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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.

Let the Newtonian gravitational potential energy of the homologous sphere be $U=-aG{M}^2/R$, where $a=3/5$ for a uniform sphere. At spherical-collapse turnaround, the kinetic energy of the bulk motion vanishes, so $E=U_{\mathrm{ta}}$. For the settled object, the virial theorem gives $2K_{\mathrm{vir}}+U_{\mathrm{vir}}=0$ and hence $E=U_{\mathrm{vir}}/2$. If energy and mass are conserved and the structure coefficient is unchanged,This is the virial radius from turnaround energy. It assumes negligible external pressure, mass loss and energy exchange. If the final structure coefficient differs, $R_{\mathrm{vir}}=(a_{\mathrm{vir}}/2a_{\mathrm{ta}})R_{\mathrm{ta}}$ instead. 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 $t_{\mathrm{coll}}=2t_{\mathrm{ta}}$.