Let's determine the mass of the Earth's atmosphere! Assume that the temperature is constant! (See the article published in the September issue.)





Translating the text as requested, while retaining the original line breaks and formatting.
In the September 1978 issue, from the barometric height formula, using the Boyle-Mariotte law $\left[\left(p / p_{0}\right)=\left(\varrho / \varrho_{0}\right)\right]$, we can directly obtain an analytical relationship for the air density measurable at height $h$:

$$
\varrho=\varrho_{0} \cdot e^{-\frac{\varrho_{0} g}{p_{0}} h}
$$

where the zero subscript denotes values measured at the Earth's surface. Since the density of the atmosphere decreases exponentially with increasing height, it is practical to divide the atmosphere into spherical shells around the Earth's center for the determination of the total mass. The thickness $\Delta h$ of the spherical shells should be chosen small enough so that the air density within them can be considered constant. In this case, the spherical shell can be treated as a cylindrical segment with a base area of $4 \pi(R+h)^{2}$ and a height of $\Delta h$, so its volume is $\Delta V=4 \pi(R+h)^{2} \Delta h$ (where $R$ is the radius of the Earth). The mass of air in it is $\Delta m=\varrho \Delta V$, and the total mass of the atmosphere is $m=\Sigma \Delta m$, where the summation should be carried out to such a height where the $\varrho / \varrho_{0}$ value becomes sufficiently small. According to equation (1), it is easily calculable that at a height of $36 \mathrm{~km}$, the air density is 1% of its surface value, while at $54 \mathrm{~km}$, it is only 0.1% of the surface value.

The calculation and summation of the partial sums $\Delta m$ is a rather tedious process: choosing $\Delta h$ to be $500 \mathrm{~m}$ and the upper boundary of the atmosphere to be $54 \mathrm{~km}$, we would need to calculate 108 partial masses. However, the task can also be solved analytically with the desired accuracy. Let $H$ denote the height of the upper boundary of the atmosphere. Then the limit of the above sum, if we refine the division of height $H$ beyond any boundary, is the following integral:

$$
m=\int_{0}^{H} \varrho_{0} e^{-\frac{\varrho_{0} g}{p_{0}} h} \cdot 4 \pi(R+h)^{2} \mathrm{~d} h
$$

The height $H$ is a tiny value compared to the radius of the Earth $(R=6400 \mathrm{~km})$. Thus, the gravitational acceleration $g$ can be considered constant at different heights in the atmosphere. With this simplification, the integral in equation (2) can be easily calculated $\left(a=\frac{\varrho_{0} g}{p_{0}}\right)$:

$$
m=-\frac{4 \pi p_{0}}{g}\left[\left\{(h+R)^{2}+\frac{2}{a}\left(\frac{1}{a}+h+R\right)\right\} e^{-a h}\right]_{0}^{H}
$$

which, considering the magnitude relationships discussed above, can be approximately calculated as:

$$
m=\frac{4 \pi R^{2} p_{0}}{g}
$$

This relationship can also be directly derived from the consideration that the weight $m g$ of the atmosphere, which is located very close to the Earth's surface, creates a pressure of $p_{0}$ on the $4 \pi R^{2}$ area of the Earth's surface. This consideration does not even use the explicit relationship for the decrease in air density (1) written above, but only that $g$ does not change significantly up to the upper boundary of the atmosphere.

Numerically: $m=5 \cdot 10^{18} \mathrm{~kg}$.