= Solution
In the synchronously <rotating reference frame>, stationary fluid obeys
$$
\nabla P=-\rho\nabla\Psi,
$$
where $\Psi$ is the sum of the two gravitational potentials and the <centrifugal potential>. The circular binary's <Kepler third law> gives $\Omega^2=G(M_1+M_2)/a^3=GM/a^3$, hence
$$
\boxed{\Psi=-\frac{GM_1}{|\mathbf r|}
-\frac{GM_2}{|\mathbf r-\mathbf a|}
-\frac{GM}{2a^3}s^2}.
$$
Uniform composition and the assumed central condensation make star 1 approximately <barotropic>, so one may define the <specific enthalpy>
$$
h(P)=\int^P\frac{dP'}{\rho(P')}.
$$
Hydrostatic balance becomes $\nabla(h+\Psi)=0$, or $h+\Psi=\text{constant}$ throughout the star. Since $h$ is a monotone function of $P$ and $\rho$, constant-$P$ and constant-$\rho$ surfaces are <equipotential surfaces> of $\Psi$.
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