The equations of hydrostatic equilibrium can be obtained from the equations of fluid dynamics (2.1)-(2.3) assuming $v_i=0$⁶¹⁶¹Actually this assumption is a little bit to stringent. For a spherical symmetric system it is enough, that the average radial component of velocity $⟨v_r⟩=0$.. This yields

	${\displaystyle \frac{\partial }{\partial t}}\rho$	$=0,$		(G.1)
	${\displaystyle \frac{\partial p}{\partial r_i}}$	$=\rho g_i.$		(G.2)

The gravitational force for a sphere with a mass profile $M(r)$ is

$g_i=-{\displaystyle \frac{\partial \mathrm{\Phi }}{\partial r_i}}=-G{\displaystyle \frac{M(r)}{r^2}},$

(G.3)

so the equation of hydrostatic equilibrium for such a configuration is

${\displaystyle \frac{1}{\rho }}{\displaystyle \frac{\partial p}{\partial r}}$

$=-G{\displaystyle \frac{M(r)}{r^2}}.$

(G.4)

Substituting the ideal gas equation $p=\rho R_{s}T$ on the left hand side leads to

${\displaystyle \frac{1}{\rho }}{\displaystyle \frac{\partial p}{\partial r}}={\displaystyle \frac{R_s}{\rho }}\left(\rho {\displaystyle \frac{\partial T}{\partial r}}+T{\displaystyle \frac{\partial \rho }{\partial r}}\right)=R_{s}T\left({\displaystyle \frac{1}{T}}{\displaystyle \frac{\partial T}{\partial r}}+{\displaystyle \frac{1}{\rho }}{\displaystyle \frac{\partial \rho }{\partial r}}\right)=R_{s}T\left({\displaystyle \frac{\partial \ln T}{\partial r}}+{\displaystyle \frac{\partial \ln \rho }{\partial r}}\right).$

(G.5)

Plugging the last result into the equation for hydrostatic equilibrium (G.4), and solving for $M(r)$ gives an useful form of the hydrostatic equilibrium equation

$M(r)$

$=-{\displaystyle \frac{R_{s}Tr}{G}}\left(r{\displaystyle \frac{\partial \ln T}{\partial r}}+r{\displaystyle \frac{\partial \ln \rho }{\partial r}}\right)=-{\displaystyle \frac{R_{s}Tr}{G}}\left({\displaystyle \frac{\partial \ln T}{\partial \ln r}}+{\displaystyle \frac{\partial \ln \rho }{\partial \ln r}}\right).$

(G.6)

If we add a turbulent pressure $p_t$ to the ideal gas equation we get for the total pressure

$p=p_{th}+p_t(l)=\rho R_{s}T+{\displaystyle \frac{1}{3}}\rho q^2(l).$

(G.7)

If we substitute this into the equation for hydrostatic equilibrium (G.4), we get an additional term due to the turbulent pressure

${\displaystyle \frac{\partial p_t(l)}{\partial r}}={\displaystyle \frac{1}{3}}\rho q^2(l)\left({\displaystyle \frac{\partial \ln q^2(l)}{\partial r}}+{\displaystyle \frac{\partial \ln \rho }{\partial r}}\right).$

(G.8)

Therefore the total gravitational mass within the radius $r$ assuming hydrostatic equilibrium including a turbulent pressure associated with a length scale $l$ is

$M(r,l)$

$=-{\displaystyle \frac{r}{G}}\left[R_s{T}_g\left({\displaystyle \frac{\partial \ln T_g}{\partial \ln r}}+{\displaystyle \frac{\partial \ln {\rho }_g}{\partial \ln r}}\right)+{\displaystyle \frac{q^2(l)}{3}}\left({\displaystyle \frac{\partial \ln q^2(l)}{\partial \ln r}}+{\displaystyle \frac{\partial \ln \rho }{\partial \ln r}}\right)\right].$

(G.9)