Kepler's third law: T² ∝ r³

Kepler's third law says that the square of a planet's orbital period is proportional to the cube of its orbital radius: T² ∝ r³. Planets further from the Sun take much longer to go round. For a circular orbit around a mass M, Newton's law of gravity turns it into an exact formula, T² = (4π²/GM) r³, which works for planets, moons and satellites alike.

This Kepler's third law calculator finds the orbital period from the central mass and the radius, or the radius from the period, and shows every step. It opens on the Earth going round the Sun.

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Kepler's Third Law Calculator

Apply Kepler's third law of planetary motion

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Kepler's third law formula

The law in its simplest form is a proportion:

For something in a circular orbit around a much heavier mass M, the full equation is:

Two things follow. The mass of the orbiting object does not appear, so a space station and a speck of dust at the same height go round in the same time. And everything orbiting the same body shares the same value of T²/r³, because that ratio only depends on M.

Kepler's third law in years and astronomical units

Measure T in years and r in astronomical units (AU, the Earth's average distance from the Sun) and, for anything orbiting the Sun, the constant becomes 1:

This is how Kepler first saw the pattern. The table uses NASA's average distances and periods for the planets. T²/r³ comes out close to 1 every time; the small differences are because the distances are averages of slightly oval orbits.

Here x is the orbital radius in AU and y is the period in years. The curve gets steeper as it goes, because T grows as r to the power 1.5: Jupiter is about 5 times further out than the Earth but takes nearly 12 times as long.

Using the ratio form

For two objects orbiting the same body, the constant cancels, so you can compare them without knowing G or M:

If one orbit is k times bigger, its period is k^1.5 times longer. Four times the radius gives 4^1.5 = 8 times the period. Try one:

Deriving Kepler's third law

The AQA A-level Physics specification asks you to derive T² ∝ r³, and the AQA data booklet does not print the result, so learn these steps. They assume a circular orbit and a central mass much heavier than the orbiting one.

1. Gravity provides the centripetal force, so set them equal: GMm/r² = mv²/r.
2. Cancel m and one r: v² = GM/r.
3. In one orbit the object travels the circumference 2πr in time T, so v = 2πr/T.
4. Substitute: 4π²r²/T² = GM/r.
5. Rearrange: T² = (4π²/GM) r³. Since 4π²/GM is constant for one central mass, T² ∝ r³.

Worked examples

These use G = 6.67 × 10⁻¹¹ N m² kg⁻², the Sun's mass 1.99 × 10³⁰ kg and the Earth's mass 5.97 × 10²⁴ kg, the values in the AQA data booklet.

Example 1: the period of Mars from the ratio

Mars orbits the Sun at an average of 228.0 million km, and the Earth at 149.6 million km. Find the orbital period of Mars.

In AU, r = 228.0 ÷ 149.6 = 1.524. Using T² = r³ for the Sun:

That is 1.88 × 365.25 = 687 days, which matches the measured value.

Example 2: the period of Jupiter from G and M

Jupiter orbits the Sun at r = 7.785 × 10¹¹ m. Find its period with the full formula.

This gives T = 3.75 × 10⁸ s. Divide by 86,400 for days and by 365.25 for years: 11.9 years. You can check it in the calculator above by typing in the same mass and radius.

Example 3: the radius of a geostationary orbit

A geostationary satellite stays over one point on the equator, so its period must equal one turn of the Earth: 23 hours 56 minutes, or 86,164 s. Find its orbital radius and its height above the surface (Earth radius 6.37 × 10⁶ m).

The height is 4.21 × 10⁷ − 6.37 × 10⁶ = 3.58 × 10⁷ m, or about 35,800 km above the ground. Using exactly 24 hours gives 4.22 × 10⁷ m, which is fine in most exams.

Example 4: the Earth's mass from the Moon

The Moon orbits the Earth at an average of 3.844 × 10⁸ m, once every 27.3 days. Estimate the mass of the Earth.

Convert the period: T = 27.3 × 86,400 = 2.359 × 10⁶ s. Rearrange for M:

That is within about 1% of the accepted 5.97 × 10²⁴ kg. This is how astronomers weigh planets, stars and even black holes: time something orbiting them.

Orbital speed

Step 2 of the derivation also gives the speed of an orbit:

For the Earth around the Sun, r = 1.496 × 10¹¹ m, so v = 2.98 × 10⁴ m/s, almost 30 km/s. Closer orbits are faster, which is why a satellite in low orbit laps the Earth in about an hour and a half while the Moon takes a month. To watch this happen, launch your own satellites in the Orbit Simulator.

Kepler's three laws

Kepler found three laws from Tycho Brahe's careful observations of the planets. NASA states them like this:

• First law: each planet moves in an ellipse with the Sun at one focus.
• Second law: the line from the Sun to a planet sweeps out equal areas in equal times, so a planet moves fastest when it is closest to the Sun.
• Third law: the squares of the orbital periods are proportional to the cubes of the semi-major axes.

The semi-major axis is half the longest width of the ellipse. For a circle it is just the radius, which is why A-level questions use r.

Common mistakes

• Using height instead of radius. r is measured from the centre of the planet. Add the planet's radius to a satellite's height first.
• Leaving T in hours or days. With G in SI units, T must be in seconds and r in metres.
• Using the satellite's mass for M. M is the mass being orbited. The orbiting mass cancels out.
• Using T² = r³ for the wrong body. The years and AU shortcut only works for orbits around the Sun.
• Forgetting the square root or cube root at the end, after rearranging for T or r.

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