Chapter 20, Problem 56A

To determine

The height of top of atmosphere.

Answer to Problem 56A

The top of atmosphere would be $h=8.33\text{km}$ .

Explanation of Solution

Given:

Surface area, $A=1{\text{m}}^2$

Weight of atmosphere, $F_g={10}^5\text{N}$

Density, $\rho =1.2{\text{kg/m}}^3$

Formula Used:

The pressure can be given as,

  $P=\rho gh$

Where, $\rho$ is density, $g$ is gravitational acceleration and $h$ is height.

The other formula of pressure is, $P=\frac{F}{A}$

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

Calculation:

The pressure is,

  $P=\frac{F_g}{A}=\frac{{10}^5\text{N}}{1{\text{m}}^2}={10}^5{\text{N/m}}^2$

Taking $g=10{\text{m/s}}^2$

Now, putting the values in formula $P=\rho gh$

  $\begin{array}{l}\Rightarrow {10}^5{\text{N/m}}^2=\left(1.2{\text{kg/m}}^3\right)\left(10{\text{m/s}}^2\right)h\\ \Rightarrow h=\frac{{10}^5}{1.2\times 10}\text{m}=8333.33\text{m}\\ \Rightarrow h=8.33\text{km}\end{array}$

Conclusion:

Therefore, the top of atmosphere would be $h=8.33\text{km}$ .