Question

Derive an expression for the binding energy of a body at rest on the Earth’s surface of a satellite.

Answer 1

1. Let,
   M = mass of the Earth
   m = mass of the satellite
   R = radius of the Earth.
2. Since the satellite is at rest on the Earth, v = 0
   ∴ Kinetic energy of satellite,
   K.E. = ${\displaystyle \frac{1}{2}{\text{mv}}^2=0}$
3. Gravitational potential at the Earth’s surface = ${\displaystyle -\frac{\text{GM}}{\text{R}}}$
   ∴ The potential energy of a satellite = Gravitational potential × mass of the satellite
   = - ${\displaystyle \frac{\text{GMm}}{\text{R}}}$
4. Total energy of satellite = T.E = P.E + K.E
   ∴ T.E. = - ${\displaystyle \frac{\text{GMm}}{\text{R}}+0=-\frac{\text{GMm}}{\text{R}}}$
5. Negative sign in the energy indicates that the satellite is bound to the Earth, due to gravitational force of attraction.
6. For the satellite to be free form Earth’s gravitational influence, its total energy should become positive. That energy is the binding energy of the satellite at rest on the surface of the Earth.
   ∴ B.E. = ${\displaystyle \frac{\text{GMm}}{\text{R}}}$