An object is allowed to fall from a height $R$ above the earth, where $R$ is the radius of earth. Its velocity when it strikes the earth's surface, ignoring air resistance, will be
Solution
<p>${U_P} = - {{GMm} \over {2R}}$</p>
<p>${U_S} = - {{GMm} \over R}$</p>
<p>$\Rightarrow$ Energy conservation</p>
<p>${1 \over 2}m{v^2} - {{GMm} \over R} = - {{GMm} \over {2R}}$</p>
<p>${v^2} = {{GM} \over R}$</p>
<p>$v = \sqrt {{{GM} \over R}} = \sqrt {gR}$</p>
About this question
Subject: Physics · Chapter: Gravitation · Topic: Gravitational Field and Potential
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