An object of mass m lying at rest at depth of ‘R/2’ from the Earth’s surface, is to be made an Earth’s satellite at a height of ‘R/2’ from the Earth’s surface. The work required to do this is ___ (M-mass of the Earth, R-radius of the Earth and neglect the rotation of earth)

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$\frac{GMm}{R}$

$\frac{17 GMm}{24 R}$

$\frac{25 GMm}{24 R}$

$\frac{15 GMm}{2 R}$

answer is B.

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Detailed Solution

$h = R / 2, r = R / 2$

Total energy at depth of R/2 is $E_{1}$

$E_{1} = V \cdot m = \frac{- GM}{2 R^{3}} [3 R^{2} - r^{2}] \cdot m$

$= \frac{- GMm}{2 R^{3}} [3 R^{2} - \frac{R^{2}}{4}]$

By doing work, let the object made as Earth’s satellite having energy $E_{2}$

$E_{2} = \frac{- GMm}{2 (R + h)} = \frac{- GMm}{3 R}$

$∴ E_{1} + Δ ω = E_{2}$

$\Rightarrow Δ ω = E_{2} - E_{1}$

$∴ Δ ω = \frac{25 GMm}{24 R}$

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