I'm trying to estimate the local density of dark matter at a given radius, $r=R_0=8\text{kpc}$. I also have values for the core radius, $a=5\text{kpc}$, and circular velocity of the Sun, $v_{\text{circ}}=220\text{kms}^{-1}$. I start with the dark matter density profile


from which I can obtain a relation for the mass interior to the radius r


$\rho_0=M(r)/(4\pi a^2(r-a\text{tan}^{-1}(r/a)+const))$

To get the local dark matter density at the given radius I assume I need to substitute the above relation into the first equation.

However, I'm also told that the dark halo contributes half of the total mass inside the Sun's orbit but I don't know how to calculate the mass interior to the Sun (probably missing something very simple here). Any help is greatly appreciated.

  • $\begingroup$ You probably should have done a definite integral to go from $\rho(r)$ to $M$, in which case, you'd not have that $const$ term, right? $\endgroup$ – Kyle Kanos May 13 '18 at 17:06
  • $\begingroup$ I'd also assume that you'd have a density profile for the matter content (e.g., Plummer or M-N models), otherwise you'd not be able to address the "50% of matter" aspect. $\endgroup$ – Kyle Kanos May 13 '18 at 17:08
  • $\begingroup$ I think it must be that the dark matter density profile given is for a spherical distribution, meaning I can assume that the halo is spherically symmetric. So, $v\text{circ}$ is related to $M(R)$ by a simple relation (I'll add an answer). $\endgroup$ – Brad May 14 '18 at 7:48

The density profile given in the question is for a spherical halo, meaning that we can assume the distribution of mass in the halo to be spherically symmetric - allowing us to use the spherically symmetric gravitational acceleration


and the centripetal acceleration for a circular orbit


Equating these and rearranging we obtain


which can be simply substituted in to the equation for $\rho_0$, which can then be substituted into the density profile given. Using the values given above I get:

$\rho(R_0)=1.85493\times10^{-21}\text{kgm}^{-3}\\\ =0.02726M_{sun}\text{pc}^{-3}$


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