= Solution
Let $D=u_1>0$ denote the front speed in the stationary upstream frame. In the <shock frame>, the upstream and downstream velocities are $-D$ and $-D/r$. Taking $M_1\to\infty$ in the <Rankine-Hugoniot conditions for a perfect gas> gives
$$
\boxed{\rho_2=\frac{\gamma+1}{\gamma-1}\rho_1,\qquad p_2=\frac{2}{\gamma+1}\rho_1D^2.}
$$
The downstream laboratory velocity is $D-D/r=2D/(\gamma+1)$, which distinguishes the gas speed from the front speed.
For the <planar blast-wave energy scaling> of a <self-similar blast wave>, integration of the total <energy density> over the shocked interval $0<z<Z$ gives the energy per unit area on this side:
$$
E_+=\int_0^Z\left(\frac{\rho u^2}{2}+\frac{p}{\gamma-1}\right)dz
=\rho_1Z\dot Z^2 I,\qquad
I=\int_0^1\left(\frac12fh^2+\frac{g}{\gamma-1}\right)d\eta.
$$
The <similarity solution> makes $I$ time-independent; a finite positive explosion energy requires $0<I<\infty$. Conservation of energy therefore gives $\dot Z=(E_+/(\rho_1I))^{1/2}Z^{-1/2}$ for the expanding front. Integrating from $Z(0)=0$,
$$
\frac23Z^{3/2}=\left(\frac{E_+}{\rho_1I}\right)^{1/2}t,
\qquad
\boxed{Z(t)=\left(\frac{9}{4I}\right)^{1/3}\left(\frac{E_+}{\rho_1}\right)^{1/3}t^{2/3}.}
$$
If $E$ denotes the one-sided energy, $C=(9/(4I))^{1/3}$. If the released energy $E$ feeds two symmetric fronts, $E_+=E/2$ and $C=(9/(8I))^{1/3}$. \b[In either convention the requested scaling is $Z=C(E/\rho_1)^{1/3}t^{2/3}$.] The constant depends on the <similarity profiles> and the energy convention; energy conservation determines the exponent without solving those profiles. The <Strong-shock Rankine-Hugoniot conditions> additionally fix $f(1)=(\gamma+1)/(\gamma-1)$ and $g(1)=h(1)=2/(\gamma+1)$.
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