= Spherical turnaround from an unperturbed Hubble velocity
{title2=$r_{\rm ta}=Mr_0/\delta M,\quad\tau\simeq\pi GM/(2G\delta M/r_0)^{3/2}$}
In an <Einstein-de Sitter universe>, give an initially spherical, pressureless region total mass $M$ and excess mass $0<\delta M\ll M$, while holding its initial expansion velocity at the unperturbed <Hubble flow> value at physical radius $r_0$. Before <shell crossing>, the outer shell has $\dot r_0^2=2G(M-\delta M)/r_0$ and conserved <specific orbital energy> $-G\delta M/r_0$. The <shell theorem> and <conservation of energy> give $\dot r^2=2GM/r-2G\delta M/r_0$, so its <cosmological turnaround> radius is $Mr_0/\delta M$. With $A=2GM$, $B=2G\delta M/r_0$ and $\epsilon=\delta M/M$, the elapsed <cosmic time> is exactly
$$
\tau=\frac{A}{B^{3/2}}\left[\frac\pi2-\arcsin\sqrt\epsilon+\sqrt{\epsilon(1-\epsilon)}\right].
$$
Replacing the square bracket by $\pi/2$ gives the displayed approximation, with relative overestimate $4\epsilon^{3/2}/(3\pi)+O(\epsilon^{5/2})$. Holding the initial expansion velocity fixed is an essential condition: a pure growing-mode <linear cosmological density perturbation> has an initial peculiar velocity as well as an excess mass, changing the binding energy at the same order.
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