 Solution

Take the initially expanding branch ${\dot{r}}_i=H_i{r}_i$; the printed squared velocity alone also permits an initially contracting branch, which would not reach a future maximum radius. Multiply the equation of motion by $\dot{r}$ and integrate. The conserved outer-shell specific orbital energy isThe initial mass follows from the background critical density:ConsequentlyIf $E$ means the energy of a test shell of mass $m$, $E=mϵ$. If it means the total energy of the entire uniform, homologously moving perturbation, direct integration instead gives $K=3M{\dot{r}}^2/10$, $U=-3G{M}^2/(5r)$ andThe common convention of writing the first integral as $E$ is therefore an energy per unit shell mass, not the total energy of the sphere.

At spherical-collapse turnaround, $\dot{r}=0$, so $ϵ=-GM/r_m$. HenceAny positive initial density contrast binds this pressureless perturbation in the stated Einstein-de Sitter universe.