 Solution

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.