Neutrons stars are known to have a radius of about 10 km to 15 km. Where that size comes from ? How to theoretically derive that number, using Newton's theory of gravity ?
I know the magnetic flux conservation : $\Phi = \pi R_0^2 \, B_0 \approx \pi R^2 \, B$, but this requires that we know the initial and final magnetic field at the surface of the star, and the star's size before the supernova. So this isn't satisfying.
Conservation of angular momentum alone isn't telling the radius neither : $S = \frac{2}{5} \, M R_0^2 \, \omega_0 = \frac{2}{5} \, M R^2 \, \omega$, since this requires that we know the star size and angular velocity before the supernova.
Using Newton's theory of gravitation and conservation of energy (or another method ?), how can we derive the theoretical size of NS ? The only input number that I could accept in the derivation is the NS angular velocity $\omega$ (and mass $M \approx 1.44 \, M_{\odot}$), since this can be found from conservation of angular momentum and simple models of supernova (if we already know the NS radius !).
Currently, the only rough argument that I know is the following : assuming an uniform sphere rotating at its maximal value to support gravity, we should have a balance relation like this : \begin{equation}\tag{1} F_{\text{centripetal}} \approx M \, \omega^2 R = F_{\text{grav}} \approx \frac{G M^2}{R^2}. \end{equation} Isolating $R$ gives \begin{equation}\tag{2} R \approx \Big( \frac{G M}{\omega^2} \Big)^{\frac{1}{3}}. \end{equation} Inserting mass $M \approx 1.44 M_{\odot}$ and period $T \approx 1 \text{ms}$ give something interesting : \begin{equation}\tag{3} R_{\text{NS}} \approx 16.9 \text{km}. \end{equation} I strongly suspect this could be improved or made more rigorous (even if it's less precise), using the conservation of energy.
EDIT : Another argument comes from the density. A neutron star has a density comparable to a nucleus or a neutron, so \begin{equation}\tag{4} \rho = \frac{3 M}{4 \pi R^3} \approx \rho_{\text{neutron}} = \frac{3 m_N}{4 \pi r_N^3}. \end{equation} This gives the star's radius (using $M \approx 1.44 \, M_{\odot}$ and $r_N \approx 10^{-15} \, \mathrm{m}$) : \begin{equation}\tag{5} R \approx \Big( \frac{M}{m_N} \Big)^{\frac{1}{3}} \, r_N \approx 12 \mathrm{km}. \end{equation}