Problem — The mass of the atmosphere

Problem

The mass of the atmosphere. Atmospheric pressure at sea level is $p_0\approx 10^5\;\text{Pa}$ and the Earth’s radius is $R\approx 6{,}4\times 10^6\;\text{m}$. (a) Argue why $p_0$ is exactly the weight of all the air above you per unit area. (b) Estimate the total mass of the atmosphere. (c) Compare with the mass of the Earth ($\sim 6\times 10^{24}\;\text{kg}$) and discuss the result.

Solution

Step 1 — Meaning of pressure (a). Consider a vertical air column of cross-section $A$ extending from the ground to the edge of the atmosphere. That column is in equilibrium: its base pushes on the ground with a force equal to its own weight $W = m_{\text{col}}\,g$. The pressure at the ground is force over area,

$p_0 = \frac{W}{A} = \frac{m_{\text{col}}\,g}{A}$

that is, exactly the weight of the overlying air per unit area. This holds at every point of the Earth’s surface.

Step 2 — From pressure to total mass (b). Summing over the whole Earth’s surface $S = 4\pi R^2$, the total weight of the atmosphere is $W_{\text{tot}} = p_0\, S$, and so the mass is

$M = \frac{W_{\text{tot}}}{g} = \frac{p_0\, S}{g} = \frac{p_0\,\cdot 4\pi R^2}{g}$

Step 3 — Numerical calculation. With $p_0 = 10^5\;\text{Pa}$, $R = 6{,}4\cdot 10^6\;\text{m}$, $g = 9{,}81\;\text{m/s}^2$:

$S = 4\pi R^2 = 4\pi\,(6{,}4\cdot 10^6)^2 = 5{,}1\cdot 10^{14}\;\text{m}^2$

$M = \frac{10^5\cdot 5{,}1\cdot 10^{14}}{9{,}81} = 5{,}2\cdot 10^{18}\;\text{kg}$

$\ev{M = \dfrac{p_0\,4\pi R^2}{g} \approx 5\cdot 10^{18}\;\text{kg}}$

Step 4 — Comparison with the Earth (c). Comparing with the Earth’s mass $M_{\oplus}\approx 6\cdot 10^{24}\;\text{kg}$:

$\frac{M}{M_{\oplus}} \approx \frac{5\cdot 10^{18}}{6\cdot 10^{24}} \approx 10^{-6}$

The atmosphere thus weighs about one millionth of the Earth’s mass: an extremely thin film compared with the mass of the solid planet, yet enough to exert at ground level the pressure of $10^5\;\text{Pa}$ (about $10\;\text{tonnes}$ per square metre).

Correction (textbook key) $\sim 10^{15}\;\text{kg}$: it is wrong by about three orders of magnitude. The correct calculation gives $M = p_0\,4\pi R^2/g \approx 5\cdot 10^{18}\;\text{kg}$ (real value $\approx 5{,}1\cdot 10^{18}\;\text{kg}$). The typo $10^{15}$ should be corrected to $\sim 5\cdot 10^{18}\;\text{kg}$.

The textbook key reports

Linked notes

Topics: Fluidostatica e fluidodinamica Concepts: Pressione Skills: Stime di Fermi