Kepler’s Third Law and the Virial Theorem

This post covers two classic results about bodies in orbit: Kepler’s third law and the virial theorem. The virial theorem matters well beyond planetary motion: it is central in statistical mechanics and appears in quantum mechanics too, for example for the quantum harmonic oscillator.

Kepler’s third law

For a circular orbit, school-level physics is enough to derive Kepler’s third law. Equating the centripetal force with gravity,

\[\begin{aligned} m r \omega^2 &= \frac{GMm}{r^2} \\ \omega^2 &= \frac{GM}{r^3} \\ \left(\frac{2 \pi}{T} \right)^2 &=\frac{GM}{r^3} \\ T^2&=\frac{4\pi^2}{GM}\,r^3. \end{aligned}\]

This is useful in practice because orbital periods are much easier to measure than distances.

Example: geostationary satellites

A geostationary satellite completes one revolution per day, so \(T \approx 24 \text{ h} = 86\,400\text{ s}\). Rearranging, \(r^3=\frac{GMT^2}{4\pi^2}\), which gives \(r \approx 4.22 \times 10^7\text{ m}\): the orbit has a radius of about \(42\,200\text{ km}\) measured from the centre of the Earth. (Strictly, one should use the sidereal day, \(86\,164\text{ s}\), which changes the answer by about 0.1%.)

Its orbital speed is \(v= r \omega \approx 3070\text{ m s}^{-1}\).

A less serious example

Let’s estimate Peter Griffin’s mass from the clip. With \(T \approx 2.75\text{ s}\) and \(r \approx 1\text{ m}\), we get \(M \approx 7.8 \times 10^{10}\text{ kg}\).

General elliptical orbits

The argument above only works for circles. Here is the general statement.

Proposition. For every elliptical orbit, \(T^2 \propto a^3\), where \(a\) is the semi-major axis.

Proof. Start from Kepler’s second law: the radius vector sweeps out area at a constant rate,

\[\dot{A}=\frac{d A}{d t}=\frac{r(r \, d \theta) / 2}{d t}=\frac{1}{2} r^2 \dot{\theta}=\frac{h}{2},\]

where \(h=r^2\dot\theta\) is the (constant) angular momentum per unit mass. Using the area of an ellipse, the period is

\[T=\frac{A}{\dot{A}}=\frac{\pi a b}{h / 2}=\frac{2 \pi a b}{h}.\]

Let \(l\) be the semi-latus rectum (shown below)

Image: Academic Kids encyclopedia.

and \(e\) the eccentricity. Then

\[a=\frac{l}{1-e^2}, \quad b=a \sqrt{1-e^2}=\frac{l}{\sqrt{1-e^2}}.\]

For an inverse-square force \(F = -k/r^2\) with \(k = GMm\), the orbit equation gives \(l=\frac{mh^2}{k}\), so

\[T=\frac{2 \pi a b}{h}=\frac{2 \pi}{h} a^2 \sqrt{1-e^2} \quad \Longrightarrow \quad T^2=\frac{(2 \pi)^2 a^3}{h^2}\, l=(2 \pi)^2 a^3\, \frac{m}{k}.\]

Hence \(T^2 / a^3=4\pi^2 / GM\), in agreement with the circular case. \(\blacksquare\)

Exercises

• Check that Kepler’s third law holds approximately for every planet in the solar system. (Secondary school) Then fit a best-fit line by minimising the residual sum of squares. (Undergraduate, year 1–2)
• State Kepler’s first and second laws. (Final year of secondary school / undergraduate year 1)

Credits

This section closely follows Prof. Arjun Berera’s notes for Introductory Dynamics (PHYS08052) at the University of Edinburgh.

The virial theorem

The virial theorem relates the time-averaged kinetic energy of a bound system of particles to its time-averaged potential energy. For gravitating bodies it states that

\[\langle V \rangle = -2 \langle K \rangle,\]

where \(K\) is the total kinetic energy and \(V\) the total gravitational potential energy.

The theorem does not say that \(\lvert V\rvert = 2K\) at every instant; it is a statement about time averages. That is what makes it surprising: the bodies in a system can follow very different orbits, some nearly circular and some highly eccentric, yet the averages always satisfy this relation, provided the system remains bound and the forces are conservative.

Intuition from a circular orbit

Consider a small star of mass \(m\) in a circular orbit of radius \(r\) around a much heavier star of mass \(M\). Its kinetic energy is

\[\begin{aligned} K&=\frac{1}{2}mv^2 = \frac{1}{2}r\left(m\frac{v^2}{r}\right) = \frac{1}{2}r\,\frac{GMm}{r^2} = \frac{GMm}{2r}, \end{aligned}\]

while its potential energy is \(V=-\frac{GMm}{r}\). So in this case \(V = -2K\) at every instant, not just on average.

Application: the Jeans mass

The virial theorem tells us when a gas cloud collapses under its own gravity. The gravitational binding energy of a uniform cloud of mass \(M\) and radius \(R\) is

\[E_{\text{grav}}=-\frac{3}{5} \frac{G M^2}{R}.\]

Each particle at temperature \(T\) has mean kinetic energy \(\frac{3}{2}kT\). If the cloud consists of \(N\) particles of mass \(m\), so that \(M=Nm\), its total thermal kinetic energy is

\[E_{\text{thermal}}=\frac{3 k T}{2} \frac{M}{m}.\]

The cloud collapses if gravity wins against the virial balance,

\[-E_{\text{grav}} > 2 E_{\text{thermal}},\]

where the factor of 2 comes from the virial theorem. This requires

\[\frac{M}{R} > \frac{5 k T}{G m}.\]

Writing \(R\) in terms of the density \(\rho\), \(R = \left(\frac{3M}{4\pi\rho}\right)^{1/3}\), turns this into a minimum mass for collapse, the Jeans mass:

\[M_J = \left(\frac{5kT}{Gm}\right)^{3/2}\left(\frac{3}{4\pi\rho}\right)^{1/2}.\]

Credits

This section closely follows the course notes for Introductory Astrophysics at the University of Edinburgh.

Changes on 25 October 2026

What was wrong.

• The introduction called the virial theorem “Virial’s theorem”, as if it were named after a person. The name comes from the Latin vis, force.
• The letter \(T\) stood both for the orbital period and for the kinetic energy.

What changed. Kinetic energy is now \(K\). The geostationary example notes that the sidereal day (86 164 s) is the more accurate period, a 0.1 % effect, and the collapse criterion is carried through to the Jeans mass. Previously titled “Two Laws Related to Planetary Revolution”.