Hello students, in this article we are going to describe the energy possess by orbiting satellite when it is revolving around the earth. We know that due to its circular motion around the earth, satellite possess the kinetic energy. Along with this satellite is always bound to earth, that means it is acted upon by the gravitational potential energy from earth. Along with this satellite needs certain minimum energy to escape from the influence of earth’s gravitational field, termed as binding energy of satellite.

**Definition of binding energy**

The minimum energy required by satellite so that it can escape from earth’s gravitational influence is called as binding energy of satellite.

**Derivation of formula for energy of orbiting satellite **

Let us find the formula for energy of orbiting satellite

Consider a satellite of mass m in a circular orbit round the earth at a height h above the surface of the earth with critical velocity v_{c}. Let M is mass and R is radius of earth. The orbiting satellite posses both potential energy and kinetic energy.

*a)***At height ‘h’ above earth surface**:

Its total energy is equal to sum of its K.E. and P.E.

T.E. = K.E + P.E. …………(A)

**Kinetic energy**: The K.E. of orbiting satellite is given by,

**Potential energy**: By definition the potential energy of unit mass of body situated in the earth’s gravitational field from the centre of the earth is equal to, product of the gravitational potential exerted by earth and mass of object.

*Potential energy = Gravitational potential × mass of object*

Negative sign indicates that the orbiting satellite is bound to the earth’s gravitational field. It is clear that if an amount of energy + GMm/2(R+h) is supplied to the satellite; it escapes from the earth’s gravitational field.

∴ B.E = + GMm/2(R+h) …….(3)

**b) At rest on earth**: When satellite is at rest on earth’s surface, K.E.= 0 and P.E. is given as,

**Points to remember:**

- Negative sign indicates that the total energy is always directed towards the centre of planet.
- It depends upon the mass of satellite, mass and radius of planet.
- It decreases as we go above the surface of earth and becomes zero at infinity.

**Numerical based on the energy of orbiting satellite**

**Ex 1:** Find the total energy of orbiting satellite of mass 5000 kg, orbiting at height of 2600 km above the surface of earth. Hence find the binding energy of satellite. (Mass of earth= 6×10^{24} kg, R=6400 km).

**Solution:**

Here, m= 5000 kg, Mass of earth, M= 6×10^{24} kg, R=6400 km=6.4×10^{6} m

G= 6.67×10^{-11} N-m^{2}/kg^{2}, R+h=6400+2600km=9000 km