 Solution

Write the lost mass fraction as $f$, with $f=0.02$, and let $\mu =GM$. Initially the circular orbit has $v_0^2=\mu /R$, zero radial velocity, and specific angular momentum $h^2=R^2{v}_0^2=\mu R$. The stipulated rapid symmetric loss leaves the satellite's instantaneous position and velocity unchanged, but reduces the gravitational parameter to ${\mu }^{\prime }=(1-f)\mu$. There is no impulsive kick from contact with ejecta in this model.

The new specific orbital energy isand $h$ remains unchanged because the post-loss force is still central. At a radial turning point,Inserting the known constants and factoring yieldsFor $0<f<1/2$ the two radii are $R$ and $R/(1-2f)$. The initial radius is the pericentre: immediately after the loss, $\ddot{r}=h^2/R^3-{\mu }^{\prime }/R^2=f\mu /R^2>0$, so the satellite begins moving outward. ThereforeEquivalently, the new semi-major axis is $49R/48$ and the orbital eccentricity is $1/49$. These values also satisfy $r_{\min }=a(1-e)$ and $r_{\max }=a(1+e)$. For loss of half the original mass the ideal trajectory becomes parabolic, and larger loss makes it unbound. Slow adiabatic loss would give a different, nearly circular expansion; it is not the instantaneous limit stipulated here.

The figure retains the same angular momentum before and after the change. Once the expelled material is far away, a spherically symmetric outer shell exerts no interior gravitational force by the shell theorem; direct ejecta interaction and the shell's finite transit time are neglected as in the source idealization.