# Question #b5921

Apr 9, 2017

Binding energy of a satellite is the energy which must be added to planet-satellite system to free the duo from their gravitational attraction.
If a satellite of mass $m$ orbits a planet having mass $M$ at a radius ${R}_{O}$ with orbital velocity $v$, the total energy of the system is given by
${E}_{\text{Total}} = G P E + K E$
$\implies {E}_{\text{Total}} = - \frac{G M m}{R} _ O + \frac{1}{2} m {v}^{2}$ .....(1)
where $G$ is Universal Gravitational constant.

We know that for circular motion
${F}_{\text{Centripetal}} = \frac{m {v}^{2}}{R} _ O$
and force of gravity is
${F}_{\text{grav}} = \frac{G M m}{R} _ {O}^{2}$

As the system is in equilibrium, the centripetal force must be balanced by the gravitational attraction force between the two. Therefore we get
$\frac{m {v}^{2}}{R} _ O = \frac{G M m}{R} _ {O}^{2}$
$\implies {v}^{2} = \frac{G M}{R} _ O$ .....(2)

Inserting this value in (1) we get
${E}_{\text{Total}} = - \frac{G M m}{R} _ O + \frac{1}{2} m \left(\frac{G M}{R} _ O\right)$
${E}_{\text{Total}} = - \frac{1}{2} \frac{G M m}{R} _ O$ ....(3)

When the planet-satellite duo are free from each others gravitational pull, (implies that ${R}_{O} = \infty$), we have

${E}_{\text{Total}} + B E = 0$
$\implies B E = - \left({E}_{\text{Total}}\right)$

Using (3) we have
$B E = \frac{1}{2} \frac{G M m}{R} _ O$