JEE MAIN - Physics (2013 (Offline) - No. 24)
What is the minimum energy required to launch a satellite of mass $$m$$ from the surface of a planet of mass $$M$$ and radius $$R$$ in a circular orbit at an altitude of $$2R$$?
$${{5GmM} \over {6R}}$$
$${{2GmM} \over {3R}}$$
$${{GmM} \over {2R}}$$
$${{GmM} \over {3R}}$$
Explanation
Energy of the satellite on the surface of the planet
Ei = 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
Ef = $${1 \over 2}m{v^2}$$ + $$\left( { - {{GMm} \over {R + 2R}}} \right)$$
In the orbital of planet, the centripetal force is provided by the gravitational force
$$\therefore$$ $${{m{v^2}} \over {R + 2R}} = {{GMm} \over {{{\left( {R + 2R} \right)}^2}}}$$
$$ \Rightarrow {v^2} = {{GM} \over {3R}}$$
$$\therefore$$ Ef = $${1 \over 2}m{v^2}$$ + $$\left( { - {{GMm} \over {R + 2R}}} \right)$$
$$ = {1 \over 2}m{{GM} \over {3R}} - {{GMm} \over {3R}}$$
= $$ - {{GMm} \over {6R}}$$
$$\therefore$$ Minimum energy required required to launch the satellite
= Ef - Ei
= $$ - {{GMm} \over {6R}}$$ - $$\left( { - {{GMm} \over R}} \right)$$
= $${{5GMm} \over {6R}}$$
Ei = 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
Ef = $${1 \over 2}m{v^2}$$ + $$\left( { - {{GMm} \over {R + 2R}}} \right)$$
In the orbital of planet, the centripetal force is provided by the gravitational force
$$\therefore$$ $${{m{v^2}} \over {R + 2R}} = {{GMm} \over {{{\left( {R + 2R} \right)}^2}}}$$
$$ \Rightarrow {v^2} = {{GM} \over {3R}}$$
$$\therefore$$ Ef = $${1 \over 2}m{v^2}$$ + $$\left( { - {{GMm} \over {R + 2R}}} \right)$$
$$ = {1 \over 2}m{{GM} \over {3R}} - {{GMm} \over {3R}}$$
= $$ - {{GMm} \over {6R}}$$
$$\therefore$$ Minimum energy required required to launch the satellite
= Ef - Ei
= $$ - {{GMm} \over {6R}}$$ - $$\left( { - {{GMm} \over R}} \right)$$
= $${{5GMm} \over {6R}}$$
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