= Solution
In the <spherical-collapse model>, a bound overdense region initially expands with the background, but its additional <Newtonian gravity> slows that expansion. Its outer shell reaches a maximum <turnaround radius>, where its radial velocity is zero, and then falls inward. A perfectly spherical pressureless solution formally collapses to zero radius. In a realistic collisionless system, deviations from that idealization, <shell crossing> and <violent relaxation> convert coherent infall into random motions, producing a roughly stationary bound object.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2012/iii/paper-60-collapse.png]
{title=Spherical expansion, turnaround, formal collapse and a schematic virialized radius}
Let the <Newtonian gravitational potential energy> of the homologous sphere be $U=-aGM^2/R$, where $a=3/5$ for a uniform sphere. At <spherical-collapse turnaround>, the <kinetic energy> of the bulk motion vanishes, so $E=U_{\rm ta}$. For the settled object, the <virial theorem> gives $2K_{\rm vir}+U_{\rm vir}=0$ and hence $E=U_{\rm vir}/2$. If energy and mass are conserved and the structure coefficient is unchanged,
$$
-\frac{aGM^2}{R_{\rm ta}}=-\frac{aGM^2}{2R_{\rm vir}},\qquad \boxed{R_{\rm vir}=\frac12R_{\rm ta}.}
$$
This is the <virial radius from turnaround energy>. It assumes negligible external pressure, mass loss and energy exchange. If the final structure coefficient differs, $R_{\rm vir}=(a_{\rm vir}/2a_{\rm ta})R_{\rm ta}$ instead. \b[Half the turnaround radius is an idealized equilibrium size, not the radius at which each collisionless particle stops moving.] The plotted settling time is schematic; the usual collapse-time convention is $t_{\rm coll}=2t_{\rm ta}$.
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