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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 $=6.4 \times 10^{6}\;m$ ; $G =6.67 \times 10^{-11}\; N m ^{2} kg ^{-2}$
Solution
Mass of the Earth, $M=6.0 \times 10^{24} kg$
Mass of the satellite, $m=200 kg$
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, $h=400 km =4 \times 10^{5} m =0.4 \times 10^{6} m$
$=\frac{1}{2} m v^{2}+\left(\frac{-G M_{c} m}{R_{e}+h}\right)$
Total energy of the satellite at height
Orbital velocity of the satellite, $v=\sqrt{\frac{ GM _{e}}{R_{e}+h}}$ $=\frac{1}{2} m\left(\frac{ G M_{e}}{R_{e}+h}\right)-\frac{ G M_{e} m}{R_{e}+h}=-\frac{1}{2}\left(\frac{ G M_{e} m}{R_{e}+h}\right)$
Total energy of height,
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)
$=\frac{1}{2} \frac{G M_{e} m}{\left(R_{e}+h\right)}$
$=\frac{1}{2} \times \frac{6.67 \times 10^{-11} \times 6.0 \times 10^{24} \times 200}{\left(6.4 \times 10^{6}+0.4 \times 10^{6}\right)}$
$=\frac{1}{2} \times \frac{6.67 \times 6 \times 2 \times 10}{6.8 \times 10^{6}}=5.9 \times 10^{9} J$
