Past exam of the astrophysical fluid dynamics course of the University of Cambridge / 2013 / Paper 1 / 1

Let $x$ measure distance down the incline and $y$ distance perpendicular to it, with the wall at $y=0$ and the free surface at $y=h$. A uniform steady falling film flow has $\mathbf{u}=u(y){\mathbf{e}}_x$. Its convective acceleration vanishes because $\mathbf{u}\cdot \nabla =u{\partial }_x$ and ${\partial }_{x}u=0$. A flat surface exposed to constant atmospheric pressure hasThus ${\partial }_{x}p=0$, and the downslope component of the Navier-Stokes equations reduces to $0=g\sin \alpha +(\eta /\rho )u^{\prime \prime }$. The printed pressure-free equation must be understood as this downslope projection: its full vector version would incorrectly omit the normal hydrostatic pressure gradient that balances $g\cos \alpha$.

The no-slip boundary condition gives $u(0)=0$. Neglecting air shear stress makes the surface shear-free, so $u^{\prime }(h)=0$. Two integrations give the uniform gravity-driven Newtonian film profileThe velocity increases up to the free surface, whereThis assumes a uniform laminar Newtonian fluid film, a stationary wall and negligible surface shear; a varying thickness or externally imposed downslope pressure gradient would alter the balance.

By reflection symmetry the vertical gravitational field of the self-gravitating incompressible slab vanishes at its mid-plane. Inside the slab, the gravitational Poisson equation therefore integrates to $d\psi /dz=4\pi G\rho z$. Treat the star as a test particle, so its own mass does not alter that field. Its vertical stellar orbit in a uniform slab obeysHence $z(t)=a\cos (\omega t)$ with $\omega =\sqrt{4\pi G\rho }$. The first mid-plane crossing occurs after one quarter of this simple harmonic motion period:Its independence of $a$ follows from the linear interior restoring force. The calculation neglects drag and changes to the slab, and it uses the infinite planar approximation rather than the field of a finite Galactic disk.

Work in the rest frame of the shock. The normal momentum flux is conserved, so $p_1+{\rho }_1{u}_1^2=p_2+{\rho }_2{u}_2^2$. In the strong-shock limit $p_1\ll {\rho }_1{u}_1^2$, while mass conservation gives ${\rho }_2{u}_2={\rho }_1{u}_1$. The given Strong-shock Rankine-Hugoniot conditions therefore yieldUsing the downstream ideal gas law $p_2/{\rho }_2=k_B{T}_2/m$ givesFor a monatomic perfect gas, $\gamma =5/3$ and the coefficient is $3/16$. This is the immediate post-shock temperature, before substantial radiative cooling; ionization, dissociation or unequal electron and ion temperatures require additional physical modeling.