= Solution
Interpret the terminal infall <speed> as the free-fall <speed> from a radius much larger than the <neutron star> radius $R_*=12\,\mathrm{km}$. In the Newtonian estimate,
$$
v_u\simeq\sqrt{\frac{2GM_*}{R_*}}
=2.00\times10^8\,\mathrm{m\,s^{-1}}\simeq0.666c,
\qquad M_*=1.8M_\odot.
$$
For fully ionized <hydrogen> with <electrons> and <protons> at a common <temperature>, there are two particles per atom and $\mu=(m_p+m_e)/2\simeq m_p/2$. The result of <1/e> then yields
$$
\boxed{k_BT_d=\frac{3GM_*m_p}{16R_*}\simeq39\,\mathrm{MeV},\qquad
T_d\simeq4.5\times10^{11}\,\mathrm K.}
$$
This is the thermal <energy> scale $k_BT_d$ per particle. In the same classical <monatomic gas> approximation, the mean random <kinetic energy> per particle is $3k_BT_d/2\simeq59\,\mathrm{MeV}$. Before the <shock wave>, the bulk <kinetic energy> per <proton> is $m_pv_u^2/2=GM_*m_p/R_*\simeq208\,\mathrm{MeV}$. Distinguishing these quantities avoids an ambiguity in “particle energy”. If only the ions support the downstream <pressure> because <electrons> have not equilibrated, $\mu\simeq m_p$ instead gives $k_BT_i\simeq78\,\mathrm{MeV}$; that is a different, two-temperature assumption.
These are estimates within the question's Newtonian, $\gamma=5/3$ model. Its self-consistency is poor here: $2GM_*/(R_*c^2)\simeq0.443$, and $39\,\mathrm{MeV}$ is much larger than the <electron> rest <energy> $0.511\,\mathrm{MeV}$. Relativistic <electrons>, electron–positron <pair production>, radiation, unequal species <temperatures>, and <general relativity> can alter the <equation of state> and <shock wave> structure. In particular, the numerical $39\,\mathrm{MeV}$ must not be presented as a precision surface <temperature> prediction.
Back to article page
