underlying gravitational potential. The density and temperature of the gas are related to the enclosed mass by the condition that the net acceleration vanishes. Thus, the hydrostatic equilibrium equation reads $$\frac{1}{\rho(x)}{\frac{\ddP(x)}{\ddx}} = g(x)
\label{eqHydro_usual}$$ where $P$ denotes the total (gas, magnetic, turbulent, cosmic-ray, and radiation) pressure, and $x$ is the relevant spatial coordinate. In spherical coordinates, $x \equiv r$, and the gravitational acceleration $$g(r) = - \frac{GM(r)}{r^2} = - \frac{4\pi G}{r^2} \int_0^r x^2 \rho(x)\ {{\rm d}}x
\label{eqGrav_sphere}$$ is determined by the total mass $M(r)$ contained within radius $r$. For a symmetric plane-parallel atmosphere, the relevant coordinate is the distance to the mid plane, $z$, and the gravitational acceleration is $$g(z) = - 2\pi G\,\Sigma(z) = - 4\pi G \int_0^z \rho(x)\ {{\rm d}}x
\label{eqGrav_plane}$$ In either case, the solution to equation (\[eqHydro\_usual\]) is $$P(x) = P_0 + \Delta P_{\rm grav}(x) \equiv P_0 + \int_0^x \rho(u)\ g(u)\ {{\rm d}}u
\label{eqHydro}$$ where $P_0$ is the central pressure. There are many situations of astrophysical interest where $\Delta P_{\rm grav}(x) \ll P_0$, and the hydrostatic equilibrium condition results in an almost isobaric atmosphere. However, depending on the total mass and the characteristic temperature of the gas, the gravitational field may be strong enough to have 