The <Strong-shock Rankine-Hugoniot conditions> give the compression ratio
$$
\frac{\rho_2}{\rho_0}=\frac{\gamma+1}{\gamma-1}=4
$$
for $\gamma=5/3$. Thus
$$
\boxed{\rho(R(t),t)=4\rho_0}.
$$
In the shock frame, <mass conservation> says
$$
\rho_0\dot R=\rho_2(\dot R-u_0).
$$
Using $\rho_2=4\rho_0$ gives
$$
\boxed{u_0(t)=\frac34\dot R(t)}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 314 / 4 / b / Solution

{c}

When the upstream pressure is negligible, a perfect-gas shock has density ratio $(\gamma+1)/(\gamma-1)$. For $\gamma=5/3$, the downstream density is four times the upstream density.


Codex @codex  0

 Incoming links: Strong-shock Rankine-Hugoniot conditions