Answer
a) $P = P_{0} \times ( 1 - \frac{\beta z}{T_{0}})^{\frac{g}{R\beta}}$
b)$P = P_{0} \times \exp( \frac{-g_{0}}{R(\beta + \frac{T_{0}}{r})} \times (\frac{1}{(\frac{T_{0}}{\beta r}+1)} \times \ln(\frac{1 + \frac{z}{r}}{1 - \frac{\beta z}{T_{0}}}) + \frac{z}{r+z} ) )$
With $r = 6,370,320m$
Work Step by Step
Assuming the atmospheric to be static, we can apply the hydrostatic law:
(1) $dP = - \rho g dz$
Assuming atmospheric air to be an ideal gas,
(2) $ \rho = \frac{m}{V} = \frac{P M_{avg}}{RT} $
Where $M_{avg}$ is the weighted average of molar weights of the components of air (28.965 $\frac{kg}{kmol}$), or simplifying it with $R^{*} = \frac{R}{M_{avg}}$
(2*) $ \rho = \frac{P}{R^{*}T} $
a)
Plugging the given equation: $T = T_{0} - \beta z$, and (2*) into (1):
$dP = - \frac{P}{R^{*}(T_{0} - \beta z)} g dz$
Integrating from the groud level ($z = 0 m, P = P_{0}$):
$\int^{P}_{P_{0}} \frac{dP}{P} = \int^{z}_{0} \frac{-gdz}{R^{*}(T_{0} - \beta z)}$
$\int^{P}_{P_{0}} \frac{dP}{P} = \frac{-g}{R^{*}\beta} \int^{z}_{0} \frac{dz}{(\frac{T_{0}}{\beta} - z)}$
$\ln{\frac{P}{P_{0}}} = \frac{-g}{R^{*}\beta} \times -\ln(\frac{\frac{T_{0}}{\beta} - z}{\frac{T_{0}}{\beta}})$
$P = P_{0} \times ( 1 - \frac{\beta z}{T_{0}})^{\frac{g}{R^{*}\beta}}$
b) For simplification, the constant 6,370,320 will be referred as $r$
The given equation is : $g = \frac{g_{0}}{(1+\frac{z}{r})^{2}}$
Similarly as in a:
$dP = - \frac{P}{R^{*}(T_{0} - \beta z)} \frac{g_{0}}{(1+\frac{z}{r})^{2}} dz$
$\int^{P}_{P_{0}} \frac{dP}{P} = \int^{z}_{0} \frac{-gdz}{R^{*}(T_{0} - \beta z)}$
$\int^{P}_{P_{0}} \frac{dP}{P} = \frac{-g_{0}}{R^{*}\beta} \int^{z}_{0} \frac{dz}{(\frac{T_{0}}{\beta} - z)(1+\frac{z}{r})^{2}}$
With $\int \frac{dz}{(\frac{T_{0}}{\beta} - z)(1+\frac{z}{r})^{2}} = \frac{1}{(\frac{T_{0}}{\beta r}+1)^{2}} \times \ln(\frac{1 + \frac{z}{r}}{\frac{T_{0}}{\beta} - z}) - \frac{1}{(\frac{T_{0}}{\beta r} + 1)(1 + \frac{z}{r})}$
$\ln{\frac{P}{P_{0}}} = \frac{-g_{0}}{R^{*}\beta} \times (\frac{1}{(\frac{T_{0}}{\beta r}+1)^{2}} \times (\ln(\frac{1 + \frac{z}{r}}{\frac{T_{0}}{\beta} - z}) - \frac{1}{(\frac{T_{0}}{\beta r} + 1)(1 + \frac{z}{r})}) - (\frac{1}{(\frac{T_{0}}{\beta r}+1)^{2}} \times \ln(\frac{1}{\frac{T_{0}}{\beta}}) - \frac{1}{\frac{T_{0}}{\beta r} + 1}))$
$P = P_{0} \times \exp( \frac{-g_{0}}{R^{*}\beta} \times (\frac{1}{(\frac{T_{0}}{\beta r}+1)^{2}} \times \ln(\frac{1 + \frac{z}{r}}{1 - \frac{\beta z}{T_{0}}}) + \frac{(1+\frac{z}{r})-1}{(\frac{T_{0}}{\beta r} + 1)(1 + \frac{z}{r})} ) )$
$P = P_{0} \times \exp( \frac{-g_{0}}{R^{*}(\beta + \frac{T_{0}}{r})} \times (\frac{1}{(\frac{T_{0}}{\beta r}+1)} \times \ln(\frac{1 + \frac{z}{r}}{1 - \frac{\beta z}{T_{0}}}) + \frac{z}{r+z} ) )$
