= Solution
\b[Supersonic communication.] A point source moving at speed $V>c_s$ outruns its small <acoustic waves>. After a time $t$, the source has moved $Vt$, while a disturbance emitted at its old position has radius $c_st$. The envelope of these spherical wavefronts is the <Mach cone>, with half-angle $\mu$ satisfying
$$
\boxed{\sin\mu=\frac{c_s}{V}=\frac1{\mathcal M},\qquad
\mathcal M=\frac{V}{c_s}>1.}
$$
Information is restricted to the downstream cone, so upstream gas cannot smoothly anticipate a supersonic obstacle. Finite compressions can also steepen because their characteristics converge. Such disturbances form a <shock wave>, across which the fluid variables change rapidly. A perfectly uniform <supersonic flow> need not contain a shock: the cone expresses the communication constraint, while a disturbance or compression supplies the reason for a discontinuity.
\b[Conserved fluxes.] Work in the <shock frame>, take velocities normal to the front, and integrate the conservation laws through a thin steady <control volume> of unit area. <Mass conservation> gives constant flux $j=\rho u$, so
$$
\boxed{\rho_2u_2=\rho_1u_1=j.}
$$
The normal <momentum> flux is the advected part $\rho u^2$ plus the pressure stress $p$. With negligible body-force impulse across the thin front,
$$
\boxed{p_2+\rho_2u_2^2=p_1+\rho_1u_1^2.}
$$
For an <ideal gas> with constant <adiabatic index> $\gamma$, the <specific internal energy> is $e=p/[(\gamma-1)\rho]$. The energy flux is
$$
j\left(e+\frac12u^2\right)+pu
=j\left(h_{\rm sp}+\frac12u^2\right),
\qquad h_{\rm sp}=e+\frac p\rho=\frac{\gamma p}{(\gamma-1)\rho},
$$
where $h_{\rm sp}$ is the <specific enthalpy>. No radiation or other heat flux escapes in this adiabatic shock model. Conservation of the energy flux, divided by the nonzero mass flux, gives
$$
\boxed{\frac{\gamma p_2}{(\gamma-1)\rho_2}+\frac12u_2^2
=\frac{\gamma p_1}{(\gamma-1)\rho_1}+\frac12u_1^2.}
$$
These are the <Rankine-Hugoniot conditions for a perfect gas>. The shock is adiabatic but irreversible: its <entropy> increases, so an isentropic relation cannot be imposed across the discontinuity.
\b[Strong-shock compression.] Put $\chi=\rho_2/\rho_1=u_1/u_2$ and neglect $p_1$ and the upstream enthalpy compared with $\rho_1u_1^2$ and $u_1^2$. Momentum conservation gives $p_2=\rho_1u_1^2(1-1/\chi)$. Substitute this into energy conservation:
$$
\frac12=\frac{\gamma}{\gamma-1}\frac{\chi-1}{\chi^2}+\frac1{2\chi^2}.
$$
Thus
$$
\chi^2-1=\frac{2\gamma}{\gamma-1}(\chi-1),
\qquad
(\chi-1)\left(\chi+1-\frac{2\gamma}{\gamma-1}\right)=0.
$$
The root $\chi=1$ describes no shock. The compressive branch of the <Strong-shock Rankine-Hugoniot conditions> is
$$
\boxed{\frac{\rho_2}{\rho_1}=\frac{u_1}{u_2}
=\frac{\gamma+1}{\gamma-1}.}
$$
It also gives $p_2=2\rho_1u_1^2/(\gamma+1)$. A <monatomic gas> has $\gamma=5/3$, hence $\chi=4$.
\b[The finite slab in its initial rest frame.] Denote the shock's lab speed by $V_s$. The upstream gas is initially stationary, so $u_1=V_s$ and the downstream shock-frame speed is $u_2=V_s/4$. Transforming to the initial slab frame gives
$$
\boxed{v_2=V_s-u_2=\frac34V_s,\qquad L_2=\frac L4.}
$$
This is <planar shock compression of a finite slab>. In the uniform post-shock approximation, breakout occurs at $t_b=L/V_s$. The original front face has then moved $(3V_s/4)t_b=3L/4$, while the shock reaches the originally stationary far face at $L$, leaving thickness $L/4$. Equivalently, the fixed mass per area $\rho L$ becomes $4\rho L_2$. The result assumes a maintained approximately constant shock speed and no important boundary rarefaction during the crossing; an unconfined finite cloud need not remain uniformly compressed.
\b[Escape estimate and the sound-speed convention.] Let $M_c$ be the slab mass, $W>0$ its binding-energy magnitude in the star's field, and $E_{\rm th}$ its initial <internal energy>. Treat the potential as nearly uniform across the slab. For the bulk-ballistic escape estimate, require
$$
E_{\rm bulk}=\frac12M_cv_2^2=\frac9{32}M_cV_s^2\gtrsim W,
\qquad
E_{\rm th}=\frac{3}{2}\frac{M_cp}{\rho}.
$$
The <Mach-number convention in shock unbinding estimates> matters here. With the usual upstream <adiabatic sound speed>, $c_{\rm ad}^2=\gamma p/\rho=5p/(3\rho)$, define $\mathcal M_{\rm ad}=V_s/c_{\rm ad}$. Then
$$
\frac{E_{\rm bulk}}{E_{\rm th}}=\frac5{16}\mathcal M_{\rm ad}^2,
\qquad
\boxed{\mathcal M_{\rm ad}^2\gtrsim\frac{16}{5}\frac{W}{E_{\rm th}}.}
$$
If instead the shock speed is normalized by the <isothermal sound speed>, $c_T^2=p/\rho$, then $\mathcal M_T^2=(5/3)\mathcal M_{\rm ad}^2$ and
$$
\boxed{\mathcal M_T^2\gtrsim\frac{16}{3}\frac{W}{E_{\rm th}}
\simeq5.3\,\frac{W}{E_{\rm th}}.}
$$
The original PDF's “around five” is consistent with the latter convention or an order-of-magnitude sound-speed estimate. It is not the exact coefficient obtained with the conventional adiabatic Mach number and the stated bulk kinetic-energy criterion. Here $W=-E_{\rm grav}$ if the potential is zero at infinity; using negative gravitational energy directly on the right would be incorrect. Subsequent conversion of shock-generated internal energy to expansion, external pressure, or radiative losses would change the physical escape threshold, so this numerical factor is not universal.
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