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 ×1024 kg; radius of the earth = 6.4 ×106 m; G = 6.67 × 10–11 N m2 kg–2. | Learn NCERT solution | Education portal Class 11 Physics | Study online Unit-8 Gravitation

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 ×1024 kg; radius of the earth = 6.4 ×106 m; G = 6.67 × 10–11 N m2 kg–2.

Mass of the Earth, M = 6.0 × 1024 kg
Mass of the satellite, m = 200 kg
Radius of the Earth, Re = 6.4 × 106 m
Universal gravitational constant, G = 6.67 × 10–11 Nm2kg–2
Height of the satellite, h = 400 km = 4 × 105 m = 0.4 ×106 m
Total energy of the satellite at height h = (1/2)mv2 + [ -GMem / (Re + h) ]
Orbital velocity of the satellite, v = [ GMe / (Re + h) ]1/2
Total energy of height, h = (1/2)GMem / (Re + h) – GMem / (Re + h)  =  -(1/2)GMem / (Re + 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) GMem / (Re + h)
= (1/2) × 6.67 × 10-11 × 6 × 1024 × 200 / (6.4 × 106 + 0.4 × 106)
= 5.9 × 109 J.