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Since the binary separation and mean motion are both one, the Kepler third law givesThe center of mass condition puts the bodies atin the rotating reference frame. Thus the sketch has $M_1$, $O$, and $M_2$ in that order along the $x$ axis, with $P=(x,y)$, andIn the inertial sketch this entire configuration rotates uniformly about $O$.

In the synchronously rotating reference frame, stationary fluid obeyswhere $\Psi$ 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

Uniform composition and the assumed central condensation make star 1 approximately barotropic, so one may define the specific enthalpyHydrostatic 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 $\Psi$.

In an Earth-fixed rotating reference frame, the horizontal displacement of a freely falling body receives contributions from Coriolis acceleration and centrifugal acceleration. A short-fall approximation substitutes the unperturbed downward velocity into the Coriolis term and the Earth's surface position into the centrifugal term. If gravitational acceleration excludes centrifugal acceleration, both contributions must be included explicitly.

A quantum vortex becomes energetically favourable in a rotating reference frame when $\Delta E-\Omega \Delta L_z<0$. For a singly quantized vortex in a long radially Thomas–Fermi condensate, the cutoff-shell estimate isIntegrating the trapped number density against $u_{\theta }^2$ produces the logarithm and constant; integrating it against angular momentum per particle gives $\Delta L_z\simeq \pi H{n}_0\hslash R_r^2/2$. Resolved-core corrections can alter the nonlogarithmic constant. This criterion compares equilibrium rotating-frame energies and does not remove a dynamical vortex-entry barrier.

A small planar velocity kick on a stellar circular orbit produces a local epicyclic motion in the source's rotating reference frame. With outward $x$, forward $y$, and initial displacements zero, Hill equations give $x/(\gamma a)=2\cos \theta (1-\cos nt)+\sin \theta \sin nt$ and $y/(\gamma a)=\cos \theta (4\sin nt-3nt)+2\sin \theta (\cos nt-1)$. Tangential kicks combine an epicycle with Keplerian shear; a radial kick gives a closed leading-order epicyclic ellipse.