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The planar Mestel disk has $F(E)\propto e^{-E/v_0^2}$, giving independent centered Gaussian distributions for $v_R,v_{\varphi }$ with velocity dispersions ${\sigma }_R={\sigma }_{\varphi }=v_0$. Its streaming velocity is zero. A maximally prograde stellar distribution instead has mean $\sqrt{2/\pi }{v}_0$ and azimuthal variance $v_0^2(1-2/\pi )$, while all second raw moments remain unchanged.

Velocity space is the space of velocity vectors at a fixed spatial position. Integrating a galactic distribution function over velocity space produces mass density; its normalized moments determine mean velocity and velocity dispersion.

Motion can weigh a system whose mass is otherwise inaccessible. For a self-gravitating isolated system in virial equilibrium, let $K$ and $U$ denote equilibrium averages and use its gravitational radius $R_g=-G_N{M}^2/U$ and a mass-weighted mean squared speed relative to its centre of mass. Then $K=M⟨v^2⟩/2$ gives $M=R_g⟨v^2⟩/G_N$. If random velocities are isotropic and mean streaming has been removed, $⟨v^2⟩=3{\sigma }_{\mathrm{los}}^2$, where ${\sigma }_{\mathrm{los}}$ is the one-dimensional line-of-sight velocity dispersion. Observations sample many objects at one epoch; replacing a time average by that sample requires a representative, approximately stationary population, beyond the trajectory identity itself.^[ref]

A practical measurement route uses an optical telescope to locate members and a spectrograph to measure Doppler shifts of identified spectral lines. Their shifts relative to the system's mean give radial velocities; angular separations plus a distance estimate constrain its size. Subtracting instrumental broadening and measurement variance, testing membership, and looking for coherent rotation or subgroups help control the inference. The projected luminous size need not equal $R_g$, and luminous tracers need not have the same distribution as all gravitating mass. Unknown velocity-anisotropy parameters, interlopers, mergers, external tidal forces and appreciable boundary stresses can bias a naive estimate. A large velocity dispersion alone does not establish virial equilibrium.

For an isolated equilibrium of equal-mass points, the virial theorem gives $⟨v^2⟩=-W/M$. A particle's instantaneous escape velocity satisfies $v_{\mathrm{esc}}^2=-2\Phi$ with zero potential at infinity. Since $W=\frac{1}{2}{\sum }_{i}m{\Phi }_i$, averaging gives $⟨v_{\mathrm{esc}}^2⟩=-4W/M$. These are three-dimensional mean-square speeds, not one-component velocity dispersions.