# How are the constants for the NFW profile determined?

I'm trying to understand the NFW profile and how it causes a flat rotation curve. $$\rho(r) = \frac{\rho_0}{\frac{r}{r_s}\left(1+\frac{r}{r_s}\right)^2}$$

I think I've got why it causes a flat rotation curve. The density seems to be inversely proportional to $r^3$ when $r >> r_s$ leading to a linearly increasing mass in proportion to radius since volume is proportional to $r^3$.

However I realized I can't really prove my statement above as I don't really understand what $r_s$ and $\rho_0$ are. Thus I was hoping to see a numerical example to get my head around the relationships.

While looking for one I came across this answer by Kyle Oman applying the NFW profile to calculate the dark matter density in the Solar System and was wondering where he got his $r_s$ and $\rho_0$ values or how one calculates them themselves.

Why does it matter what their values are? $r_s$ is some characteristic scale length of the system where it switches from a $r^{-1}$ dependence to a $r^{-3}$ dependence and $\rho_0$ is a normalisation constant that ensures that the total mass within some radius is correct.

The values of these constants are obtained by fitting the function to an observed density profile or modelling the rotation curve of a galaxy.

If $$\rho(r) = \frac{\rho_0}{\frac{r}{r_s}(1+\frac{r}{r_s})^2}$$

Then the mass of a shell of thickness $\Delta r$ at radius $r$ is $$\Delta M = \frac{\rho_0}{\frac{r}{r_s}(1+\frac{r}{r_s})^2}\ 4\pi r^2 \Delta r\ .$$

When $r \ll r_s$ $$\Delta M \sim 4\pi r_s r \rho_0\ \Delta r$$ and by integrating shells, the total mass inside radius $r$ increases as $r^2$.

When $r\gg r_s$ $$\Delta M \sim 4\pi \frac{r_s^{3}}{r} \rho_0\ \Delta r$$ and the total mass within $r$ increases as $\ln r$.

Typical values found for the Milky Way dark halo are $r_h \simeq 12$ kpc and $\rho_0 \simeq 10^{-2}$ $M_{\odot}/{\rm pc}^{3}$. See for example Sofue (2012) who does exactly the calculation that Kyle Oman did in the Physics SE question that you referred to.