Let $g=|\nabla\Psi|$ on the equipotential $S(\Psi)$. The volume between neighboring equipotentials separated by $d\Psi$ is
$$
dV=\int_S\frac{d\Psi}{g}\,dS
=S\langle g^{-1}\rangle d\Psi.
$$
Because density is constant on each equipotential,
$$
dm=\rho,dV.
$$
The <volume-equivalent radius> is defined by $V=4\pi r'^3/3$, so $dV=4\pi r'^2dr'$. It follows immediately that
$$
\boxed{\frac{dm}{dr'}=4\pi r'^2\rho}.
$$

Hydrostatic balance gives $dP/d\Psi=-\rho$, while
$$
\frac{dm}{d\Psi}=\rho S\langle g^{-1}\rangle.
$$
Consequently
$$
\frac{dP}{dm}=-\frac1{S\langle g^{-1}\rangle}.
$$
Defining the geometric correction
$$
\boxed{f_P=\frac{4\pi r'^4}{GmS\langle g^{-1}\rangle}}
$$
puts this in the one-dimensional stellar-structure form
$$
\boxed{\frac{dP}{dm}=-f_P\frac{Gm}{4\pi r'^4}}.
$$
For a sphere, $S=4\pi r'^2$ and $g=Gm/r'^2$, so $f_P=1$ as required.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 317 / 4 / iii / Solution

{title2=$r'$}

The volume-equivalent radius of a closed surface enclosing volume $V$ is defined by $V=4\pi r'^3/3$.


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