# Barometric formula with variable gravity

Recently, I have wondered about what would be the atmospheric pressure as a function of altitude in a planet that only consists of gas. The equation that does just that is the barometric formula (see https://en.wikipedia.org/wiki/Barometric_formula) but the gravitational acceleration is usually considered to be constant, which is certainly not the case in this thought experiment. I've tried to solve the differential equations, but I didn't managed to make it.

The gravitational acceleration as a function of altitude $$d$$ (from the center of the planet) is: $$g(d)=\frac{4 \pi G}{d^2}\int_{0}^{d}R^2 \rho(R) \mathrm{d}R$$ Where $$\rho(R)$$ is the density of air at a distance $$R$$ from the center of the planet

The below is taken from section 3 of J. Patrick Harrington's Notes on Hydrostatic Equilibrium, which I recommend reading for more detail.

For a spherical distribution of gas where the gravitational force is due to the gas itself, the equation of hydrostatic equilibrium is

$$\frac{dP}{dr}=-\frac{GM_r\rho}{r^2}, \tag{1}$$

where $$M_r$$ is the mass of gas contained within a sphere of radius $$r$$.

The variation of $$M_r$$ with radius is given by

$$\frac{dM_r}{dr}=4\pi r^2\rho \tag{2}.$$

Thus, after differentiating (1) and use of (2) we get

$$\frac{1}{r^2}\frac{d}{dr}\left[\frac{r^2}{\rho}\frac{dP}{dr}\right]=-4\pi G\rho. \tag{3}$$

In general, $$P=P(\rho,T)$$, so we need to know the variation of $$T$$ with $$r$$ to achieve closure. If we assume that the gas is isothermal and ideal, then $$P=c_i^2\rho$$, where $$c_i$$ is the constant isothermal speed of sound, and (3) becomes

$$\frac{c_i^2}{r^2}\frac{d}{dr}\left[\frac{r^2}{\rho}\frac{d\rho}{dr}\right]=-4\pi G\rho. \tag{4}$$

In general, equation (4) does not have an analytic solution for our boundary conditions and indeed is badly behaved at $$r=0$$, necessitating a series solution from $$r=0.01$$, say, to start a numerical integration.

However, there is an analytic solution if we let $$\rho(0)\to\infty$$. In this case, called the 'singular isothermal sphere',

$$\rho(r)=\frac{c_i^2}{2\pi G}\frac{1}{r^2}.$$

There's a load of interesting discussion and further exploration of the problem in the notes I linked.