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The signed instantaneous contribution of apsidal precession to azimuthal motion during isotropic stellar mass loss is $\dot{\varpi }/\dot{\theta }=-(\dot{\mu }/\mu n)(1-e^2{)}^{3/2}\sin f/[e(1+e\cos f{)}^2]$. Its maximum absolute magnitude is small when $n|\mu /\dot{\mu }|\gg F(e)$, where $F(e)\sim 1/e$ near a circular orbit and $F(e)\to 3\sqrt{3}/4$ as $e\to 1$. The low-orbital eccentricity coordinate singularity can be avoided by evolving the eccentricity vector.

Time averaging over a Kepler orbit gives $⟨\cos f⟩=-e$, as follows from $dt=(1-e\cos E)dE/n$ and $\cos f=(\cos E-e)/(1-e\cos E)$. Consequently slow isotropic stellar mass loss preserves the secular orbital eccentricity. Conserved specific angular momentum then gives $\mu a=\mathrm{constant}$, so pericentre, apocentre and semimajor axis expand inversely with the remaining gravitational parameter.

Isotropic stellar mass loss changes the central gravitational parameter $\mu (t)$ while preserving the central direction of the force, provided the escaping material causes no recoil or drag on the orbiting body. The relative equation is $\ddot{\mathbf{r}}=-\mu (t)\mathbf{r}/r^3$. Specific angular momentum remains exactly constant; specific orbital energy obeys $\dot{C}=-\dot{\mu }/r$. Orbital expansion is adiabatic only when the mass-loss law varies slowly on the orbital period.

For isotropic stellar mass loss the eccentricity vector obeys $\dot{\mathbf{e}}=-(\dot{\mu }/\mu )(\mathbf{e}+\hat{\mathbf{r}})$. With true anomaly $f$ and $e>0$, this gives $\dot{e}=-(\dot{\mu }/\mu )(e+\cos f)$ and $\dot{\varpi }=-(\dot{\mu }/\mu )\sin f/e$. The osculating pericentre satisfies $\dot{q}/q=-(\dot{\mu }/\mu )(1-\cos f)/(1+e)$, hence cannot decrease during mass loss and is instantaneously stationary at pericentre. The vector form is nonsingular at a circular orbit.

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.