Chapter 10, Problem 74GP

To determine

To calculate: The total mass of the earth’s atmosphere.

Answer to Problem 74GP

The total mass of the earth’s atmosphere is $5.29\times {10}^{18}\text{kg}$ .

Explanation of Solution

Given:

The value of atmospheric pressure is $P=1.013\times {10}^5\raisebox{1ex}{$\text{N}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{2}}$}\right.$

Formula used:

The earth’s surface area is $4\pi R^2$ . The force pushing inward over this area amounts to,

  $F=P_{o}A=P_o\left(4\pi R^2\right)$

And,

The pressure is the force per unit area, which is given as,

  $P=\frac{F}{A}$

Where, P is the pressure and F is the force

Calculation:

The air pressure due to the atmospheric pressure, with the area equal to the earth’s surface area, it is given as,

  $P=\frac{F}{A}=mg\Rightarrow m=\frac{PA}{g}$

Now, substitute the values in the formula, then the mass is,

  $\begin{array}{l}m=\frac{PA}{g}=\frac{1.013\times {10}^5\raisebox{1ex}{$\text{N}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{2}}$}\right.\times 4\pi r_{\text{earth}}^{\text{2}}}{9.80\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{${\text{sec}}^{\text{2}}$}\right.}\\ =\frac{1.013\times {10}^5\raisebox{1ex}{$\text{N}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{2}}$}\right.\times 4\times 3.14\times {\left(6.38\times {10}^6\text{m}\right)}^2}{9.80\raisebox{1ex}{$\text{m}$}\!\left/ \!\raisebox{-1ex}{${\text{sec}}^{\text{2}}$}\right.}\\ =5.29\times {10}^{18}\text{kg}\end{array}$

Conclusion:

The total mass of the earth’s atmosphere is found to be $5.29\times {10}^{18}\text{kg}$ .