Introduce the <dimensionless variables>
$$
\widetilde z=\frac zH,
\qquad
\rho=\frac\Sigma H\widetilde\rho(\widetilde z),
\qquad
p=\frac PH\widetilde p(\widetilde z).
$$
Using $P=\Sigma H^2\Omega_z^2$, vertical <hydrostatic equilibrium> becomes the parameter-free equation
$$
\boxed{\frac{d\widetilde p}{d\widetilde z}
=-\widetilde\rho\,\widetilde z},
$$
with normalizations $\int\widetilde\rho\,d\widetilde z=1$ and $\int\widetilde p\,d\widetilde z=1$. For an <isothermal atmosphere>, $p=c_s^2\rho$, and these conventions give the <Gaussian distribution>
$$
\boxed{
\widetilde\rho(\widetilde z)=\widetilde p(\widetilde z)
=\frac1{\sqrt{2\pi}}e^{-\widetilde z^2/2}}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 321 / 3 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-321/3/c/solution