[Math] Calculate the mass of the earth’s atmosphere give the density of air.

Question

Assume the desity of air $\rho$ is given by

$\rho(r)=\rho_0$$e^{-(r-R_0)/h_0}$ for $r\ge R_0$

where $r$ is the distance from the centre of the earth, $R_0$ is the radius of the earth in meters, $\rho_0=1.2kg/m^3$ and $h_0=10^4m$

Assuming the atmosphere extends to infinity, calculate the mass of the portion of the earth's atmosphere north of the equator and south of $30^\circ$N latitude.

How do I even start this problem? Do I need to convert it into spherical coordinates? But then what limits do I use for the integration?

Answer 1

You don't have to do integrals! Divide atmospheric pressure A = 101.3 kPa by g = 9.8 m/s² to give the mass per unit area (kg/m²). Multiply this by the area of the earth and you're done. (Assumptions: g is a constant over the height of the atmosphere; g independent of latitude; neglect the mass of the air displaced by the volume of the land about sea level.)

ADDENDUM: Also you can use the fact that 1 atmosphere = 760 Torr = 15 lb-force/in² to estimate the mass of the atmosphere per unit area as 0.76m ρ_Hg or 15 lb/in² (ρ_Hg = density of mercury = 13.53 metric tonnes/m³).