Outside the star, $p=\rho=0$. The first <Tolman–Oppenheimer–Volkoff equation> gives $m(r)=M$, and the second gives
$$
\Phi'(r)=\frac{M}{r(r-2M)}
=\frac12\frac d{dr}\log\left(1-\frac{2M}{r}\right).
$$
Asymptotic flatness fixes the integration constant, so $e^{2\Phi}=1-2M/r$. The exterior line element is therefore the <Schwarzschild metric> with mass $M=m(R)$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 311 / 1 / a / i / Solution

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The Tolman–Oppenheimer–Volkoff equation governs a static, spherically symmetric <perfect fluid in general relativity>. In geometrized units, if $m(r)$ is the mass inside areal radius $r$, then
$$
\frac{dm}{dr}=4\pi r^2\rho,
\qquad
\frac{dp}{dr}=-\frac{(\rho+p)(m+4\pi r^3p)}{r(r-2m)}.
$$

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