## Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson

# Chapter 15 - Exercises and Problems: 71

5,683 meters

#### Work Step by Step

a) Since pressure and density are directly proportional, we use the result from the previous problem to find: $\rho = \rho_0e^{\frac{-h}{h_0}}$ b) We know that the mass at a certain height in the atmosphere is given by: $M = \int_0^h dm$ We need this integral to be in terms of h so that we can find the desired height, so we obtain: $dm = \rho dV = (R_e+h)^2 e^{\frac{-h}{h_0}} \rho_0 dh$ (Note, above we substitute the equation from part a in for the value of $\rho$). Taking the integral of this quantity and plugging in $\frac{1}{2}M$ for the value of M on the left side of the equation, we obtain: $\frac{1}{2}=|1-e^{\frac{-h}{h_0}}|$ $h=|h_0ln(\frac{1}{2})|$ $|8,200ln(\frac{1}{2})|=\fbox{5,683 meters}$

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