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Newton's Second Law

A body starts...

Question

A body starts from rest from a point distance $R_{0}$ from the centre of the earth of mass M, radius R.. The velocity acquired by the body when it reaches the surface of the earth will be

A

2 G M (\frac{1}{R} - \frac{1}{R_{0}})

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B

\sqrt{2 G M (\frac{1}{R_{0}} - \frac{1}{R})}

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C

G M (\frac{1}{R} - \frac{1}{R_{0}})

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D

2 G M \sqrt{(\frac{1}{R} - \frac{1}{R_{0}})}

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Solution

The correct option is B $\sqrt{2 G M (\frac{1}{R_{0}} - \frac{1}{R})}$
$P . E . = \int_{R_{0}}^{R} \frac{G M m}{r^{2}} d r = - G M m [\frac{1}{R} - \frac{1}{R_{0}}]$
The K.E. acquired by the body at the surface $= \frac{1}{2} m v^{2}$
$∴ \frac{1}{2} m v^{2} = - G M m [\frac{1}{R} - \frac{1}{R_{0}}]$
$v = \sqrt{2 G M (\frac{1}{R_{0}} - \frac{1}{R})}$

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