= Solution
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 gives
$$
\frac{d}{dt}\left(\frac{v^2}{2}-\frac{\mu}{r}\right)=0.
$$
Taking 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, gives
$$
\dot r=\frac{\mu}{h}e\sin f,\qquad r\dot\theta=\frac{\mu}{h}(1+e\cos f).
$$
Substitution into the <specific orbital energy> yields
$$
C=\frac{\mu^2}{2h^2}\left[e^2\sin^2f+(1+e\cos f)^2-2(1+e\cos f)\right]
=\boxed{\frac{\mu^2}{2h^2}(e^2-1)}.
$$
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. Now
$$
\dot C=-\frac{\dot\mu}{r}.
$$
The same algebraic relation between $C,h,e$ holds for the instantaneous <osculating orbital elements>. Differentiating it and using $h^2/(\mu r)=1+e\cos f$ gives
$$
e\dot e=\frac{h^2}{\mu^2}\dot C-\frac{2Ch^2}{\mu^3}\dot\mu
=-\frac{\dot\mu}{\mu}e(e+\cos f),
$$
and, for $e>0$,
$$
\boxed{\dot e=-\frac{\dot\mu}{\mu}(e+\cos f)}.
$$
A particularly useful nonsingular description uses the <eccentricity vector>
$$
\mathbf e=\frac{\mathbf v\times\mathbf h}{\mu}-\widehat{\mathbf r},\qquad
\boxed{\dot{\mathbf e}=-\frac{\dot\mu}{\mu}(\mathbf e+\widehat{\mathbf r})}.
$$
The 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
$$
\boxed{\dot\varpi=-\frac{\dot\mu}{\mu}\frac{\sin f}{e}}.
$$
The osculating <pericentre> is $q=h^2/[\mu(1+e)]$. Therefore
$$
\boxed{\frac{\dot q}{q}=-\frac{\dot\mu}{\mu}\frac{1-\cos f}{1+e}}.
$$
For <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 $n=\sqrt{\mu/a^3}$, then
$$
\cos f=\frac{\cos E-e}{1-e\cos E},\qquad dt=\frac{1-e\cos E}{n}\,dE,\qquad P=\frac{2\pi}{n}.
$$
It follows immediately that
$$
\langle\cos f\rangle=\frac1{2\pi}\int_0^{2\pi}(\cos E-e)\,dE=-e,
\qquad \boxed{\langle\dot e\rangle=0}.
$$
Averaging 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> give
$$
\boxed{\mu a=\text{constant},\qquad a\propto\mu^{-1}}.
$$
The <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 by
$$
\boxed{\frac{\dot\varpi}{\dot\theta}
=-\frac{\dot\mu}{\mu}\frac{r^2}{he}\sin f
=-\frac{\dot\mu}{\mu n}\frac{(1-e^2)^{3/2}\sin f}{e(1+e\cos f)^2}}.
$$
It 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 $0<e<1$ is
$$
n\tau_\mu\gg F(e),\qquad F(e)=\frac{(1-e^2)^{3/2}}e\max_f\frac{|\sin f|}{(1+e\cos f)^2}.
$$
The maximizing cosine is $c=(1-\sqrt{1+8e^2})/(2e)$: differentiating the last factor gives $c+2e-ec^2=0$. Thus one can evaluate $F(e)$ explicitly by putting $\sqrt{1-c^2}/(1+ec)^2$ into the formula. Its useful limits are
$$
F(e)\sim e^{-1}\quad(e\to0),\qquad F(e)\to\frac{3\sqrt3}{4}\quad(e\to1).
$$
\b[For <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 $n\tau_\mu e\gg1$.] 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.
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