### Q.19:- 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 ×10^{24} kg; radius of the earth = 6.4 ×10^{6} m; G = 6.67 × 10^{–11} N m^{2} kg^{–2}.

**Answer:-**Mass of the Earth, *M* = 6.0 × 10^{24} kg

Mass of the satellite, *m* = 200 kg

Radius of the Earth, *R*_{e} = 6.4 × 10^{6} m

Universal gravitational constant, G = 6.67 × 10^{–11} Nm^{2}kg^{–2}

Height of the satellite, *h* = 400 km = 4 × 10^{5} m = 0.4 ×10^{6} m

Total energy of the satellite at height *h* = (1/2)*mv*^{2} + [ -G*M*_{e}*m* / (*R*_{e} + *h*) ]

Orbital velocity of the satellite, *v* = [ G*M*_{e} / (*R*_{e} + *h*) ]^{1/2}

Total energy of height, *h* = (1/2)G*M*_{e}*m* / (*R*_{e} + *h*) – G*M*_{e}*m* / (*R*_{e} + *h*) = -(1/2)G*M*_{e}*m* / (*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)

= (1/2) G*M*_{e}*m* / (*R*_{e} + *h*)

= (1/2) × 6.67 × 10^{-11} × 6 × 10^{24} × 200 / (6.4 × 10^{6} + 0.4 × 10^{6})

= 5.9 × 10^{9} J.