A satellite moves in a circular orbit of radius 2R about the Earth, where R=6400 km is the Earth's radius. A short tangential retro-impulse reduces…

The retro-impulse point is the apogee of the new ellipse, $r_a=2R$, and the perigee is $r_p=R$. Angular momentum conservation gives $v_p R=v_a(2R)$, so $v_p=2v_a$. Energy conservation gives $\tfrac12 v_a^2-\dfrac{GM}{2R}=\tfrac12 v_p^2-\dfrac{GM}{R}$. Substituting $v_p=2v_a$ gives $\tfrac32v_a^2=\dfrac{GM}{2R}$, so $v_a^2=\dfrac{GM}{3R}=\dfrac{gR}{3}$ with $GM=gR^2$. Then $v_p=2\sqrt{\dfrac{gR}{3}}=2\sqrt{\dfrac{9.8\times6.4\times10^{6}}{3}}=9.14\times10^{3}\ \mathrm{m\,s^{-1}}$. This exceeds the circular-orbit speed at $R$ ($7.92$ km/s) as it must for an ellipse, a useful sanity check. The retro-impulse reduces kinetic energy at apogee only, so the new orbit shares its apogee point with the old circle. The semi-major axis is $1.5R$, giving the same result through the vis-viva equation $v^2=GM(2/r-1/a)$.