 Solution

For a uniform sphere of radius $R$, write $M(r)=M(r/R{)}^3$ and $dM=3M{r}^{2}dr/R^3$. The self-gravitational potential energy is assembled by adding shells:The cosmological constant produces the external potential per unit mass ${\Phi }_{\Lambda }=-\Lambda r^2/6$, since $-\nabla {\Phi }_{\Lambda }=\Lambda \mathbf{r}/3$ in units $c=1$. Its energy isThere is no additional factor $1/2$ for this fixed external potential, unlike the pair-counting factor for gravitational self-energy. At the maximum radius $R=r_m$, the kinetic energy vanishes, soThis is the turnaround energy in uniform-sphere collapse with a cosmological constant. The original PDF's shell kinetic term is ${\dot{r}}^2/2$, not the $r^2/2$ produced by the TeX transcription.