A satellite moves in a circular orbit of radius r0 around a planet with period 120 min. A small tangential thrust suddenly reduces its speed to 0.8…

Use units $GM=r_0=1$, so the circular speed is $1$ and the new speed at $r_0$ is $v=0.8$. Since the velocity at $r_0$ is perpendicular to the radius and the orbit is lower in energy, $r_0$ is the apogee. At perigee $r_p=x r_0$: angular momentum gives $v_p=0.8/x$ and energy gives $\tfrac12(0.64)-1=\tfrac12\dfrac{0.64}{x^2}-\dfrac1x$, i.e. $0.68x^2-x+0.32=0$ with roots $x=1$ (apogee itself) and $x=0.4706$. The semi-major axis is $a=\dfrac{1+0.4706}{2}r_0=0.7353\,r_0$. By Kepler's third law $T\propto a^{3/2}$, so $T=120\times0.7353^{1.5}=120\times0.6305=75.66\ \mathrm{min}$. A common error is to treat $0.8$ as the energy ratio or to take $a=r_0$.