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For a circular orbit the origin of mean anomaly is arbitrary because there is no distinguished pericentre. Choose $M=0$ at the ascending node, so $M$ is also the argument of latitude. If a different periapsis convention is used, replace $M$ below by $M+\omega$. The orbital-frame rotation from inclination and node givesIn particular the crossing at $M=0$ has increasing $z$ for prograde $n>0$ and $0<I<\pi$, confirming the ascending node convention.

Put $ϵ=a/a_{\mathrm{pl}}$ and $c=\cos \psi =\hat{\mathbf{r}}\cdot {\hat{\mathbf{r}}}_{\star }$. The binomial series expansion of the direct disturbing function isThe indirect disturbing function subtracts $G{M}_{\star }ac/a_{\mathrm{pl}}^2$, cancelling the dipole term. The constant $G{M}_{\star }/a_{\mathrm{pl}}$ has no spatial derivative and does not perturb the particle. After omitting that constant, the leading stellar quadrupole disturbing function isThus the PDF's equality is understood modulo a dynamically irrelevant additive constant and higher multipoles.

Define $\ell =M+\Omega$ and let ${\ell }_{\star }$ be the star's orbital longitude as seen from the planet. Its orbital plane has $I_{\mathrm{pl}},{\Omega }_{\mathrm{pl}}$. The vector from planet to star is the negative of the planet's heliocentric vector: if ${\ell }_{\mathrm{pl}}$ describes that latter vector, use ${\ell }_{\star }={\ell }_{\mathrm{pl}}+\pi$. This sign matters in $\cos \psi$ but drops out of ${\cos }^2\psi$. Expanding the two rotated unit vectors givesFor example, the particle's horizontal correction is $-(I^2/2)\sin (\ell -\Omega )(-\sin \Omega ,\cos \Omega )$, while its vertical component is $I\sin (\ell -\Omega )$; their dot products with the corresponding stellar components produce these terms. No cubic terms occur because simultaneous reversal of both small orbital inclinations leaves the dot product unchanged.

The two mean longitudes are uniform for the circular Kepler orbits, so orbit averaging is uniform in both $\ell$ and ${\ell }_{\star }$. Set $c_0=\cos (\ell -{\ell }_{\star })$ and $c_2=\cos \psi -c_0$ through quadratic order. The needed averages follow by integrating sine and cosine products over $[0,2\pi {]}^2$:For the last identity, expand $c_0=\cos \ell \cos {\ell }_{\star }+\sin \ell \sin {\ell }_{\star }$ and factor the two integrations, each using $⟨{\sin }^2\ell ⟩=⟨{\cos }^2\ell ⟩=1/2$. Squaring $c_0+c_2$ now yieldswhere the error includes all fourth-order combinations of the two orbital inclinations. ThereforeMultiplying this by $G{M}_{\star }{a}^2/a_{\mathrm{pl}}^3$ gives the requested secular disturbing function. Equivalently, the exact angular factor for two circular rings is $(3{\cos }^2{I}_m-1)/8$, where $I_m$ is their mutual inclination; its small-angle expansion gives the same result. This is the double-averaged circular tidal potential.

For aligned equatorial and planetary orbital planes, set $I_{\mathrm{pl}}=0$. The secular stellar quadrupole disturbing function then has derivative ${\partial }_I{\overline{\mathcal{R}}}_{\star }=-3G{M}_{\star }{a}^{2}I/(4a_{\mathrm{pl}}^3)$. The supplied Lagrange planetary equations for nodal precession givesThe planetary quadrupole coefficient $J_2>0$ contributes ${\partial }_I{\overline{\mathcal{R}}}_{J_2}=-3G{M}_{\mathrm{pl}}{J}_2{R}_{\mathrm{pl}}^{2}I/(2a^3)$, henceBoth are retrograde nodal precession rates. Equating them cancels $n$ and gives the Laplace radiusThe planetary quadrupole dominates inside $a_L$ and the stellar tide outside it. These are small-inclination rates for planet-bound Kepler orbits in the hierarchical regime $a\ll a_{\mathrm{pl}}$. At exactly $I=0$, the node itself is undefined; the rates are the limiting precession frequencies of an infinitesimal tilt.

The central equation must be read as $\ddot{\mathbf{r}}=-\mu \mathbf{r}/r^3$; its printed left-hand expression lacks $=0$. Initially $\mu$ is constant. Dotting this equation with the velocity givesTaking its cross product with $\mathbf{r}$ gives $d(\mathbf{r}\times \mathbf{v})/dt=0$. In the fixed orbital plane the magnitude is $h=r^2\dot{\theta }$. These are conservation of specific orbital energy and specific angular momentum.

