In convective equilibrium prove the following result Question: In convective equilibrium show that pressure $p$ at any height $z$ above the earth's surface is
$$p=p_0\left[ 1-\dfrac{\gamma-1}{\gamma}\left(\dfrac{1}{\rho_0^{\gamma-1}}\right)\dfrac{gz}{k}\right]^{\frac{\gamma}{\gamma-1}}$$ 
My approach: Suppose the gravity $g$ is constant. Then the pressure equation becomes
\begin{equation}
dp=-\rho g\, dz \tag{1}
\end{equation}
In adiabatic condition,
\begin{equation}
p=k \rho^{\gamma}\tag{2}
\end{equation}
This implies 
\begin{equation}
dp=k\gamma \rho^{\gamma-1}\, d\rho\tag{3}
\end{equation}
From (1) and (3),
\begin{equation}
k\gamma \rho^{\gamma-1}d\rho=-\rho g \,dz
\end{equation}
Integrating and simplifying, 
$\dfrac{p}{\rho}\left(\dfrac{\gamma}{\gamma-1}\right)=C-gz$, $c=$ constant of integration.
From $p=\rho RT$,
\begin{equation}
RT\left(\dfrac{\gamma}{\gamma-1}\right)=C-gz \tag{4}
\end{equation} 
At sea level, $z=0, T=T_0$, So we have
$$\dfrac{T}{T_0}=1-\dfrac{\gamma-1}{\gamma}\cdot \dfrac{gz}{RT_0}$$ 
How can I show the required result?
 A: As you have done, integrating $k\gamma \rho^{\gamma-1}d\rho=-\rho g \,dz$ results in 
$$\dfrac{p}{\rho}\left(\dfrac{\gamma}{\gamma-1}\right)=C-gz,$$
where $C$ is a constant of integration. Equivalently, since $p = k\rho^\gamma$ we have 
$$k\rho^{\gamma-1}\left(\dfrac{\gamma}{\gamma-1}\right)=\dfrac{p}{\rho}\left(\dfrac{\gamma}{\gamma-1}\right)=C-gz,$$
Multiplying both sides by $\frac{\gamma-1}{k\gamma}$ we get 
$$\tag{1}\rho^{\gamma-1} = \dfrac{\gamma-1}{\gamma} \frac{C}{k}- \dfrac{\gamma-1}{\gamma}\dfrac{gz}{k}$$
Since $\left.\rho\right|_{z = 0} = \rho_0$ at sea level, substitution into (1) yields
$$\rho_0^{\gamma-1} =  \dfrac{\gamma-1}{\gamma} \frac{C}{k} $$
Whence,
$$\tag{2}\rho^{\gamma-1} = \rho_0^{\gamma-1} - \dfrac{\gamma-1}{\gamma}\dfrac{gz}{k} = \rho_0^{\gamma-1} \left[1 - \dfrac{\gamma-1}{\gamma}\left(\frac{1}{\rho_0^{\gamma-1}} \right)\dfrac{gz}{k} \right]$$
Note that $p = k\rho^\gamma$ implies $\rho = \frac{p^{1/\gamma}}{k^{1/\gamma}}$ and  $\rho_0 = \frac{p_0^{1/\gamma}}{k^{1/\gamma}}$. Substituting into (2) we get 
$$\frac{p^{\frac{\gamma-1}{\gamma}}}{k^\frac{\gamma-1}{\gamma}}=   \frac{p_0^{\frac{\gamma-1}{\gamma}}}{k^\frac{\gamma-1}{\gamma}}\left[1 - \dfrac{\gamma-1}{\gamma}\left(\frac{1}{\rho_0^{\gamma-1}} \right)\dfrac{gz}{k}\right]$$ 
Multiplying both sides by $k^\frac{\gamma-1}{\gamma}$ and then raising both sides to the power $\frac{\gamma}{\gamma-1}$ yields the desired result
$$p =   p_0\left[1 - \dfrac{\gamma-1}{\gamma}\left(\frac{1}{\rho_0^{\gamma-1}} \right)\dfrac{gz}{k}\right]^{\frac{\gamma}{\gamma-1}}$$
