Question

A satellite of mass $250kg$ is orbiting the Earth a height of $500km$ above the surface of Earth. How much energy must be expended to rocket the satellite out of the gravitational influence of the Earth ? Given mass of the Earth $=6.0\times {10}^{24}kg$radius of the Earth = 6400 km ; and G = 6.67 x ${10}^{-11}N{m}^{2}k{g}^{-2}$

Answer 1

Given,

Mass of the Earth, $M=6.0\times {10}^{24}kg$

Mass of the satellite, $m=200kg$

Radius of the Earth, $R_e=6.4\times {10}^{6}m$

Universal gravitational constant, $G=6.67\times {10}^{-11}{Nm}^2{kg}^{-2}$

Height of the satellite, $h=400km=4\times {10}^{5}m=0.4\times {10}^{6}m$

Total energy of the satellite at height, $E=\frac{1}{2}m{v}^2+\left(\frac{-G{M}_{e}m}{R_e+h}\right)$

Orbital velocity of the satellite, $V=$$\sqrt{\frac{G{M}_e}{R_e+h}}$

Total energy of height, $E=\frac{1}{2}m\left(\frac{\text{G}{\text{M}}_e}{R_{\text{e}}+h}\right)-\frac{\text{G}{M}_{\text{e}}m}{R_{\text{e}}+h}=-\frac{1}{2}\left(\frac{\text{G}{M}_{\text{e}}m}{R_{\text{e}}+h}\right)$

The negative sign indicates that the satellite is bound to the Earth. This is called bound energy of the satellite

Energy required to send the satellite out of its orbit =-(Bound energy)

$=\frac{1}{2}\frac{G{M}_{e}m}{(R_e+h)}$

$=\frac{1}{2}\times \frac{6.67\times {10}^{-11}\times 6.0\times {10}^{24}\times 200}{(6.4\times {10}^6+0.4\times {10}^6)}$

$=\frac{1}{2}\times \frac{6.67\times 6\times 2\times 10}{6.8\times {10}^6}=5.9\times {10}^{9}J$

Hence, $5.9\times {10}^{9}J$ given to satellite.