A satellite moves in a circular orbit of radius 2R round the Earth, where R = 6400 km is the…

Step 1. First orbit. Gravity gives the centripetal force: $\dfrac{v_1^2}{2R} = \dfrac{GM}{(2R)^2}$, so $v_1 = \sqrt{\dfrac{gR}{2}} = 5657\ \text{m/s}$.

Step 2. The ends of the ellipse. Let $v_p$ be the speed at the near end ($2R$) and $v_a$ at the far end ($4R$). At both ends the velocity is at right angles to the radius, so angular momentum gives $v_p \times 2R = v_a \times 4R$. Energy gives $\tfrac12 v_p^2 - \dfrac{GM}{2R} = \tfrac12 v_a^2 - \dfrac{GM}{4R}$. Put $v_a = v_p/2$:

Step 3. The two firings. First: from $5657\ \text{m/s}$ to $6532\ \text{m/s}$, an increase of $875\ \text{m/s}$. The circular speed at $4R$ is $\sqrt{\dfrac{gR}{4}} = 4000\ \text{m/s}$, so the second firing raises the speed from $3266\ \text{m/s}$ to $4000\ \text{m/s}$, by $734\ \text{m/s}$.

Step 4. Time. The ellipse has semi-major axis $a = 3R$. By Kepler's third law its period is $2\pi\sqrt{\dfrac{a^3}{GM}}$, and the trip is half of that: