2
Start with the original equation. Let's first write the Hydrostatic equation:
$$\frac{\partial p}{\partial z}=-\rho g$$
So let's prove that $$-g=c_p\theta\frac{\partial \pi}{\partial z}$$
If we use the product rule, we observe $$-g=c_p(\frac{\partial \theta \pi}{\partial z}-\pi\frac{\partial \theta}{\partial z})$$
Since $\pi=\frac{T}{\theta}$, we can ...
1
After looking back into this I found my error to lie in the long equation
$\dfrac{\partial \pi}{\partial z}=\left(\frac{1}{p_0}\right)^{\frac{R}{c_p}}\dfrac{\partial}{\partial z}\exp\left(\frac{R}{c_p}\log(p)\right)=\left(\frac{1}{p_0}\right)^{\frac{R}{c_p}}* \frac{T}{\theta}*{\frac{R}{c_p}}*\frac{1}{p}*\dfrac{\partial p}{\partial z}$.
I was wrong, ...
Only top voted, non community-wiki answers of a minimum length are eligible
