Question

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 energy must be expended to rocket the satellite out of the earths gravitational influence ?

Answer 1

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, $G=6.67\times {10}^{-11}N{m}^2{kg}^{-2}$

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

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

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

Total energy of height, $h=\frac{1}{2}m\sqrt{\frac{G{M}_e}{R_e+h}}-\sqrt{\frac{G{M}_{em}}{R_e+h}}=-\frac{1}{2}\frac{-G{M}_{em}}{R_e+h}$

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)

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

$=\frac{1}{2}\times \frac{6.67\times {10}^{-11}\times 6\times {10}^{24}\times 200}{(6.4\times {10}^6)+(0.4\ast {10}^5)}$

$=5.9\times {10}^{9}J$