Energy required to move a body of mass $m$ from an orbit of radius $2 R$ to $3 R$ is (where $M =$ mass of earth, $R =$ radius of earth)

\frac{G M m}{12 R^{2}}

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\frac{G M m}{3 R^{2}}

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\frac{G M m}{8 R}

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\frac{G M m}{6 R}

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Solution

The correct option is D $\frac{G M m}{6 R}$
Potential energy in a orbit of radius $r$ is given by $U = \frac{- G M m}{r}$
Thus initial potential energy $U_{1} = \frac{- G M m}{2 R}$
Also, final potential energy $U_{2} = \frac{- G M m}{3 R}$
Energy spent $=$ increase in potential energy $= U_{2} - U_{1}$
$= - \frac{G M m}{3 R} - (- \frac{G M m}{2 R}) = \frac{G M m}{6 R}$ .

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Energy required to move a body of mass m from an orbit of radius 2R to 3R is (where M= mass of earth, R= radius of earth)

The correct option is D GMm6RPotential energy in a orbit of radius r is given by U=−GMmrThus initial potential energy U1=−GMm2RAlso, final potential energy U2=−GMm3REnergy spent = increase in potential energy =U2−U1=−GMm3R−(−GMm2R)=GMm6R.

Energy required to move a body of mass $m$ from an orbit of radius $2 R$ to $3 R$ is (where $M =$ mass of earth, $R =$ radius of earth)