from a third body (usually Earth) — for example, the time from one opposition of Mars to the next, or from one full moon to the next. For Mars, the sidereal period is 1.881 years but the synodic period is 2.135 years, because Earth is also moving. The relation is 1/S = |1/P₁ − 1/P₂| where P₁ and P₂ are the two sidereal periods. Kepler's law deals in sidereal periods only.
From data. Johannes Kepler published the third law in his Harmonices Mundi in 1619, after two decades of work fitting Tycho Brahe's exceptionally precise naked-eye observations of Mars. He had already published the first two laws (elliptical orbits and equal areas in equal times) in 1609, and the third followed as a numerical pattern he recognised while searching for harmonic relationships between the planets. It was Newton, seventy years later, who showed that all three of Kepler's laws follow from a single inverse-square gravitational force plus his own laws of motion. Kepler's achievement was to extract a clean mathematical law from noisy data; Newton's was to explain why it must be true.
Yes, but with the central mass being the planet, not the Sun. For a moon of Jupiter, M is Jupiter's mass in solar masses (0.000954 M☉), and the units must be chosen consistently — or you can work in kilometres and seconds with the full form T² = 4π²a³/(GM). A more practical approach for moons is to compute the ratio relative to a known moon: T/T_Io = (a/a_Io)3/2. This avoids having to know G and Jupiter's mass to full precision. All of the Galilean moons of Jupiter obey Kepler's third law to within the accuracy of the observations, which was one of Galileo's arguments against the Ptolemaic system.
Yes, and this is one of its most powerful applications. Rearranging the general form gives M = 4π²a³/(GT²), so measuring a body's orbital period and semi-major axis yields the mass of the central object directly. This is how the mass of the Sun, every planet with a moon, every binary star system, and the supermassive black holes at the centres of galaxies have been measured. For the Milky Way's central black hole, astronomers tracked the star S2 through one complete orbit: a = 1,020 AU, T = 16.05 years, giving M ≈ 4.15 × 10⁶ M☉. The same formula, one equation, applied at vastly different scales.
The astronomical unit was originally defined as the mean Earth–Sun distance, but it was redefined in 2012 by the International Astronomical Union to be exactly 149,597,870,700 metres — a fixed number, not a measured one. The modern definition makes the AU a purely conventional unit, decoupled from the actual Earth–Sun distance, which varies throughout the year because Earth's orbit is elliptical (ranging from 0.983 AU at perihelion to 1.017 AU at aphelion). The mean distance is exactly 1 AU by definition. The speed of light in these units is 1 AU per 499.005 seconds, a fact that is used constantly in solar system astronomy to convert between light-travel times and distances.
Variables
| Symbol | Quantity | Unit |
|---|---|---|
| T | Orbital period | years |
| a | Semi-major axis | AU |
What it means
This is the simplified version of Kepler’s third law for bodies orbiting the Sun. When period T is measured in Earth years and semi‑major axis a in astronomical units (AU), the relation T² = a³ holds exactly (within the approximation of negligible mass). This is used throughout solar system astronomy to determine orbital distances from periods, and vice versa. It also extends to other systems when the mass is known. Understanding this is a cornerstone of celestial mechanics and is used in planetary science and mission planning.
Worked example
Kepler's Third Law (AU‑Year) – Two Detailed Examples
Real‑World| Parameter | Value |
|---|---|
| a (AU) | 1.524 |
| Parameter | Value |
|---|---|
| a | 2.5 |
Common mistakes
- Kepler’s third law (simplified): T² = a³ – for objects orbiting the Sun.
- T: Orbital period in Earth years.
- a: Semi‑major axis in Astronomical Units (AU).
- Sun only: This form applies only to objects orbiting the Sun – not other stars.
- Units: If masses differ, use the full form: T² = (4π²/GM) a³.
Applications
Kepler's third law in its simplified form for the Solar System, T² = a³ (with T in years and a in AU), is a practical approximation for planetary motion. This is used by students and amateur astronomers to compute orbital periods of asteroids and comets when their semi‑major axes are known. It is also used in exoplanet studies when the host star is solar‑like, though with a mass term. This simple relation is a cornerstone of planetary science and space mission planning.
- Calculation of orbital periods of planets, asteroids, and comets
- Estimation of semi‑major axis from observed period
- Introductory astronomy education
- Mission planning for Solar System exploration
- Verification of Keplerian motion in student labs
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