A body of mass m falls from a height R above | vedclass

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Question

according to conservation of energy

$\frac{-\mathrm{GMm}}{2 \mathrm{R}}+0=-\frac{\mathrm{GMm}}{\mathrm{R}}+\frac{1}{2} \mathrm{mv}^{2}$

$\Righta

Vedclass · Accepted Answer

$\sqrt {gR} $

Vedclass · Answer

according to conservation of energy

$\frac{-\mathrm{GMm}}{2 \mathrm{R}}+0=-\frac{\mathrm{GMm}}{\mathrm{R}}+\frac{1}{2} \mathrm{mv}^{2}$

$\Rightarrow \frac{\mathrm{GMm}}{2 \mathrm{R}}=\frac{1}{2} \mathrm{mv}^{2} \Rightarrow \mathrm{v}^{2}=\frac{\mathrm{GM}}{\mathrm{R}}=\frac{\mathrm{GM}}{\mathrm{R}^{2}} 	imes \mathrm{R}$

$\Rightarrow \quad \mathrm{v}=\sqrt{\mathrm{gR}}$

A body of mass $m$ falls from a height $R$ above the surface of the earth, where $R$ is the radius of the earth. What is the velocity attained by the body on reaching the ground? (Acceleration due to gravity on the surface of the earth is $g$)

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