Isothermal Atmosphere

The vertical thickness of the atmosphere is only a few tens of kilometers, and is, therefore, much less than the radius of the Earth, which is about $6000\,{\rm km}$ . Consequently, it is a good approximation to treat the atmosphere as a relatively thin layer, covering the surface of the Earth, in which the pressure and density are only functions of altitude above ground level, $z$ , and the gravitational potential energy per unit mass takes the form ${\mit\Psi}=g\,z$ , where $g$ is the acceleration due to gravity at $z=0$ . It follows from Equation (13.1) that

$$\frac{dp}{dz} = -\rho\,g.$$ (13.5)

Now, in an isothermal atmosphere, in which the temperature, $T$ , is assumed not to vary with height, the ideal gas equation of state (1.84) yields [cf., Equation (13.3)]

$$\frac{p}{\rho} = \frac{R\,T}{M}.$$ (13.6)

The previous two equations can be combined to give

$$\frac{dp}{dz} = -\frac{g\,M}{R\,T}\,p.$$ (13.7)

Hence, we obtain

$$p(z) = p_0\,\exp(-z/H),$$ (13.8)

where $p_0\simeq 10^5\,{\rm N\,m^{-2}}$ is atmospheric pressure at ground level, and

$$H = \frac{R\,T}{g\,M}$$ (13.9)

is known as the isothermal scale height of the atmosphere. Using the values $T=273\,{\rm K}$ ( $0^\circ \,{\rm C}$ ), $M = 29\times 10^{-3}\,{\rm kg}$ , and $g=9.8\,{\rm m\,s^{-2}}$ , which are typical of the Earth's atmosphere (at ground level), as well as $R= 8.315\,{\rm J\,mol^{-1}\,K^{-1}}$ , we find that $H= 7.99\,{\rm km}$ . Equations (13.6) and (13.8) yield

$$\rho(z)= \rho_0\,\exp(-z/H),$$ (13.10)

where $\rho_0= p_0/(g\,H)$ is the atmospheric mass density at $z=0$ . According to Equations (13.8) and (13.10), in an isothermal atmosphere, the pressure and density both decrease exponentially with increasing altitude, falling to $37\%$ of their values at ground level when $z=H$ , and to only $5\%$ of these values when $z=3\,H$ .