= Solution
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 $\varepsilon=-GM/(2A)$, and the cycloidal parametrization is
$$
R=A(1-\cos\theta),\qquad t=B(\theta-\sin\theta),\qquad A^3=GMB^2.
$$
Substitution in $\dot R^2/2-GM/R=\varepsilon$ verifies this parametrization. At $\theta=\pi$, $R_{\rm ta}=2A$ and $t_{\rm ta}=\pi B$; formal collapse occurs at $\theta=2\pi$, so $t_{\rm coll}=2t_{\rm ta}$.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2003/iii/paper-66-spherical-collapse.png]
{title=Expansion, turnaround and idealized virialization of a spherical overdensity, compared with formal cold collapse}
For a uniform sphere, $W=-3GM^2/(5R)$. At turnaround the coherent kinetic energy vanishes, so $E=W_{\rm ta}$. At the final equilibrium, the <virial theorem> gives $2T_{\rm vir}+W_{\rm vir}=0$, whence $E=W_{\rm vir}/2$. <Conservation of energy> therefore implies
$$
\boxed{R_{\rm vir}=\frac{R_{\rm ta}}2,\qquad\rho_{\rm vir}=8\rho_{\rm ta}.}
$$
This <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 $\bar\rho(t)=1/(6\pi Gt^2)$ in an <Einstein-de Sitter universe>. The turnaround sphere has
$$
\rho_{\rm ta}=\frac{3M}{4\pi(2A)^3}=\frac3{32\pi GB^2},
\qquad
\frac{\rho_{\rm ta}}{\bar\rho(t_{\rm ta})}=\frac{9\pi^2}{16}.
$$
Since the background density falls by four between $t_{\rm ta}$ and $t_{\rm coll}$, the <Einstein-de Sitter virial overdensity> is
$$
\boxed{\frac{\rho_{\rm vir}}{\bar\rho(t_{\rm coll})}=32\frac{9\pi^2}{16}=18\pi^2.}
$$
Finally $\bar\rho(z)=\rho_0(1+z)^3$, so assigning the nominal collapse redshift gives
$$
\boxed{\rho_{\rm vir}(z_{\rm coll})=18\pi^2\rho_0(1+z_{\rm coll})^3.}
$$
The 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.
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