A satellite moves in a circular orbit of radius 4R about a planet of radius R, with period 8.0 hours. A tangential thruster fires briefly and reduces…

Right after the burn the satellite is at its farthest point (apogee), since the speed is below circular speed and the velocity is still tangential. For a Kepler orbit the energy gives $\tfrac12v^2-\dfrac{GM}{r}=-\dfrac{GM}{2a}$. With $r=4R$ and $v^2=0.64\,\dfrac{GM}{r}$, we get $\dfrac{1}{2a}=\dfrac1r-\dfrac{v^2}{2GM}=\dfrac1r(1-0.32)$, so $a=\dfrac{r}{1.36}=0.7353\,r$. Check that the perigee is outside the planet: $r_p=2a-r=0.4706\,r=1.88R>R$, so it is safe. Kepler's third law gives $T\propto a^{3/2}$, so $T_{new}=8\times(0.7353)^{3/2}=8\times0.6305=5.04$ hours. The semi-major axis is smaller than the original radius, so the period shortens, as expected after a retro burn.