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 influence? Mass of the satellite $=200kg$; mass of the earth $=6.0\times {10}^{24}kg$; radius of the earth $=6.4\times {10}^{6}m$; G = $6.67\times {10}^{-11}N{m}^{2}k{g}^{-2}$.

Answer 1

Mass of the Earth, $M=6.0\times {10}^{24}kg$
$m=200kg$
$R_e=6.4\times {10}^{6}m$
$G=6.67\times {10}^{-11}N{m}^{2}k{g}^{-2}$
Height of the satellite,$h=400km=4\times {10}^{5}m$

Total energy of the satellite at height h$=\left(1/2\right)m{v}^2+(-G\frac{M_{e}m}{(R_e+h)})$
Orbital velocity of the satellite, $v=\sqrt{\frac{G{M}_e}{R_e+h}}$

Total energy at height h $=\frac{1}{2}\frac{G{M}_{e}m}{R_e+h}-\frac{G{M}_{e}m}{R_e+h}$

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

The negative sign indicates that the satellite is bound to the Earth.

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

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

$=\frac{6.67\times {10}^{-11}\times 6\times {10}^{24}\times 200}{2(6.4\times {10}^6+4\times {10}^5)}$

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

If the satellite just escapes from the gravitational field, then total energy of the satellite is zero. Therefore, we have to supply $5.9\times {10}^{9}J$ of energy to just escape it.