Planets dance in the sky following a set of elegant rules: learn how to calculate Kepler's third law with CalcTool to see part of the choreography above us.

Reading our article, you will learn:

  • What are Kepler's laws of planetary motion, with a focus on his third law;
  • How to calculate Kepler's third law: equation and units.
  • How to use Kepler's third law: example of application of celestial mechanics.

Gearing of the celestial spheres: Kepler's law and fundamentals of orbital mechanics

Free from the atmosphere, where friction is nothing but a distant concept, planets and stars move following only the rules dictated by gravity.

The movement of two bodies bounded by the gravitational force alone, the so-called two body problem, is a staple of orbital mechanics. Physicists love it since it has exact solutions, something that is rarely true for a three-body problem. These same solutions came to scientists after centuries of observations. It's on the observations that Kepler, a German astronomer (and much else, as it was common back then), at the beginning of the XVIIth century, formulated the famous Kepler's law of planetary motion.

Kepler's law came to the world too soon: Newton's theory for the gravitational force was still some years away, with it, their explanation and complete understanding. With a lot of intuition and intelligence, Kepler supposed and modeled his laws so that years after his death, the newly discovered force fitted his result as a glove.

🙋 The gravitational force is a complex and fundamental topic of physics: we talked about it in detail on our gravitational force calculator.

Kepler formulated three laws:

  • The first Kepler's law: planets move in elliptic orbits, with the Sun occupying one of the foci.
  • The second Kepler's law: a segment joining a planet and the Sun covers equal areas in equal intervals of time.
  • The third Kepler's law: a planet's orbital period squared is proportional to the cube of its orbit's semi-major axis.

Let's analyze in detail Kepler's third law: from calculations to examples and much more.

Kepler's third law equation: how to calculate Kepler's third law

The original formulation of Kepler's third law — and the others — relied on the Copernican heliocentric model and the extremely precise observations by Tycho Brahe.

"The square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit."

Kepler himself could see the moons of Jupiter moving according to its law. What can we see in this law?

  • The elliptical nature of an orbit; the semi-major axis is half the longer axis departing the center of the orbit;
  • The orbital period; a planet takes a fixed amount of time to complete a revolution around its star.
  • There is a connection between period and distance from the star.

When Newton came up with his genial understanding of the gravitational force, Kepler's law finally found the mathematical formalism they needed. Newton's version of Kepler's third law requires us to begin by equating the centripetal force and the gravitational force:

mωr2=GmMr2m\cdot \omega \cdot r^2 =G\cdot \frac{m\cdot M}{r^2}

🙋 If you need to refresh your knowledge of the centripetal force, head to the centripetal force calculator

Where:

  • mm — The mass of the planet;
  • MM — The mass of the star;
  • rr — The semi-major axis;
  • GG — The gravitational constant; and
  • ω\omega — The angular velocity of the planet.

Let's write the angular speed in terms of the orbital period:

mr(2πT)2=GmMr2m\cdot r\cdot \left(\frac{2\cdot \pi}{T}\right)^2=G\cdot \frac{m\cdot M}{r^2}

And rearrange in a suitable way:

(2πT)2=GmMmr3\left(\frac{2\cdot \pi}{T}\right)^2 = G\cdot \frac{\cancel{m}\cdot M}{\cancel{m}\cdot r^3}

In a few steps, the Newton's version of Kepler's third law gave us the mathematical relationship we wanted:

T2r3T^2\propto r^3

Obviously, the true equation for Kepler's third law is more than just this proportionality relation. We must remember to add the proper units to Kepler's third law. Knowing that:

  • The period has unit [s][\text{s}]; and
  • The semi-major axis ahs unit [m][\text{m}].

We can assume the presence of a constant kk with units [s2/m3][\text{s}^2/\text{m}^3]. From Newton's version of Kepler's third law, we can say that:

k=(2π)2GMk = \frac{\left(2\cdot \pi\right)^2}{G\cdot M}

The gravitational constant is G=6.6743×1011 m3/(kgs2)G= 6.6743 \times 10^{-11}\ \text{m}^3/\left(\text{kg}\cdot\text{s}^2\right): here you can see the correct units for Kepler's third law finally coming in place.

Let's see an example of Kepler's third law equation.

How to use Kepler's third law: a heavenly example

We can try to calculate Kepler's third law in an example. The constant we met in the last section is valid for each orbit centered on the same object: we will focus on two objects orbiting the Sun. This time we chose Earth and its twin sister, Venus.

Earth has a pretty low eccentricity (meaning its orbit is almost circular). The parameters of its orbit are:

  • Semi-major axis: r=149.60×106 kmr = 149.60\times 10^6\ \text{km}; and
  • Period: T=365.256 daysT = 365.256\ \text{days}.

🙋 We have an entire tool dedicated to our planet's orbit: visit the Earth's orbit calculator to learn more about it!

We apply Kepler's third law equation to find the value of the constant relating semi-major axis and period:

k=T2r3=(365.256 days)2(149.60×106 km)3=3.98474×1020 days2km3\begin{align*} k &= \frac{T^2}{r^3} = \frac{\left( 365.256\ \text{days}\right)^2}{\left(149.60\times 10^6 \ \text{km}\right)^3}\\ \\ &= 3.98474\times10^{-20}\ \frac{\text{days}^2}{\text{km}^3} \end{align*}

Don't mind the unit, but remember to be consistent. Let's find Venus's semi-major axis using the calculated Kepler's third law's constant and its orbital period (fairly easy to measure from Earth):

r=(224.701 days)23.98474×1020 days2km33=108.21×106 km\begin{align*} r& = \sqrt[3]{\frac{\left( 224.701\ \text{days} \right)^2}{3.98474\times10^{-20}\ \frac{\text{days}^2}{\text{km}^3}}} \\ \\ &= 108.21\times10^{6}\ \text{km} \end{align*}

That is, effectively, the distance of Venus's apogee.

Davide Borchia
Star mass (M)
Suns
Planet period (T)
yrs
Semi-major axis (a)
au
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