## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

The atmospheric pressure at the earth's surface would be equal to $\rho~g~h$, where $h$ is the total thickness of the atmosphere. Therefore; $\rho~g~h = P_0$ $h = \frac{P_0}{\rho~g}$ $h = \frac{1.013\times 10^5~N/m^2}{(1.3~kg/m^3)(9.80~m/s^2)}$ $h = 7950~m = 7.95~km$ The thickness of the atmosphere would be 7.95 km