# Atmospheric pressure as a function of altitude

Consider a planet with a thin atmosphere. The planet is assumed to have radius $$r_0$$ and mass M.

The gravitational field outside the planet is given by: $$\vec g = - \frac {GM} {r^2} \vec e_r$$ where G is Newton's gravitational constant. The atmosphere near the surface of the planet can be considered as a stationary linearly compressible fluid, which means that a relationship $$ρ = ρ_0 + α(p - p_0)$$ applies between the density of the fluid $$ρ$$ and the pressure $$p$$. Here $$ρ_0$$ and $$p_0$$ are the density of the atmosphere respective pressure at the planetary surface.

i) Set up a differential equation that describes how the pressure $$p$$ varies with the distance r from the center of the planet. Hint: You can base your reasoning on static equilibrium and Archimedes' principle.

ii)Calculate how the atmospheric pressure p and the density of the atmosphere $$ρ$$ depend on r.

Assume that pressure and density only depend on r.

I am not quite sure how to start on i) since equilibrium in the radial direction is given by $$-m\frac {GM} {r^2} + ρVg = 0$$ and $$ρV = m$$ the previous expression is just zero and that doesn't give me anything

• Perhaps you would be better off posting this on physics.stackexchange.com – Filthyscrub Sep 14 '20 at 15:36
• $\frac{dp}{dx}=\rho g$ – Aditya Dwivedi Sep 14 '20 at 15:36
• Answer to Filthyscrub: Hi, this problem came up in a math course so i don't think you need more than basic knowledge in physics that is why i posted it here. Answer to Aditya Dwivedi: I can understand why 𝑑𝑝/𝑑𝑥=𝜌𝑔 would hold in an ideal fluid but why does it hold in this case when we are talking about pressure in the atmosphere? Also i don't understand how g comes into the picture then – kpopgirl Sep 14 '20 at 19:05

Imagine a box of small height hanging in the air. The basis of the box has area A and height dr. The pressure at the bottom of the box is $$p$$. The pressure at the top of the box is $$p+dp$$. Consider the equilibrium of the box. You have force $$pA$$ pushing box upwards and force $$(p+dp)A$$ pushing box downwards. You have also the weight of the air in the box which is $$\rho g Adr$$ (product of mass contained in the box and acceleration of gravity). Equation of equilibrium is:

$$pA-\rho g A dr-(p+dp)A=0$$

$$dp=-\rho g dr=-(\rho_0+\alpha (p-p_0))\frac {GM}{r^2}dr$$

$$\frac{dp}{\rho_0+\alpha (p-p_0)}=-\frac {GM}{r^2}dr$$

$$\int_{p_0}^p\frac{dp}{\rho_0+\alpha (p-p_0)}=\int_{r_0}^r-\frac {GM}{r^2}dr$$

$$\frac{1}{\alpha}\ln[\rho_0+\alpha(p-p_0)]|_{p_0}^p=\frac{GM}r|_{r_0}^r$$

$$\frac{1}{\alpha}\ln[\rho_0+\alpha(p-p_0)]-\frac{1}{\alpha}\ln\rho_0=GM(\frac 1r - \frac 1{r_0})$$

$$\ln[1+\frac{\alpha(p-p_0)}{\rho_0}]=\alpha GM(\frac 1r - \frac 1{r_0})$$

$$\frac{\alpha(p-p_0)}{\rho_0}=e^{\alpha GM(\frac 1r - \frac 1{r_0})}-1$$

$$p=\frac{\rho_0}{\alpha}[e^{\alpha GM(\frac 1r - \frac 1{r_0})}-1]+p_0$$

$$p(r)=p_0-\frac{\rho_0}{\alpha}[1-e^{-\alpha GM(\frac 1{r_0}-\frac 1r)}]$$

In reality, atmosphere of Earth is modelled a little bit differently. Starting point is temperature of air as a function of height, equation which is made empirically from measurements, not relation between pressure and density of air. But as you said, this is a problem from mathematics and it has very little to do with some real physics of atmosphere.