6.4 Energy of a satellite

  Gravitation

Consider a satellite of mass $m$ revolving in a circular path of radius $r$ around the earth.

Potential energy of the system is, $$U = - \frac{{GMm}}{r}$$

Kinetic energy of the satellite is, $$K = \frac{1}{2}m{v^2}$$$$K = \frac{1}{2}m\left( {\frac{{GM}}{r}} \right)\quad \quad \left( {{\text{As, }}\frac{{m{v^2}}}{r} = \frac{{GMm}}{{{r^2}}}} \right)$$$$K = \frac{{GMm}}{{2R}}$$
Total energy of the system is, $$T = K + U$$$$T = \frac{{GMm}}{{2r}} + \left( { - \frac{{GMm}}{r}} \right)$$$$T = - \frac{{GMm}}{{2r}}$$
Negative energy show that the system is closed.

Note:

1. The relation between potential energy and kinetic energy is given by,

$$K = \frac{{\left| U \right|}}{2}$$
2. The relation between potential energy and total energy is given by, $$U=2T$$
3. Graph of potential, kinetic and total energy vs $r$ is given by,

4. The total energy is constant. The farther the satellite is from the earth, the greater is its total energy.