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Let $g=|\nabla \Psi |$ on the equipotential $S(\Psi )$. The volume between neighboring equipotentials separated by $d\Psi$ isBecause density is constant on each equipotential,The volume-equivalent radius is defined by $V=4\pi r^{\prime 3}/3$, so $dV=4\pi r^{\prime 2}d{r}^{\prime }$. It follows immediately that

Hydrostatic balance gives $dP/d\Psi =-\rho$, whileConsequentlyDefining the geometric correctionputs this in the one-dimensional stellar-structure formFor a sphere, $S=4\pi r^{\prime 2}$ and $g=Gm/r^{\prime 2}$, so $f_P=1$ as required.

A Roche lobe is the region around one member of a circular, synchronously rotating binary star bounded by the critical effective potential surface through the inner Lagrange point. The two lobes meet at that saddle point. The construction includes the gravitational effects of both stars and the centrifugal contribution in the rotating frame; it is not simply a sphere at a fixed distance from one star.

A star that expands to fill this region can lose gas through Roche-lobe overflow. A volume-equivalent radius provides a convenient single size for the nonspherical lobe, allowing comparison with a model stellar radius. Changing the binary mass ratio or orbital separation changes that size, so orbital evolution can initiate transfer even without intrinsic stellar expansion.