= Solution
Immediately upstream, free fall from rest at infinity gives $u_1^2=2GM_0/r_0$. Treat the surface shock as locally planar and stationary, with negligible upstream <pressure>. The <Strong-shock Rankine-Hugoniot conditions> for a <monatomic gas>, $\gamma=5/3$, give $\rho_2/\rho_1=4$ and $u_2=u_1/4$. <momentum conservation> then gives
$$
p_2=\rho_1u_1^2-\rho_2u_2^2=\frac34\rho_1u_1^2,\qquad\frac{p_2}{\rho_2}=\frac3{16}u_1^2.
$$
Use the source's gas-constant convention $p=\rho(R^*/\mu)T$. The immediate postshock <temperature> is
$$
\boxed{T_2=\frac\mu{R^*}\frac3{16}\frac{2GM_0}{r_0}=\frac\mu{R^*}\frac38\frac{GM_0}{r_0}}.
$$
Here adiabatic means that the jump does not lose appreciable heat to the surroundings; the shock is nevertheless irreversible and raises <entropy>. The gas still has <velocity> $u_1/4$ immediately downstream, so setting all upstream <kinetic energy> equal to downstream internal <energy> would give the wrong coefficient. Later settling, <radiative cooling> or changing ionization can alter the <temperature>; they are outside this immediate, fixed-composition perfect-gas jump.
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