= Solution
\b[Hydrostatic equilibrium with self-gravity.] A resting <fluid> obeys $\nabla p=-\rho\nabla\Psi$. Spherical symmetry and the <Poisson equation for Newtonian gravity> give
$$
\frac{dp}{dr}=-\rho\frac{d\Psi}{dr},\qquad
\frac1{r^2}\frac{d}{dr}\left(r^2\frac{d\Psi}{dr}\right)=4\pi G\rho.
$$
Eliminate the <Newtonian gravitational potential> derivative to obtain
$$
\boxed{\frac{d}{dr}\left(\frac{r^2}{\rho}\frac{dp}{dr}\right)
=-4\pi Gr^2\rho.}
$$
Equivalently, the <shell theorem> gives $\Psi'=GM(r)/r^2$ and $M'(r)=4\pi r^2\rho$, which leads to the same <hydrostatic equilibrium> equation.
\b[A uniform-density liquid.] With constant $\rho$, $M(r)=4\pi\rho r^3/3$. There is no central point mass, so the interior <gravitational acceleration> magnitude is $4\pi G\rho r/3$. Thus
$$
\frac{dp}{dr}=-\frac{4\pi G\rho^2}{3}r.
$$
Use $p(R)=0$ at a vacuum boundary, neglecting surface tension. The <uniform-density planetary pressure profile> is
$$
\boxed{p(r)=\frac{2\pi G\rho^2}{3}(R^2-r^2),\quad 0\le r\le R.}
$$
The central value can also be written
$$
\boxed{p_c=\frac{2\pi G\rho^2R^2}{3}=\frac{3GM^2}{8\pi R^4}.}
$$
It has units pascals and decreases smoothly from its maximum at the centre to zero at the surface.
\b[An isothermal gas.] Here $c$ is the constant <isothermal sound speed>, with $p=c^2\rho$. Try $p=A/r^2$, so $\rho=A/(c^2r^2)$. The two sides of the equilibrium equation become
$$
\frac{d}{dr}\left(\frac{r^2}{\rho}\frac{dp}{dr}\right)
=\frac{d}{dr}(-2c^2r)=-2c^2,
\qquad -4\pi Gr^2\rho=-\frac{4\pi GA}{c^2}.
$$
Thus $A=c^4/(2\pi G)$, giving the <singular isothermal sphere>,
$$
\boxed{p(r)=\frac{c^4}{2\pi Gr^2},\qquad
\rho(r)=\frac{c^2}{2\pi Gr^2},\qquad
M(r)=\frac{2c^2r}{G}.}
$$
This does \b[not describe a finite gaseous sphere in vacuum]. Both <pressure> and <mass density> diverge at the centre, and $p(R)>0$ for every finite outer radius $R$, so it cannot meet $p=0$ there. Extending it to infinity removes the finite-radius pressure jump but makes the total mass infinite. It is useful as an idealized profile over an intermediate radial range, or as a truncated model with nonzero external pressure. The <vacuum boundary obstruction for an isothermal sphere> also applies to the regular, nonsingular isothermal solution; removing the central singularity does not produce a finite vacuum surface.
\b[Steady outflow.] In the wind calculation, take the central mass $M$ to be fixed and neglect the wind's own <self-gravity>. Constant <mass flow rate> and the <isothermal equation of state> give
$$
4\pi r^2\rho u=\dot M,\qquad
\frac1\rho\frac{d\rho}{dr}=-\frac2r-\frac1u\frac{du}{dr},\qquad p=c^2\rho.
$$
The steady radial <Euler equation> is
$$
u\frac{du}{dr}=-c^2\frac1\rho\frac{d\rho}{dr}-\frac{GM}{r^2}.
$$
Substituting the density derivative gives
$$
\boxed{\left(u-\frac{c^2}{u}\right)\frac{du}{dr}
=\frac{2c^2}{r}-\frac{GM}{r^2}.}
$$
This is the <isothermal Parker wind> equation. A smooth <sonic point> requires both sides to vanish when $u=c$, so
$$
\boxed{r_s=\frac{GM}{2c^2},\qquad u(r_s)=c.}
$$
Maintaining constant <temperature> during expansion generally requires heat input; it is an isothermal model, not the adiabatic blast-wave model of Question 5.
\b[The escaping branch at large radius.] Integrate the wind equation, using dimensionless logarithm arguments:
$$
\frac{u^2}{2}-c^2\ln\frac uc
=2c^2\ln\frac r{r_s}+\frac{GM}{r}+C_E.
$$
Set $x=r/r_s$ and $w=u^2/c^2$. Then
$$
w-\ln w=4\ln x+\frac4x+C_0.
$$
For the accelerating <transonic branch>, $w(1)=1$, which fixes $C_0=-3$. At $x\gg1$ on its supersonic branch, $w\to\infty$ and $\ln w$ is smaller than $w$. Consequently
$$
w\sim4\ln x,\qquad
\boxed{u(r)\sim2c\sqrt{\ln(r/r_s)}.}
$$
A first correction makes the ordering explicit:
$$
w=4\ln x+\ln(4\ln x)-3+o(1).
$$
This is the <large-radius asymptotics of an isothermal Parker wind>. The leading expression also holds for other supersonic branches with finite $C_0$.
The source implicitly selects an escaping wind. Positive steady outward flow alone does not determine that branch: a <subsonic isothermal breeze> also satisfies the differential equation. Its $w\ll1$ limit gives $w\sim e^{-C_0}x^{-4}$, hence $u\propto r^{-2}$ and $\rho\to\rho_\infty>0$. Such a breeze needs nonzero pressure at infinity. Selecting a regular outflow from a subsonic base into vanishing ambient pressure instead gives the accelerating transonic wind used above.
\Image[/past-exam-of-the-astrophysical-fluid-dynamics-course-of-the-university-of-cambridge/2019-q6-isothermal-wind.png]
{title=Original numerical branches of the integrated isothermal wind equation, showing the escaping wind, its logarithmic asymptote, and a pressure-confined subsonic breeze}
The asymptotic speed has the same units as $c$; the logarithm is dimensionless. Its unlimited growth is a formal feature of indefinite isothermal heating and a prescribed point-mass potential. Real heating, surrounding pressure, and eventually the neglected wind mass limit that extrapolation.
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