Question

A satellite orbits the earth at a height R from the surface. How
much energy must be expended to rocket the satellite out of
earths gravitational influence ?

Answer 1

Dear Student,
Let Mass of the Earth be M_e

Mass of the satellite be m

Radius of the Earth be R_e

Universal gravitational constant, G = 6.67 × 10^–11 Nm²kg^–2

Height of the satellite = R

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

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

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

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)

$=\frac{1}{2}\left(\frac{G{M}_{e}m}{R_e+R}\right)$
Regards