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Kepler's third law of planetary motion (Definition)

Kepler's first two laws of planetary motion describe the shape of a planet's orbit and allow us to calculate the speed of its motion at any point in the orbit. Kepler was pleased to have discovered such fundamental rules, but they did not satisfy his quest to fully understand planetary motions. He wanted to know why the orbits of the planets were spaced as they are and to find a mathematical pattern in their movements - a "harmony of the spheres" as he called it. For many years he worked to discover mathematical relationships governing planetary spacing and the time each planet took to go around the Sun.

In1619, Kepler discovered a basic relationship to relate the planets' orbits to their relative distances from the Sun. We define a planet's orbital period, $P$, as the time it takes a planet to travel once around the Sun. Also, recall that a planet's semimajor axis, $a$, is equal to its average distance from the Sun. The relationship, now known as Kepler's third law, says that a planet's orbital period squared is proportional to the semimajor axis of its orbit cubed, or

$\displaystyle P^2 \propto a^3. $

When $P$ (the orbital period) is measured in years, and $a$ is expressed in a quantity known as an textbfastronomical unit (AU), the two sides of the formula are not only proportional but equal. One AU is the average distance between Earth and the Sun and is approximately equal to $1.5 \times 10^8$ kilometers. In these units,

$\displaystyle P^2 = a^3 $

Kepler's third law applies to all objects orbiting the Sun, including Earth, and provides a means for calculating their relative distances from the Sun from the time they take to orbit. Let's look at a specific example to illustrate how useful Kepler's third law is.

For instance, suppose you time how long Mars takes to go around the Sun (in Earth years). Kepler's third law can then be used to calculate Mars' average distance from the Sun. Mars' orbital period ($1.88$ Earth years) squared, or $P^2$, is $1.88^2 = 3.53$, and according to the equation for Kepler's third law, this equals the cube of its semimajor axis, or $a^3$. So what number must be cubed to give $3.53$? The answer is $1.52$ (since $1.52 \times 1.52 \times 1.52 = 3.53$). Thus, Mars' semimajor axis in astronomical units must be $1.52$ AU. In other words, to go around the Sun in a little less than two years, Mars must be about $50 \%$ (half again) as far from the Sun as Earth is.

This article is a derivative work of the creative commons share alike with attribution in [1].

Bibliography

[1] Fraknoi, Andrew, David Morrison, and Sidney Wolff. The Sky Above. In Astronomy 2e. Houston, Texas : OpenStax, 2022. The Sky Above



"Kepler's third law of planetary motion" is owned by bloftin.

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See Also: Kepler's first two laws of planetary motion, motion in central-force field, Kepler's three laws of planetary motion summarized

Other names:  Kepler's third law

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example calculating orbital periods (Example) by bloftin
applying Kepler's third law (Application) by bloftin

Cross-references: work, objects, formula, motion, speed, Kepler's first two laws of planetary motion
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This is version 1 of Kepler's third law of planetary motion, born on 2025-03-01.
Object id is 968, canonical name is KeplersThirdLawOfPlanetaryMotion.
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Physics Classification: 45.50.Pk (Celestial mechanics )
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