The swept shell has $M_s=Ar$, where $A=2f_{\rm gas}\sigma^2/G$. Its steady radial <thin-shell momentum equation> is
$$
v\frac d{dr}(Arv)=F_{\rm rad}-\frac{4f_{\rm gas}\sigma^4}{G}.
$$
Let $r_0$ be the launch radius and impose $v(r_0)=0$.

In the single-scattering regime, define
$$
\Gamma_{\rm ss}=\frac{L}{L_{\rm crit,s.s.}}>1.
$$
The force difference is constant, and integration gives
$$
\boxed{v_{\rm ss}^2(r)=2\sigma^2(\Gamma_{\rm ss}-1)
\left[1-\left(\frac{r_0}r\right)^2\right].}
$$
The shell accelerates monotonically, but continuous sweeping of $M_s\propto r$ makes the speed approach the finite maximum
$$
\boxed{v_{\rm ss,max}=v_{\rm ss}(\infty)
=\sqrt2\sigma\sqrt{\Gamma_{\rm ss}-1}.}
$$

For the optically thin ultraviolet regime, define the launch Eddington factor
$$
\Gamma_{\rm UV}=\frac{L}{L_{\rm crit,UV}(r_0)}>1.
$$
Since $F_{\rm rad}=\tau_{\rm UV}L/c\propto r^{-1}$ while the shell's gravitational force is constant, integration gives
$$
\boxed{v_{\rm UV}^2(r)=2\sigma^2\left[
2\Gamma_{\rm UV}\frac{r_0}r-1
+(1-2\Gamma_{\rm UV})\left(\frac{r_0}r\right)^2
\right].}
$$
It initially accelerates, reaches its maximum at
$$
\boxed{r_{\max}=r_0\left(2-\Gamma_{\rm UV}^{-1}\right),
\qquad
v_{\max}=\sqrt2\sigma
\frac{\Gamma_{\rm UV}-1}{\sqrt{2\Gamma_{\rm UV}-1}},}
$$
and then decelerates, formally stalling at $r=r_0(2\Gamma_{\rm UV}-1)$. The contrast is physical: single-scattering transfers the same $L/c$ to the shell at every radius, whereas an ultraviolet-thin shell intercepts a fraction proportional to its declining optical depth, so the isothermal host's gravity eventually wins.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 347 / 3 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-347/3/c/solution