A satellite of earth of mass 'm' taken from orbital radius 2R to 3R, then minimum work done

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7.Gravitation

hard

A satellite of earth of mass $'m'$ taken from orbital radius $2R$ to $3R,$ then minimum work done

A

$\frac{{GMm}}{{6R}}$

B

$\frac{{GMm}}{{12R}}$

C

$\frac{{GMm}}{{24R}}$

D

$\frac{{GMm}}{{3R}}$

Solution

Centripetal acceleration of the satellite is provided by the gravitational force exerted by earth.

$\Longrightarrow \frac{m v^{2}}{r}=\frac{G M m}{r^{2}}$

$\Longrightarrow v^{2}=\frac{G M}{r}$

Total energy of the system $=\mathrm{K.E.}+\mathrm{G.P.E.}=\frac{1}{2} m v^{2}-\frac{G M m}{r}$

$=-\frac{G M m}{2 r}$

Therefore energy required to jump from orbital radius of $2 \mathrm{R}$ to orbital radius of $3 \mathrm{R}$ is given

$E=\frac{G M m}{2}\left(\frac{1}{2 R}-\frac{1}{3 R}\right)$

$E=\frac{G M m}{12 R}$

Standard 11

Physics

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