Because $H\ll R$ and $M_{\rm env}\ll M$, the envelope has nearly constant area $4\pi R^2$ and nearly constant <surface gravity of a star> $g=GM/R^2$. Plane-parallel <hydrostatic equilibrium> is
$$
\frac{dP}{dz}=-\rho g.
$$
Since $d\Sigma/dz=-\rho$, one has $dP/d\Sigma=g$. Taking the pressure to vanish with $\Sigma$ at the surface gives
$$
\boxed{P=g\Sigma}.
$$

The spherical stellar energy equation is $dL_r/dr=4\pi r^2\rho\epsilon$. With $r\simeq R$ and $F=L_r/(4\pi R^2)$,
$$
\frac{dF}{dr}=\rho\epsilon.
$$
Division by $d\Sigma/dr=-\rho$ gives
$$
\boxed{\frac{dF}{d\Sigma}=-\epsilon}.
$$
Likewise, <radiative diffusion in a star> gives
$$
\frac{dT}{dr}=-\frac{3\kappa\rho L_r}{16\pi ac r^2T^3}
\simeq-\frac{3\kappa\rho F}{4acT^3},
$$
and therefore
$$
\boxed{\frac{dT}{d\Sigma}=\frac{3\kappa F}{4acT^3}}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 317 / 3 / i / Solution

{title2=$g$}

The surface gravity of a spherical star of mass $M$ and radius $R$ is $g=GM/R^2$ in Newtonian gravity.


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 Incoming links: Surface gravity of a star