Answer:
Approximately [tex]2.4 \times 10^{7}\; {\rm J}[/tex].
Explanation:
When an object of mass [tex]m[/tex] is moving at a speed of [tex]v[/tex], the kinetic energy of that object will be [tex](1/2)\, m\, v^{2}[/tex].
Kinetic energy of this satellite when in orbit:
[tex]\begin{aligned}\text{KE} &= \frac{1}{2}\, m\, v^{2} \\ &= \frac{1}{2}\times 30\; {\rm kg} \times (800\; {\rm m\cdot s^{-1}}) \\ &\approx 9.60 \times 10^{6}\; {\rm J} \end{aligned}[/tex].
The gravitational field near the surface of the earth is approximately uniform: [tex]g \approx 9.81\; {\rm m\cdot s^{-2}}[/tex]. In a uniform gravitational field, raising an object of mass [tex]m[/tex] by a height of [tex]h[/tex] will require [tex]m\, g\, h[/tex] of work.
Apply unit conversion and ensure that the height of the satellite is measured in standard units (meters):
[tex]h = 50\; {\rm km} = 50000\; {\rm m}[/tex].
Work required to bring this satellite from a ground to a height of [tex]50000\; {\rm m}[/tex]:
[tex]\begin{aligned}m\, g\, h &= 30\; {\rm kg} \times 9.81\; {\rm m\cdot s^{-2}} \times 50000\; {\rm m} \\ &\approx 1.47 \times 10^{7}\; {\rm J}\end{aligned}[/tex].
Work required in total will be approximately:
[tex]9.60 \times 10^{6}\; {\rm J} + 1.47\times 10^{7}\: {\rm J} \approx 2.4 \times 10^{7}\; {\rm J}[/tex].
