= Solution
Take the initially expanding branch $\dot r_i=H_ir_i$; 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> is
$$
\epsilon=\frac12\dot r^2-\frac{GM}{r}.
$$
The initial <mass> follows from the background <critical density>:
$$
M=\frac{4\pi}{3}r_i^3\rho_{m,i}(1+\delta_i)
=\frac{H_i^2r_i^3}{2G}(1+\delta_i).
$$
Consequently
$$
\boxed{\epsilon=-\frac12H_i^2r_i^2\delta_i<0.}
$$
If $E$ means the <energy> of a test shell of <mass> $m$, $E=m\epsilon$. If it means the total <energy> of the entire uniform, homologously moving perturbation, direct integration instead gives $K=3M\dot r^2/10$, $U=-3GM^2/(5r)$ and
$$
\boxed{E_{\rm sphere}=\frac35M\epsilon=-\frac3{10}MH_i^2r_i^2\delta_i.}
$$
The common convention of writing the first integral as $E$ is therefore an <energy> \i[per unit shell mass], not the total <energy> of the sphere.
At <spherical-collapse turnaround>, $\dot r=0$, so $\epsilon=-GM/r_m$. Hence
$$
\boxed{r_m=\frac{GM}{|\epsilon|}=r_i\frac{1+\delta_i}{\delta_i}.}
$$
Any positive initial <density contrast> binds this pressureless perturbation in the stated <Einstein-de Sitter universe>.
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