QJEEM13Phac04: JEE Main 2013 Physics - Class 11 JEEM13Phac04 Solution: What is the minimum energy required to launch a satellite of…

Energy of the satellite on the surface of the planet

E_i = K.E + P.E = 0 + $\left( { - {{GMm} \over R}} \right)$ = ${ - {{GMm} \over R}}$

Energy of the satellite at 2R distance from the surface of the planet while moving with velocity v

E_f = ${1 \over 2}m{v^2}$ + $\left( { - {{GMm} \over {R + 2R}}} \right)$

In the orbital of planet, the centripetal force is provided by the gravitational force

$\therefore$ ${{m{v^2}} \over {R + 2R}} = {{GMm} \over {{{\left( {R + 2R} \right)}^2}}}$

$\Rightarrow {v^2} = {{GM} \over {3R}}$

$\therefore$ E_f = ${1 \over 2}m{v^2}$ + $\left( { - {{GMm} \over {R + 2R}}} \right)$

$= {1 \over 2}m{{GM} \over {3R}} - {{GMm} \over {3R}}$

= $- {{GMm} \over {6R}}$

$\therefore$ Minimum energy required required to launch the satellite

= E_f - E_i

= $- {{GMm} \over {6R}}$ - $\left( { - {{GMm} \over R}} \right)$

= ${{5GMm} \over {6R}}$