Answer
$1.99 \times 10^{10} \ J$
Work Step by Step
The work needed to increase the separation from a distance $r_1$ to a distance $r_2$ can be calculated as:
$ Work \ done =\int_{r_1}^{r_2} \dfrac{G Mm}{r^2} \ dr \\=[\dfrac{-G Mm}{r^2}]_{r_1}^{r_2} \\=G Mm (\dfrac{1}{r_1}-\dfrac{1}{r_2})$
The satellite covers the distance from $r_1= r_e$to $r_2 =R_e +12,00,000$. Plug in the given data:
$W= G Mm (\dfrac{1}{r_1}-\dfrac{1}{r_2}) \\=(6.67 \times 10^{-11} )(5.98 \times 10^{24}) (2000) $
By using a calculator, the required result is:
$W \approx 1.99 \times 10^{10} \ J$
