# Calculate the mass of the earth's atmosphere give the density of air.

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

$$\rho(r)=\rho_0e^{-(r-R_0)/h_0}$$ 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?

• There's a small typo here that Stackexchange won't let me fix. Maybe the OP can. It's not rho that's equal to 1.2KG/M**3, it's rho nought. Jun 4, 2019 at 18:48

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 ρHg or 15 lb/in2Hg = density of mercury = 13.53 metric tonnes/m3).

The problem is simplified by the fact that the density is independent of the latitude and longitude. So we can compute the total mass and the multiply this by the fraction of area in the specified region over the total area.

The total enclosed volume at radius $r$ is $V(r) = {4 \over 3} \pi r^3$, hence we have ${d V(r) \over dr} = 4 \pi r^2$ (the surface area).

The mass of air between radius $r$ and $r+\delta$ is approximately $m(r+\delta)-m(r) \approx\rho(r) {d V(r) \over dr} \delta$, and so we see that ${d m(r) \over dr} = \rho(r) {d V(r) \over dr}$, from which we obtain $m(\infty)-m(R_0) = \int_{R_0}^\infty \rho(r) {d V(r) \over dr} dr = 4 \pi \int_{R_0}^\infty \rho(r) r^2 dr$.

This gives the total mass of air. To find the portion above the specified area, we need to find the fraction of the Earth's surface area represented by the area between $0^∘$ and $30^∘N$.

The area between $0$ and $\phi$ is given by $\int_0^\phi (2 \pi R_0 \cos \alpha) R_0 d \alpha = 2 \pi R_0^2 \sin \phi$, from which we get the fraction between $0^∘$ and $30^∘N$ to be ${ \sin {\pi \over 6} \over 2} = {1 \over 4}$.

Hence the mass of air above the specified region is $\pi \int_{R_0}^\infty \rho(r) r^2 dr$.

Assuming I haven't made a mistake, this gives:

$\pi \int_{R_0}^\infty \rho(r) r^2 dr = \rho_0 \pi h_0(R_0^2+2 R_0 h_0 + 2 h_0^2)$.

The limits of iteration in spherical coordinates could be

• $r>R_0$ (the exterior of the earth)
• $0\le\theta<2\,\pi$ (all around the earth)
• =$\pi/6<\phi<\pi/2$ (between 30ºN and the equator)
• How do I incorporate $\theta$ and $\phi$ into the original equation? Feb 24, 2014 at 17:13
• The density $\rho$ is independent of $\theta$ and $\phi$. Feb 24, 2014 at 17:14
• Do I do a triple integral with respect to $dV$ in spherical coordinates? Feb 24, 2014 at 17:28
• Yes, that is what you must do. Feb 24, 2014 at 17:37
• When I integrate with respect to r first and put the limits of $R_0$ and $\infty$ then definite integral is not defined. How can I resolve this? Feb 24, 2014 at 18:00

To calculate the mass of the atmosphere above the earth's surface you do not need to know the density variation of air vertically nor the extent of the atmosphere above the earth. All one needs to know is the pressure at sea level and the value of acceleration due to gravity (g) at sea level (and an assumption that this does not vary over the depth of the atmosphere, which is about 50 km).

Use this formula P = m" g ; pressure is the weight/m^2 of the atmosphere at sea level. p= 1.01325 N/m^2; g = 9.8 m/s^2, m" = 1.0339E4 kg/m^2 Radius of earth (R) = 6.372E3 km; Surface area of earth = 4 Pi R^2 = 5.1E8 km^2 Therefore mass of atmosphere over the earth = 1.0339E4 x 5.1E14 = 5.274E18 kg

• I wholeheartedly agree with the last comment. Additionally, in $50\,Km$ the variation of $g$ is about the $0.4\%$, so it is not completely negligible (at least if we want an estimation with four significative figures). Mar 23, 2018 at 20:46
• Very good estimate! Slight improvement, knowing 30% of earth is land, averaged 840m above water, with air pressure 90% of sea level. 70% + 30%*90% = 97%. Mass of air ≈ 5.12E18 kg Aug 26, 2019 at 1:54