Strong-shock accretion temperature ($k_{\mathrm{B}}{T}_{\mathrm{d}}=3\mu v_{\mathrm{u}}^2/16$)

A strong, nonradiative shock in a monatomic ideal gas gives $p_{\mathrm{d}}=3{\rho }_{\mathrm{u}}{v}_{\mathrm{u}}^2/4$ and ${\rho }_{\mathrm{d}}=4{\rho }_{\mathrm{u}}$. The ideal gas law therefore gives the temperature in the title, with $\mu$ the mean particle mass. If the upstream speed is a Newtonian free-fall speed, $k_{\mathrm{B}}{T}_{\mathrm{d}}=3\mu GM/(8R)$. Mass alone does not determine this temperature without a radius and composition; electron-ion disequilibrium and radiation can also make the emergent spectrum differ from the ideal gas temperature.