A satellite of mass $200 ext{ kg}$ orbits the Earth at a height of $400 ext{ km}$ above the surfac

Question

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

Professor PC Thomas Classes · Answer

The total energy of an orbiting satellite is $E = -\frac{GMm}{2(R_E + h)}$. To escape, it needs to reach zero total energy. The energy to be expended is $\Delta E = 0 - E = \frac{GMm}{2(R_E + h)}$. Substituting the values gives $\Delta E = \frac{6.67 	imes 10^{-11} 	imes 6.0 	imes 10^{24} 	imes 200}{2 	imes (6.4 	imes 10^6 + 0.4 	imes 10^6)} \approx 5.88 	imes 10^9	ext{ J} \approx 5.9 	imes 10^9	ext{ J}$.