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

The shell equation is conservative because its radial acceleration is the negative derivative of the specific potential energyMultiplying the equation of motion by $\dot{r}$ givesHenceThis is energy per unit shell mass. The total energy of a uniform sphere requires integrating the energies of its shells.

For a sphere of physical radius $R$, the shell theorem gives $M(r)=M(r/R{)}^3$ and $dM=3M{r}^{2}dr/R^3$. Assemble the sphere shell by shell to obtain its Newtonian gravitational potential energy:The cosmological constant contributes a prescribed one-body potential $-\Lambda r^2/6$, rather than a pair interaction, so there is no additional factor $1/2$:Evaluating at the turnaround radius givesAs in the question, $\Lambda$ here includes the factor $c^2$ that would multiply the geometrical cosmological constant in SI units.

For the final equilibrium apply the scalar virial theorem. The gravitational virial is $W_G$, whereas the cosmological constant force contributesThus, neglecting a surface-pressure term,At fixed $R$, a positive cosmological constant reduces the kinetic energy needed for virial equilibrium because its outward acceleration partially offsets self-gravity.

For the uniform-sphere collapse with a cosmological constant, assume a uniform final sphere of the same mass, with no energy loss or escaping matter. PutThe sphere has no bulk kinetic energy at turnaround, so $E_{\mathrm{ta}}=-A-B$. The final virial theorem givesSince ${\rho }_{\mathrm{ta}}=3M/(4\pi r_{\mathrm{ta}}^3)$,Conservation of energy therefore givesSelect the collapsing branch continuous from $x=1/2$ at $\eta =0$, rather than an unrelated root of the cubic equation.

For $|\eta |\ll 1$, the root lies close to $1/2$. Linearize $x^3$ there:Substitution into the cubic equation gives $(\eta /2-2)x+1-\eta /2\simeq 0$, whenceThe rational expression is an approximation, not an exact solution of the cubic equation. In fact both it and the exact collapsing root agree through order ${\eta }^2$; their first difference is of order ${\eta }^3$.

With zero cosmological constant, spherical collapse gives $r_f=r_{\mathrm{ta}}/2$. For small positive $\Lambda$, the same sphere with the same turnaround radius reaches a smaller final radius. This does not mean that repulsion promotes collapse: as the sphere contracts, $W_{\Lambda }$ becomes less negative, so some released gravitational binding energy pays the energetic cost of moving inward against the repulsive force. Combined with the modified virial theorem, energy conservation requires a more compact endpoint. Positive $\Lambda$ can prevent turnaround altogether; an inward acceleration at $r_{\mathrm{ta}}$ requiresThe perturbative radius formula applies well inside this limit, for spheres that actually turn around. A small negative cosmological constant instead gives $r_f/r_{\mathrm{ta}}>1/2$ at fixed turnaround data.

The shell feels only radial gravity and the radial force due to the cosmological constant, so its torque vanishes and its specific angular momentum $L=r^2\dot{\theta }$ is conserved. Multiplyingby $\dot{r}$ and integrating gives the conserved specific orbital energy

For a uniform sphere, assembling concentric shells gives its gravitational potential energyThe cosmological-constant potential per unit mass is $-\Lambda r^2/6$. Since $⟨r^2⟩=3r_{\mathrm{ta}}^2/5$ in a uniform sphere,

The scalar virial theorem weights a potential homogeneous of degree $n$ by $-nW$. Gravity has degree $-1$ and the $\Lambda$ potential degree $2$, so the final state obeysAt turnaround $T=0$, while the virial relation gives $E_f=W_{G,f}/2+2W_{\Lambda ,f}$. Conservation of energy, together with $M=4\pi {\rho }_{\mathrm{ta}}{r}_{\mathrm{ta}}^3/3$, then gives, for $x=r_f/r_{\mathrm{ta}}$ and $\eta =\Lambda /(4\pi G{\rho }_{\mathrm{ta}})$,The root connected continuously to the $\Lambda =0$ solution has $x=1/2-\eta /8+O({\eta }^2)$, equivalentlyto first order in $\eta$. With $\Lambda =0$, virialization occurs at half the turnaround radius. At fixed turnaround state, positive $\Lambda$ makes this equilibrium root slightly smaller because its repulsive quadratic potential enters both energy conservation and the virial relation; sufficiently strong repulsion instead prevents a bound virialized state. Negative $\Lambda$ shifts the root in the opposite direction.

An Einstein-de Sitter universe has $a\propto t^{2/3}$, $H=2/(3t)$, and homogeneous density $\rho =1/(6\pi G{t}^2)$. HenceBefore the perturbation appreciably changes the motion, every shell follows the Hubble flow,

By Newton's shell theorem, only matter currently inside a shell contributes to its radial acceleration. Labelling matter by initial radius givesBefore shell crossing, the enclosed mass of a given shell is $M_i$. Its initial specific energy isAt the turnaround radius, $E_i=-G{M}_i/r_{\ast }$, soThe radial Kepler solution from the Big Bang to apocenter givesAfter turnaround a shell falls inward. Shell crossing makes its enclosed mass time dependent; shells then oscillate through the center, form caustics near successive apocenters, and phase mix into a halo.

For self-similar secondary infall, writeSubstitution in the shell equation givesUsing $G{t}_{\ast }^2/r_{\ast }^3={\pi }^2/(8M_i)$,

Now imposeThe case $ϵ=0$ is a scale-independent fractional overdensity, for which all shells turn around together in the ideal model. The case $ϵ=1$ has constant excess mass $\delta M_i=M_0$, corresponding to a central point-mass seed outside its core.

Since $r_{\ast }/r_i=(M_i/M_0{)}^{ϵ}$, while $r_i^3=(9/2)G{t}_i^2{M}_i$, the turnaround formulas giveThe shell turning at time $t$ therefore hasFor a shell with $\tau =t/t_{\ast }(M_i)$,Consequently its autonomous similarity equation isThe exponent $2/(3ϵ)$ follows directly from the preceding mass ratio; the printed $2ϵ/3$ in the final displayed equation is inconsistent with the supplied initial condition and preceding requested result. The corrected nonlinear equation is integrated numerically after turnaround.