If a bound system of particles is in steady (time-averaged) equilibrium, the virial theorem relates its total kinetic energy \(K\) and the force field acting on the particles. For inverse–square gravity it reduces to a remarkably simple constraint between kinetic and potential energies.

We start from the scalar virial \(G \equiv \sum_i \mathbf{r}_i \!\cdot\! \mathbf{p}_i\). Differentiating twice and time-averaging over many dynamical times for a system whose size does not grow without bound (\(\langle \ddot{G}\rangle = 0\)) gives $$ \left\langle \sum_i \mathbf{r}_i \!\cdot\! \mathbf{F}_i \right\rangle + 2\langle K\rangle = 0. $$

\begin{equation} \boxed{\,2\langle K\rangle + \left\langle \sum_i \mathbf{r}_i \!\cdot\! \mathbf{F}_i \right\rangle = 0\,} \tag*{} \end{equation}

For Newtonian gravity between pairs \(i,j\), each term \(\mathbf{r}_i\!\cdot\!\mathbf{F}_{ij}= -G m_i m_j/r_{ij}\). Summing over all pairs yields the gravitational potential energy $$ \Omega \equiv -\sum_{i<j}\frac{G m_i m_j}{r_{ij}}. $$ Hence, for self-gravitating systems,

\begin{equation} \boxed{\,2\langle K\rangle + \langle \Omega\rangle = 0\,} \tag*{} \end{equation}

—a result that holds for any bound, steady system interacting via \(1/r^2\) gravity (angles denote time averages).

Example (circular orbit): a satellite of mass \(m\) around a mass \(M\) at radius \(r\) has \(v^2=GM/r\). Then $$ E_k=\tfrac12 m v^2=\tfrac12 \frac{G M m}{r},\qquad E_p=-\frac{G M m}{r}, $$ so \begin{equation} \boxed{\,E_k = -\tfrac12 E_p\,} \tag*{} \end{equation} in agreement with \(2K+\Omega=0\).

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Derivation from hydrostatic equilibrium (stellar virial theorem).

For a spherical star in hydrostatic balance, $$ \frac{dP}{dr}=-\rho(r)\frac{G M(r)}{r^2}. $$ Multiply by the volume element \(4\pi r^3\,dr\) and integrate from \(0\) to \(R\): $$ \int_0^R 4\pi r^3 \frac{dP}{dr}\,dr = -\int_0^R 4\pi r^3 \rho(r)\frac{G M(r)}{r^2}\,dr . $$ Integrate the left side by parts: $$ \int_0^R 4\pi r^3 dP =\Big[4\pi r^3 P\Big]_0^R - \int_0^R 12\pi r^2 P\,dr \approx -3\!\int_0^R 4\pi r^2 P\,dr = -3\!\int P\,dV , $$ where the surface term \((4\pi R^3 P(R))\) vanishes for negligible external pressure.

Rewrite the right side using \(dM=4\pi r^2\rho\,dr\): $$ -\int_0^R 4\pi r^3 \rho\frac{G M}{r^2}\,dr = -\int_0^R \frac{G M(r)}{r}\,dM(r) = \Omega , $$ because \( \Omega = -\int_0^R \frac{G M(r)}{r}\,dM(r) \) is the gravitational binding energy of a sphere.

Equating both sides gives the pressure–gravity identity \begin{equation} \boxed{\,3\!\int P\,dV + \Omega = 0\,} \tag*{} \end{equation}

For an ideal gas, the internal (thermal) kinetic energy is $$ K = \tfrac{3}{2}\!\int P\,dV . $$ Insert this in the previous box: \begin{equation} \boxed{\,2K + \Omega = 0\,} \tag*{} \end{equation} — the virial theorem recovered directly from hydrostatic equilibrium.

The virial theorem ties global energetics to local balance:

• In any bound, steady, self-gravitating system, kinetic energy is half the magnitude of the (negative) potential energy.
• From hydrostatic equilibrium, the identity \(3\int P dV + \Omega=0\) shows that pressure support exactly balances gravity in aggregate; with an ideal gas this becomes \(2K+\Omega=0\).
• Heating a star (increasing \(K\)) makes \(\Omega\) more negative; the star expands or contracts until the virial relation is restored—central to stellar stability and pre-main-sequence contraction.

1. Starting from \( \frac{dP}{dr}=-\rho GM(r)/r^2 \), reproduce the step \( 3\int P\,dV + \Omega=0 \) (show the integration by parts and the \(dM\) substitution).
2. Use \(2K+\Omega=0\) to explain why removing energy from a self-gravitating gas can make it hotter (negative heat capacity).
3. For a cluster of \(N\) equal-mass stars with 1D velocity dispersion \(\sigma\), show that \( M \approx \frac{\eta\, \sigma^2 R}{G} \) (state a reasonable \(\eta\) from geometry) and discuss observational uses of this virial mass.