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Use conservation of energy for an isolated spherical perturbation of fixed mass $M$, with negligible initial random motion and no significant surface-pressure term. For a homologous density profile its Newtonian gravitational potential energy is $U=-\alpha G{M}^2/R$, with the same positive structure coefficient $\alpha$ at both stages; a uniform sphere has $\alpha =3/5$.

At spherical-collapse turnaround, coherent radial motion momentarily vanishes, so $E=U_{\max }=-\alpha G{M}^2/R_{\max }$. In the final virialized state, the virial theorem gives $2T_{\mathrm{vir}}+U_{\mathrm{vir}}=0$, hence $E=T_{\mathrm{vir}}+U_{\mathrm{vir}}=U_{\mathrm{vir}}/2$. ThereforeThis is the virial radius from turnaround energy in the spherical-collapse model. It is an idealized virial-radius estimate rather than the radius of the formal pressureless point collapse. If the profile coefficient changes, the same argument instead gives $R_{\mathrm{vir}}/R_{\max }={\alpha }_{\mathrm{vir}}/(2{\alpha }_{\max })$. Significant energy loss, mass loss or boundary work also changes the result.

An initially overdense spherical region first expands approximately with the background, but its self-gravity slows it more strongly. It reaches turnaround of spherical collapse at its maximum radius, recollapses, and forms a bound approximately stationary object when infall energy is redistributed into random motions. An exactly cold, uniform pressureless solution would instead collapse to zero radius; virialization replaces that formal endpoint.

For fixed mass $M$ and no cosmological constant, integrate $\ddot{R}=-GM/R^2$ for a bound shell. Its specific energy can be written $\epsilon =-GM/(2A)$, and the cycloidal parametrization isSubstitution in ${\dot{R}}^2/2-GM/R=\epsilon$ verifies this parametrization. At $\theta =\pi$, $R_{\mathrm{ta}}=2A$ and $t_{\mathrm{ta}}=\pi B$; formal collapse occurs at $\theta =2\pi$, so $t_{\mathrm{coll}}=2t_{\mathrm{ta}}$.

For a uniform sphere, $W=-3G{M}^2/(5R)$. At turnaround the coherent kinetic energy vanishes, so $E=W_{\mathrm{ta}}$. At the final equilibrium, the virial theorem gives $2T_{\mathrm{vir}}+W_{\mathrm{vir}}=0$, whence $E=W_{\mathrm{vir}}/2$. Conservation of energy therefore impliesThis virial radius from turnaround energy assumes conserved mass and energy, negligible surface-pressure/tidal terms, and the same gravitational profile coefficient. The sketch's settling segment is qualitative; the cycloid is the exact pre-virialization cold model, not a dissipative relaxation solution.

To compare with the cosmological density, use $\bar{\rho }(t)=1/(6\pi G{t}^2)$ in an Einstein-de Sitter universe. The turnaround sphere hasSince the background density falls by four between $t_{\mathrm{ta}}$ and $t_{\mathrm{coll}}$, the Einstein-de Sitter virial overdensity isFinally $\bar{\rho }(z)={\rho }_0(1+z{)}^3$, so assigning the nominal collapse redshift givesThe density ratio eight compares the same object's physical densities at turnaround and virialization; $18{\pi }^2$ compares its final density with the background at the later collapse epoch.

The regional mass density is $3H_0^2{\Omega }_m/(8\pi G{a}^3)$, while the Einstein-de Sitter universe has ${\rho }_{\mathrm{bg}}=1/(6\pi G{t}^2)$. Substituting the parametric spherical collapse from part (a) yieldsAt spherical-collapse turnaround, $\theta =\pi$, and henceFor an isolated uniform sphere, the Newtonian gravitational potential energy is $U=-3G{M}^2/(5R)$. At spherical-collapse turnaround the coherent kinetic energy is zero, so conservation of energy fixes $E=U_m$. At virial equilibrium, the virial theorem gives $2T_v+U_v=0$, hence $E=U_v/2$. Equating these energies proves the virial radius from turnaround energy:This uses the same mass and gravitational structure coefficient before and after collapse and neglects boundary pressure, mass loss and external energy exchange.

If virial equilibrium is assigned to $2t_m$, the sphere's density is eight times its turnaround density, whereas the background mass density has decreased by four. ConsequentlyThe often quoted value $18{\pi }^2$ is a density ratio; the density contrast defined as $(\rho -{\rho }_{\mathrm{bg}})/{\rho }_{\mathrm{bg}}$ is smaller by one. This finite virialized estimate and the infinite cold-collapse contrast in part (b) describe different physical endpoints.

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 $t_{\mathrm{coll}}=2t_{\mathrm{ta}}$.