Derive an expression for energy of satellite.

Explanation

The total energy of the satellite is the sum of its kinetic energy and the gravitational potential energy. The potential energy of the satellite is,
` U = - (GM_s M_E)/((R_E + h))` ...(1)
Here `M_s` -mass of the satellite, `M_E` -mass of the Earth, `R_E` -radius of the Earth. The Kinetic energy of the satellite is
`K.E = 1/2 M_(s) v^2` .....(2)
Here v is the orbital speed of the satellite and is equal to
` v = sqrt( (GM_E)/((R_E + h)) `
Substituting the value of v in (2) the kinetic energy of the satellite becomes,
`K.E = 1/2 ( GM_E M_s)/((R_E + h))`
Therefore the total energy of the satellite is
`E = 1/2 (GM_E M_s)/((R_E + h)) - (GM_s M_E)/((R_E + h))`
` E = (GM_s M_E)/((R_E + h)) `
The total energy implies that the satellite is bound to the Earth by means of the attractive gravitational force. Note: As h approaches `oo` , the total energy tends to zero. Its physical meaning is that the satellite is completely free from the influence of Earth.s gravity and is not bound to Earth at large distance.