Kepler's Third Law

Equivalent forms

T^2 \propto a^3

\frac{T^2}{a^3} = \frac{4\pi^2}{GM}

An empirical pattern from 17th-century stargazing turned out to be a direct consequence of inverse-square gravity.

Symbol	Name	SI unit	Dimension	Range
$T$	Orbital periodoutput Time to complete one orbit	s	T	0 – 10000000000
$G$	Gravitational constant Universal gravitational constant	m³·kg⁻¹·s⁻²	M^-1L^3T^-2	6.674e-11 – 6.674e-11
$M$	Central mass Mass of the dominant body (e.g., the Sun)	kg	M	1e+28 – 1e+32
$a$	Semi-major axis Average orbital radius (semi-major axis of the elliptical orbit)	m	L	10000000000 – 10000000000000

Unit systems

Symbol	SI	CGS	Imperial
$T$	s	s	s
$G$	m³·kg⁻¹·s⁻²	cm³·g⁻¹·s⁻²	ft³·slug⁻¹·s⁻²
$M$	kg	g	slug
$a$	m	cm	ft

Where it holds

Valid for bound two-body orbits where one body dominates (M >> m). For comparable masses, replace M with the total M + m.

Dimensional analysis

T^{2} \to [T^{2}]

a^{3}/(GM) \to [L^{3}] / ([M^{-1}L^{3}T^{-2}][M]) = [L^{3}]/[L^{3}T^{-2}] = [T^{2}]

Both sides are

[T^{2}]

.

\checkmark

Discovery

Johannes Kepler · 1619

Kepler published the third law in 'Harmonices Mundi' after years of analyzing Tycho Brahe's planetary data. Newton later derived it from his law of universal gravitation.

Try this

Why does Jupiter take 12 Earth years to orbit the Sun?

Given Jupiter's orbital radius a = 5.20 AU, predict its orbital period T using Kepler's Third Law T² = (4π²/GM)a³ for the Sun's mass M = 1.989e30 kg.

Research status: stable

Real-world applications

Predicting satellite orbital periods (GPS, ISS)
Mass determination of stars hosting exoplanets
Solar System ephemerides
Galactic dynamics and dark matter detection

Common misconceptions

It is not just a statement about the Solar System — it works for any inverse-square gravitational orbit
The 'a' is the semi-major axis, not the perihelion distance
Equal areas in equal times (2nd law) holds even when $T^{2} \propto a^{3} (3rd$ law) — they're complementary

Experimental verification

Confirmed by every planet, moon, and exoplanet observed. Used to determine the masses of stars hosting exoplanets and the mass of the Milky Way's central black hole.

Derivation

For a circular orbit, gravity provides centripetal force:

GMm/a^{2} = m\omega ^{2}a

, giving

\omega ^{2} = GM

/

a^{3}

.

With

T = 2\pi /\omega

,

T^{2} = 4\pi ^{2}a^{3}/(GM)

.

The result extends to ellipses with a as the semi-major axis.

Limiting cases

a \to 0

⟶

T \to 0

Tighter orbits close to the central body have vanishing periods (Keplerian shear).

a \to \infty

⟶

T \to \infty

An infinite orbital radius takes infinitely long to traverse.

M \to \infty

⟶

T \to 0

A more massive central body pulls orbits faster, shrinking the period.

What if…

What if the orbit radius doubles?

T scales by $2^1.5 \approx 2.83\times$ . An orbit at 2 AU takes about 2.83 years (vs. $1\,\mathrm{AU} = 1\,\mathrm{yr})$ .

What if the central mass doubles?

T shrinks $by \sqrt{2} \approx 1.414\times$ . The same orbit completes $\sim 41$ % faster.

What if the orbit is highly elliptical?

Use the semi-major axis (not the perihelion or aphelion). The period only depends on a, not on eccentricity.

1

Jupiter's period

Given ·

G:: 6.674e-11
M:: 1.989e+30
a:: 778500000000

Find · T

Steps

$a^{3} = (7.785e11)^{3} \approx 4.718e35$
$GM = 6.674e-11 \times 1.989e30 \approx 1.327e20$
$T^{2} = 4\pi ^{2} \times 4.718e35 / 1.327e20 \approx 1.404e17$
$T = \sqrt{1.404}e17 \approx 3.74e_{8} s \approx 11.86\,\mathrm{yr}$

Result ·

T^{2} = 4\pi ^{2} \times (7.785e11)^{3} / (6.674e-11 \times 1.989e30) \approx 1.404e17\,\mathrm{s}^{2}

;

T \approx 3.74e_{8} s \approx 11.86

years

2

Low-Earth-orbit satellite

Given ·

G:: 6.674e-11
M:: 5.972e+24
a:: 6771000

Find · T

Steps

$a^{3} \approx 3.103e20$
GM (Earth $) \approx 3.986e14$
$T^{2} = 4\pi ^{2} \times 3.103e20 / 3.986e14 \approx 3.07e_{7}$
$T = \sqrt{3.07}e_{7} \approx 5540\,\mathrm{s} \approx 92$ minutes

Result ·

T^{2} = 4\pi ^{2} \times (6.771e_{6})^{3} / (6.674e-11 \times 5.972e24) \approx 3.07e_{7} s^{2}

;

T \approx 5540\,\mathrm{s} \approx 92 \min

Newton's Law of Universal Gravitation→Centripetal Acceleration→Escape Velocity→