Kepler's second law covers one planet. In 1619 he found a rule that links them all: the farther a planet is from the Sun, the longer its year, in a precise way. It uses the fractional exponents you know: $P = a^{3/2}$.
Why it works
Measure the period $P$ in years and the semi-major axis $a$ in AU, and for every planet $P^2 = a^3$. Two things stretch an outer planet's year. Its path is longer, in proportion to $a$. And gravity is weaker out there, so it moves more slowly: for circular orbits, speed goes as $1/\sqrt{a}$. Time is path divided by speed: $P \propto a \div \frac{1}{\sqrt{a}} = a^{3/2}$, so $P^2 \propto a^3$. Earth sets the units: $1$ AU and $1$ year make the constant $1$.
Key idea
Kepler's third law: $P^2 = a^3$ ($P$ in years, $a$ in AU, for bodies orbiting the Sun). So $P = a^{3/2}$ and $a = P^{2/3}$. Twice as far means $2^{3/2} \approx 2.8$ times as long.
| Planet | $a$ (AU) | $P$ (yr) | $a^3$ | $P^2$ |
|---|---|---|---|---|
| Mercury | 0.387 | 0.241 | 0.058 | 0.058 |
| Earth | 1.000 | 1.000 | 1.00 | 1.00 |
| Mars | 1.524 | 1.881 | 3.54 | 3.54 |
| Jupiter | 5.203 | 11.86 | 140.8 | 140.7 |
| Saturn | 9.537 | 29.46 | 867 | 868 |
Worked examples
Worked example: Jupiter's year
Jupiter's orbit has $a = 5.20$ AU. Find its period.
- $P^2 = a^3 = 5.20^3 = 140.6$.
- $P = \sqrt{140.6} = 11.9$ years.
Answer: About $11.9$ years.
Worked example: From period to distance
An asteroid orbits the Sun every $8$ years. How far out is it?
- $a^3 = P^2 = 64$.
- $a = \sqrt[3]{64} = 4$ AU: beyond the main asteroid belt, a little more than three-quarters of the way out to Jupiter.
Answer: $4$ AU.
Worked example: Days to years
Venus orbits in $224.7$ days. Find $a$.
- Convert to years: $224.7 \div 365.25 = 0.6152$ yr.
- $a^3 = 0.6152^2 = 0.3785$, so $a = \sqrt[3]{0.3785} = 0.723$ AU.
Answer: $0.723$ AU.
Common mistake
A tempting guess is that a planet twice as far out takes twice as long. That forgets the second effect: it also moves more slowly. Check with $P^2 = a^3$: $a = 2$ gives $P = \sqrt{8} \approx 2.83$ years, not 2. Mars ($1.52$ AU) takes $1.88$ years, not $1.52$.
This form works only for objects orbiting the Sun, in years and AU. Newton's version works around any star or planet. Launch your own orbits in the Orbit Lab.