= Strong-shock accretion temperature
{title2=$k_{\mathrm B}T_{\mathrm d}=3\mu v_{\mathrm u}^2/16$}
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>.
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