= Solution
\b[Orbital energy.] Treat the star as stationary to leading order in $m/M$, with Newtonian <gravitational parameter> $\mu=GM$. For <velocity> components in <plane polar coordinates>, $v^2=\dot r^2+r^2\dot\phi^2$. The <kinetic energy> is $mv^2/2$. Integrating <Newton's law of universal gravitation> with zero <gravitational potential energy> at infinite separation gives $U(r)=-GMm/r$. Thus
$$
\boxed{E=\frac12m\dot r^2+\frac12mr^2\dot\phi^2-\frac{GMm}{r}
=\frac12m\dot r^2+\frac{J^2}{2mr^2}-\frac{GMm}{r},}
$$
where $J=mr^2\dot\phi$ is the satellite's <angular momentum>. The first term is radial <kinetic energy>; the second is transverse <kinetic energy>, also the centrifugal contribution to the radial <effective potential>; the last is the attractive <gravitational potential energy>. These are not three independent sources of energy: the centrifugal term is already part of the ordinary <kinetic energy>.
For an <elliptic Kepler orbit> of <semi-major axis> $a$ and <orbital eccentricity> $e$, the <polar equation of a Kepler orbit> $r=p/(1+e\cos\phi)$ has <semilatus rectum> $p=J^2/(GMm^2)=a(1-e^2)$. Substituting $\dot r=(Je/(mp))\sin\phi$ into the preceding expression reduces it to
$$
E=\frac{GMm}{2p}(e^2-1),
\qquad
\boxed{E=-\frac{GMm}{2a}.}
$$
For a <circular Kepler orbit>, $a=R$, $v_0^2=GM/R$, $K=GMm/(2R)$, and $U=-GMm/R$. Negative total <energy> describes a <bound orbit>.
\b[Rapid, symmetric mass loss.] Assume the mass is removed on a timescale short compared with the satellite's <orbital period>, without an impulsive kick or appreciable wind <drag> on the satellite. The post-loss gravitational parameter is $\mu'=G(1-f)M$ with $f=0.01$. At the change, $r=R$, $\dot r=0$, and $v=v_0=\sqrt{GM/R}$ are unchanged. Spherical symmetry keeps the <force> central, so the <specific angular momentum> $j=J/m=\sqrt{GMR}$ is conserved. <Orbital energy> is not conserved across the changing gravitational field:
$$
\varepsilon'=\frac{E'}m=\frac12\frac{GM}{R}-\frac{G(1-f)M}{R}
=-\frac{GM}{2R}(1-2f).
$$
The new <semi-major axis> is consequently
$$
a'=-\frac{\mu'}{2\varepsilon'}=\frac{1-f}{1-2f}R.
$$
The original <speed> now exceeds the new circular <speed> at $R$, and the outward radial <acceleration> is $v_0^2/R-\mu'/R^2=fGM/R^2>0$. Thus the release point is the new <pericentre>, not the <apocentre>. The <impulsive orbital expansion under isotropic mass loss> gives
$$
r_{\min}=R,\qquad r_{\max}=2a'-R=\frac{R}{1-2f},\qquad e'=\frac{f}{1-f}.
$$
For the stated one-percent loss,
$$
\boxed{r_{\min}=R,\qquad r_{\max}=\frac{50}{49}R\simeq1.02041R,\qquad e'=\frac1{99}.}
$$
An independent <angular momentum> check is $j^2=\mu'a'(1-e'^2)=GMR$. The impulsive orbit stays bound for $f<1/2$; loss of half the central mass is the zero-<orbital energy> limit.
\b[What the wind timescale changes.] The source specifies a spherically symmetric wind but does not give its duration. The preceding nearest and furthest distances require the rapid-loss assumption. In the opposite, slow limit, <adiabatic orbital expansion under isotropic mass loss> preserves an initially circular orbit to leading adiabatic order. Conserved <specific angular momentum> then gives $j^2=G M(t)r(t)$, hence $M(t)r(t)=MR$. After the same loss,
$$
\boxed{r_{\min}\simeq r_{\max}\simeq\frac{R}{1-f}=\frac{100}{99}R\simeq1.01010R\quad\text{(adiabatic limit)}.}
$$
For a finite intermediate timescale, the orbit depends on the mass-loss history; the percentage alone does not select a unique pair of turning distances. Neglecting wind recoil and wind <drag> is common to both idealizations.
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