Third Law of Kepler

Kepler's Laws of Planetary Motion

In the 1500s astronomer, Nicolaus Copernicus hypothesized that the Sun was stationary at the center of our universe, with the other planets, such as Earth, orbiting around it with circular revolutions. One of the major improvements to this model was originated by Johannes Kepler, who used astronomical observations to show that instead of having circular orbits, the planets actually followed an elliptical motion around the Sun.

These orbital trajectories were described by Kepler's Laws, which we will be discussing and expanding upon in this article. Even though these were originally formulated for the planets in the Solar System orbiting the Sun, they can be applied universally. They can be applied to exoplanets around other stars, as well as describing the motion of a satellite orbiting a planet. However, if we want to apply Kepler's Laws we must make sure that the center of mass of the satellite-planet system is inside the central body and that the satellite is much less massive than the planet's center of mass.

Kepler's First Law states that planets move in elliptical orbits with the sun at its focus.

Fig. 2 - Elipses have two focus points as well as a semi-major axis and a semi-minor axis.

In a circular orbit, the satellite will move at the same speed throughout the orbit. However, according to Kepler's Second Law, in an elliptical orbit, a satellite travels faster when it is nearer the planet and moves slower when farthest away from the planet. Fig. 3 - A visual representation of Kepler's Second Law. Notice how equal areas of space are swept across equal periods of time.

Third Law of Kepler: Definition

Kepler was so amazed by his finding that he stated the following in his book Harmonies of the World (1619):

The planets in the solar system exhibit different orbital periods. For example, let's compare the orbital periods of Venus and Mercury. Venus' orbital period is 224 Earth days and Mercury's orbital period is 88 Earth days. We are familiar with describing orbital periods in Earth days, but we have to remember that days last 24 hours only on planet Earth. Some planets even have days longer than a year. An example of this is the planet Venus. It takes 224 Earth days for Venus to orbit the Sun. In comparison, it takes 243 Earth days for Venus to complete a rotation about its axis.

Why is it that different planets have different orbital periods? If we look at the distances of the respective planets to the Sun, we see that Mercury is the closest planet to the Sun. It, therefore, has the shortest orbital period of the planets.

Third Law of Kepler: Formula

Kepler's Third Law explains that the period for a satellite to orbit Earth increases rapidly with the radius of its orbit. For an elliptical orbit, Kepler's Third Law can be expressed as follows,

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

where \(T\) is the orbit's period in seconds \(\mathrm s\), \(G\) is the gravitational constant \(6.67\times10^{-11}\;\frac{\mathrm N\;\mathrm m^2}{\mathrm{kg}^2}\), \(M\) is the planet's mass in \(\text{kg}\), and \(a\) is the semi-major axis in \(\text{m}\). The semi-major axis is equal in length to half of the longest diameter of an ellipse. In the case of a circular orbit, the semi-major axis can be replaced with the radius of the orbit.

Fig. 4 - This is a diagram visualizing Kepler's Third Law. According to Kepler's Third Law, the ratio of the square of the orbital period with the cube of the semi-major axis of its orbit is a constant.

Proving the Third Law of Kepler

This relationship may be obtained in the case of a circular orbit using Newton's Law of Gravitation. To find it, we can equal the centripetal force of the satellite to the gravitational force between the planet and the orbiting object.

Newton's Law of Gravitation is given by

$$F=\frac{GMm}{r^2},$$

\(G\) is the gravitational constant \(6.67\times10^{-11}\;\frac{\mathrm N\;\mathrm m^2}{\mathrm{kg}^2}\), \(M\) is the planet's mass in kilograms \(\mathrm{kg}\), \(m\) is the orbiting object's mass in kilograms \(\mathrm{kg}\), and \(r\) is the distance between them in meters \(\mathrm m\).

The centripetal force is expressed below:

$$F=m\omega^2r,$$

where we define \(m\) as the orbiting object's mass in kilograms \(\mathrm{kg}\), while \(\omega\) is the angular velocity in radians per second \(\frac{\mathrm{rad}}{\mathrm s}\), and \(r\) is the distance between the planet and the orbiting object in meters \(\mathrm m\).

Now we equal the centripetal force of the satellite to the gravitational force between the planet and the Sun,

$$\frac{GMm}{r^2}=m\omega^2r.$$

We can express the angular velocity in terms of the orbital period and then rearrange the equation:

$$\begin{align*}\frac{GMm}{r^2}&=mr\left(\frac{2\pi}T\right)^2,\\mr\frac{4\pi^2}{T^2}&=\frac{GMm}{r^2}.\end{align*}$$

If we solve for \(T^2\), we find the expression for the Third Law of Kepler:

$$T^2=\left(\frac{4\pi^2}{GM}\right)r^3.$$

Since the value in the parenthesis is a constant, we can see that \(T^2\) is proportional to \(r^3\).

Instead of circles, a more comprehensive derivation may be made using generic elliptical orbits. As a result, a circle's radius \(r\) is replaced by the semi-major axis \(a\) of an ellipse.

As a result, the formula of Kepler's Third Law for elliptical orbits is given by

$$T^2=\left(\frac{4\pi^2}{GM}\right)a^3.$$