Answer
The work $W$ required to move the satellite is
$W \approx 6.19 \times {10^9}{\rm{joules}}$
Work Step by Step
The earth's gravitational field is given by
${\bf{F}} = - G{M_e}\dfrac{{{{\bf{e}}_r}}}{{{r^2}}}$,
where $r$ is the distance from the center of the earth, $G \approx 6.67 \times {10^{ - 11}}$ ${m^3}k{g^{ - 1}}{s^{ - 2}}$ and ${M_e} \approx 5.98 \times {10^{24}}$ kg is the mass of the earth.
The radius of the earth is approximately $6.4 \times {10^6}$ meters, so the satellite must be moved from ${r_1} = 4 \times {10^6} + 6.4 \times {10^6} = 10.4 \times {10^6}$ meters to ${r_2} = 6 \times {10^6} + 6.4 \times {10^6} = 12.4 \times {10^6}$ meters.
The force on the satellite is $m{\bf{F}} = 1000{\bf{F}}$. So the work $W$ required to move the satellite is
$W = - \mathop \smallint \limits_{{r_1}}^{{r_2}} m{\bf{F}}\cdot{\rm{d}}{\bf{r}} = 1000\mathop \smallint \limits_{10.4 \times {{10}^6}}^{12.4 \times {{10}^6}} G{M_e}\dfrac{{{{\bf{e}}_r}}}{{{r^2}}}\cdot{\rm{d}}{\bf{r}}$
$W = 1000G{M_e}\mathop \smallint \limits_{10.4 \times {{10}^6}}^{12.4 \times {{10}^6}} \dfrac{1}{{{r^2}}}\cdot{\rm{d}}r$
$W = - \dfrac{{1000G{M_e}}}{r}|_{10.4 \times {{10}^6}}^{12.4 \times {{10}^6}}$
$W = - 1000\cdot6.67 \times {10^{ - 11}}\cdot5.98 \times {10^{24}}\left( {\frac{1}{{12.4 \times {{10}^6}}} - \frac{1}{{10.4 \times {{10}^6}}}} \right)$
$W \approx 6.19 \times {10^9}{\rm{joules}}$
