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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.