1609, in the book Astronomia Nova. They state that:

1) Planets move in elliptical orbits about the Sun, which sits at one focus.

and 2) a planet’s orbital velocity varies over the course of one revolution such that a line

drawn between the planet and the Sun will sweep out equal area in equal time, wherever

in the orbit you are.

Kepler published his Third Law ten years later, in his 1619 book Harmonices Mundi, or The

Harmony of the World. By sifting through observations of how the planets move in their orbits, he

noticed something curious. The time it takes for a planet to orbit the Sun once (its orbital

period, P) is related to its average distance from the Sun (its semi-major axis, a) in a

very specific way:The square of the period is proportional to the semi-major axis cubed.

In fact, if we choose our units carefully, with the period in years (the time it takes

for Earth to go once around the Sun) and the semi-major axis in astronomical units, or

AU (the distance between the Earth and the Sun, about 150 million kilometers), the proportionality

becomes an exact equation: the square of the period is equal to the semi-major axis cubed.

Let’s try it out!

Earth’s semi-major axis is 1 AU (because that’s the definition of an astronomical

unit). Mars is farther from the Sun; it has a semi-major axis of 1.524 AU. We can use

Kepler’s Third Law to figure out how long it takes Mars to complete one orbit around

the Sun - that is, how many Earth years make up one Mars year.

The semi-major axis, 1.524 AU, cubed equals 3.54, and the square root of that gives us

Mars’ orbital period: 1.88 Earth years.

Sure enough, Mars takes 687 Earth days to orbit the Sun, which is 1.88 Earth years!

And this works for all of the planets in our solar system.

But what if we don’t want to use astronomical units and Earth years, but something more

conventional like kilometers? Then we have to unpack Kepler’s equation a little bit

more. In modern notation, it becomes: P squared equals 4 pi squared over G (the

gravitational constant) times M-Sun, all multiplied by a to the third power.

That term before the a cubed is the constant of proportionality in Kepler’s original equation.

M-sun is the mass of the Sun, and to be completely accurate, the equation should say M-Sun plus

m-planet, the mass of the Sun and planet together. But because the Sun is so much more massive

than even the biggest planet in our solar system, Jupiter (The mass of the sun is more

than a thousand times the mass of Jupiter!), the extra mass added by m-planet is tiny compared

to M-sun and doesn’t make much of a difference.

G is the gravitational constant, which might look familiar if you’ve ever seen Newton’s

Law of Gravitation, which states that the force due to gravity equals G times big M

times little m, all over the distance squared.

Actually, you can derive Kepler’s Third Law from Newton’s Law of Gravitation (although,

of course, Kepler didn’t know that!) by setting the force of gravity equal to the

centripetal force, the force the planet feels because it’s revolving around the Sun. Let’s

consider a circular orbit first, because that’s the simplest case. The centripetal force depends

on the planet’s speed, v, which for a circular orbit is constant:

F centripetal equals m v squared over r.

Setting these two forces equal to each other gives us a relationship that simplifies to

v equals the square root of G times M over r:

Now, since we’re considering a circular orbit at the moment, we also know that there’s

a relationship between the speed, v, and the orbital period, P. The planet has to go around

the whole circumference of the orbit (2pi r) in the time P, so the speed must be equal

to 2 pi r over P:

And if we substitute that into our force equation from before, and use the semi-major axis,

a, as our distance from the Sun (now we’re back to elliptical orbits), we have

2 pi a over P equals the square root of G times M over a:

After some rearranging, we’ve gotten back to Kepler’s Third Law! That is,

P squared equals 4 pi squared over G times M-Sun,

all multiplied by a cubed.

This expression is especially informative because it relates two values that you can

measure with a telescope - distance to the Sun and orbital period - to the mass of the

Sun, and that’s how we know how heavy the Sun is! Even better, it works for all kinds

of orbiting systems. You can find the mass of the Earth, for example, by rewriting the

equation using the distance and orbital period of the Moon:

The Moon’s orbital period, P-Moon, is 27.3 days, or 2,358,720 seconds and the semi-major

axis of the Moon’s orbit, aMoon, is 384,400,000 m.

G, as usual, is 6.67 × 10^-11 meters cubed per kilogram per second squared.

Solving the equation using these parameters

yields an estimate for the mass of the Earth of

6 times ten to the 24 kilograms.

Any planet with a moon can easily be “weighed” this way,

and it’s all thanks to Kepler’s Third Law!