# A satellite orbits the earth at a height of 400 km above the surface.

Question:

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

Solution:

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

Mass of the satellite, $m=200 \mathrm{~kg}$

Radius of the Earth, $R_{\mathrm{e}}=6.4 \times 10^{6} \mathrm{~m}$

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

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

Total energy of the satellite at height $h=\frac{1}{2} m v^{2}+\left(\frac{-\mathrm{G} M_{c} m}{R_{\mathrm{e}}+h}\right)$

Orbital velocity of the satellite, $v=\sqrt{\frac{G M_{e}}{R_{\mathrm{e}}+h}}$

Total energy of height, $h=\frac{1}{2} m\left(\frac{\mathrm{G} M_{\mathrm{e}}}{R_{\mathrm{e}}+h}\right)-\frac{\mathrm{G} M_{\mathrm{e}} m}{R_{\mathrm{e}}+h}=-\frac{1}{2}\left(\frac{\mathrm{G} M_{\mathrm{e}} m}{R_{\mathrm{c}}+h}\right)$

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)

$=\frac{1}{2} \frac{\mathrm{G} M_{\mathrm{e}} m}{\left(R_{\mathrm{e}}+h\right)}$

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

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