Answer
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})|$
$|8,200ln(\frac{1}{2})|=\fbox{5,683 meters}$
