Atmospheric physics 5: Barometric formula

By combining the hydrostatic equation with the general gas equation we can eliminate the density $\rho$ $\begin{aligned} & & P &= \rho \frac{RT}{M} \\ \Rightarrow & & \rho &= \frac{M}{RT} P \\ \Rightarrow & & \frac{dP}{dz} &= -\rho g = -\frac{M g}{R T} P \end{aligned}$ Since the temperature is a constant $T = T_0$ , the pressure $P$ is the only unknown, which depends on the alititude $z$ . Thus, we haven a differential equation for the pressure, which can be solved by separation of variables: $\begin{aligned} & & \frac{dP}{dz} &= -\frac{M g}{R T} P \\ \Rightarrow & & \frac{dP}{P} &= -\frac{M g}{R T} dz \\ \Rightarrow & & \int_{P_0}^{P(z)} \frac{dP}{P} &= -\frac{M g}{R T} \int_{0}^{z} dz \\ \Rightarrow & & \log \frac{P(z)}{P_0} &= -\frac{M g}{R T} z \\ \Rightarrow & & P(z) &= P_0 \exp \left[ -\frac{M g}{R T} z \right] \end{aligned}$ Thus, the solution is an exponential function $P = P_0 e^{-z/H}$ with the scale height $H = RT/M g$ .