$A$ satellite orbits the Earth at a height of $400 \; km$ above the surface. How much energy must be expended to rocket the satellite out of the Earth's gravitational influence? (Mass of the satellite $= 200 \; kg$; mass of the Earth $= 6.0 \times 10^{24} \; kg$; radius of the Earth $= 6.4 \times 10^{6} \; m$; $G = 6.67 \times 10^{-11} \; N m^{2} kg^{-2}$)

(N/A) The total energy of a satellite in orbit is the sum of its kinetic energy and potential energy.
Total Energy $E = K + U = \frac{1}{2} m v^{2} - \frac{G M m}{R_{e} + h}$.
Since the orbital velocity $v = \sqrt{\frac{G M}{R_{e} + h}}$,the kinetic energy is $K = \frac{1}{2} m \left( \frac{G M}{R_{e} + h} \right)$.
Thus,$E = \frac{G M m}{2(R_{e} + h)} - \frac{G M m}{R_{e} + h} = -\frac{G M m}{2(R_{e} + h)}$.
The energy required to escape the gravitational influence is the negative of the total energy: $E_{req} = -E = \frac{G M m}{2(R_{e} + h)}$.
Given: $G = 6.67 \times 10^{-11} \; N m^{2} kg^{-2}$,$M = 6.0 \times 10^{24} \; kg$,$m = 200 \; kg$,$R_{e} = 6.4 \times 10^{6} \; m$,$h = 0.4 \times 10^{6} \; m$.
$R_{e} + h = 6.8 \times 10^{6} \; m$.
$E_{req} = \frac{6.67 \times 10^{-11} \times 6.0 \times 10^{24} \times 200}{2 \times 6.8 \times 10^{6}} = \frac{80.04 \times 10^{14}}{13.6 \times 10^{6}} \approx 5.885 \times 10^{8} \; J \approx 5.9 \times 10^{8} \; J$.