Question #b5921

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Question #b5921

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Answer 1 · 2017-04-09T12:14:40

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 $m$ $m$ orbits a planet having mass $M$ $M$ at a radius $R_{O}$ $R_O$ with orbital velocity $v$ $v$ , the total energy of the system is given by
$E_{Total} = G P E + K E$ $E_"Total"=GPE+KE$
$\Rightarrow E_{Total} = - \frac{G M m}{R_{O}} + \frac{1}{2} m v^{2}$ $=>E_"Total"=-(GMm)/R_O+1/2mv^2$ .....(1)
where $G$ $G$ is Universal Gravitational constant.

We know that for circular motion
$F_{Centripetal} = \frac{m v^{2}}{R_{O}}$ $F_"Centripetal" = (m v^2) / R_O$
and force of gravity is
$F_{grav} = \frac{G M m}{R_{O}^{2}}$ $F_"grav" = ( G Mm ) / R_O^2$

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

Inserting this value in (1) we get
$E_{Total} = - \frac{G M m}{R_{O}} + \frac{1}{2} m (\frac{G M}{R_{O}})$ $E_"Total"=-(GMm)/R_O+1/2m( ( G M ) / R_O)$
$E_{Total} = - \frac{1}{2} \frac{G M m}{R_{O}}$ $E_"Total"=-1/2(GMm)/R_O$ ....(3)

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

$E_{Total} + B E = 0$ $E_"Total"+BE=0$
$\Rightarrow B E = - (E_{Total})$ $=>BE=-(E_"Total")$

Using (3) we have
$B E = \frac{1}{2} \frac{G M m}{R_{O}}$ $BE=1/2(GMm)/R_O$

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Question #b5921

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