Answer
a) $W=2.36\times10^{11}J$
b) $W=-2.36\times10^{11}J$
Work Step by Step
(a) The work-energy theorem states that $$W=\frac{1}{2}mv_f^2-\frac{1}{2}mv_0^2$$
Here, we know $m_{satellite}=7420kg$, initial speed $v_0=v_{apogee}=2820m/s$ and $v_f=v_{perigee}=8450m/s$, so we can calculate the work done by gravitational force: $$W=2.65\times10^{11}-2.95\times10^{10}=2.36\times10^{11}J$$
(b) This case is similar to (a), except now $v_0=8450m/s$ and $v_f=2820m/s$. So, $$W=\frac{1}{2}mv_f^2-\frac{1}{2}mv_0^2=2.95\times10^{10}-2.65\times10^{11}=-2.36\times10^{11}J$$
