Chapter 20, Problem 57A

Average atmospheric pressure at Earth’s surface is $1.01\times {10}^{5}N/m^2$. Earth’s radius is $6.37\times {10}^{6}m$. Show that the total weight of Earth’s atmosphere is about $5.15\times {10}^{19}N$.

To determine

To show: The total weight of Earth’s atmosphere is about $5.15\times {10}^{19}\text{N}$ .

Explanation of Solution

Given:

Atmospheric pressure at Earth’s surface, $P=1.01\times {10}^5{\text{N/m}}^2$

Radius of Earth, $R=6.37\times {10}^6\text{m}$

Formula Used:

The pressure is given as,

  $P=\frac{F}{A}$

Where, $F$ is force and $A$ is area in contact.

Calculation:

The total surface area in contact is,

  $\begin{array}{l}A=4\pi R^2\\ =4\pi {\left(6.37\times {10}^6\text{m}\right)}^2\\ =4\pi \times \left(6.37\times {10}^6\right){\text{m}}^2\end{array}$

So, on substituting the values

  $\begin{array}{l}P=\frac{F}{A}\\ 1.01\times {10}^5{\text{N/m}}^2=\frac{F}{4\pi \times \left(6.37\times {10}^6\right){\text{m}}^2}\\ \Rightarrow F=5.15\times {10}^{19}\text{N}\end{array}$

Conclusion:

Hence, it is proved that weight of atmosphere is $5.15\times {10}^{19}\text{N}$ .