Let $f=\theta -\varpi$. Differentiating the polar equation of the Kepler orbit, with its elements initially constant, givesSubstitution into the specific orbital energy yields

For isotropic stellar mass loss with no recoil or drag, the force is still central, so $\mathbf{h}$ remains exactly constant, although $C$ does not. NowThe same algebraic relation between $C,h,e$ holds for the instantaneous osculating orbital elements. Differentiating it and using $h^2/(\mu r)=1+e\cos f$ givesand, for $e>0$,A particularly useful nonsingular description uses the eccentricity vectorThe vector equation remains meaningful at a circular orbit, where a longitude of pericentre is undefined. Its components along and perpendicular to $\mathbf{e}$ give the scalar formula above and

The osculating pericentre is $q=h^2/[\mu (1+e)]$. ThereforeFor mass loss, $\dot{\mu }<0$, this is nonnegative: pericentre does not decrease. It is positive except at pericentre itself, where $\cos f=1$ and $\dot{q}=0$. Thus a strictly opposite sign at every instant is not literally true; a pericentre passage with nonzero mass-loss rate is a counterexample to the strict wording. The intended monotonicity follows exactly, without requiring slow mass loss.

For adiabatic orbital expansion under isotropic mass loss, average over the unperturbed Kepler orbit and hold $\dot{\mu }/\mu$ constant to leading order during that orbit. If $E$ is the eccentric anomaly and $n=\sqrt{\mu /a^3}$, thenIt follows immediately thatAveraging with uniform true anomaly instead of uniform time would give the wrong result. Since $\mu a(1-e^2)=h^2$, conserved $h$ and constant secular orbital eccentricity giveThe pericentre and apocentre expand in the same secular proportion. These are leading adiabatic invariants, rather than exact invariants for arbitrary time-dependent $\mu$.

For the instantaneous precession fraction write $\theta =f+\varpi$ and $\dot{\theta }=h/r^2$. The adiabatic apsidal condition is controlled byIt is a signed fraction: apsidal advance contributes positively on the outward leg during mass loss, and negatively on the inward leg. Let ${\tau }_{\mu }=|\mu /\dot{\mu }|$. Ordinary orbit averaging requires ${\tau }_{\mu }\gg P$, together with negligible variation of the mass-loss law over one orbit. If the pericentre direction is also to vary negligibly relative to the instantaneous azimuthal motion, the maximum absolute fraction must be small. A precise condition for $0<e<1$ isThe maximizing cosine is $c=(1-\sqrt{1+8e^2})/(2e)$: differentiating the last factor gives $c+2e-e{c}^2=0$. Thus one can evaluate $F(e)$ explicitly by putting $\sqrt{1-c^2}/(1+ec{)}^2$ into the formula. Its useful limits areFor orbital eccentricity of order unity, mass loss much slower than an orbital period suffices; near a circular orbit, keeping the pericentre direction slowly varying additionally requires $n{\tau }_{\mu }e\gg 1$. The latter divergence is a singularity of the pericentre coordinate, not a divergence of the eccentricity vector dynamics. Indeed, at frozen $\dot{\mu }/\mu$, integrating the scalar evolution to first order gives $\delta e=-(\dot{\mu }/\mu )(1-e^2)\sin E/n$ up to an integration constant. Its absolute amplitude stays small for slow mass loss even when the scalar longitude of pericentre becomes unsuitable.

The dimensionless coefficient $J_2$ describes the leading axisymmetric departure of a planet's external Newtonian gravitational potential from a monopole: $\Phi =-G{M}_p/r+[G{M}_p{J}_2{R}_p^2/r^3]P_2(\cos \vartheta )$, where $\vartheta$ is colatitude from the spin axis. An oblate mass distribution has positive $J_2$. The disturbing function is minus the nonspherical potential term. On a circular inclined Kepler orbit, $\cos \vartheta =\sin I\sin M$, so orbit averaging gives ${\overline{\mathcal{R}}}_{J_2}=G{M}_p{J}_2{R}_p^2{a}^{-3}(1/2-3{\sin }^{2}I/4)$. The exact circular nodal rate is $-3n{J}_2(R_p/a{)}^2\cos I/2$; its small-tilt limit is $-3n{J}_2(R_p/a{)}^2/2$.