JEE MAIN - Physics (2023 - 30th January Evening Shift - No. 3)
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
$\sqrt{\frac{g R}{2}}$
$\sqrt{g R}$
$\sqrt{2 g R}$
$2 \sqrt{g R}$
Explanation
$${U_P} = - {{GMm} \over {2R}}$$
$${U_S} = - {{GMm} \over R}$$
$$\Rightarrow$$ Energy conservation
$${1 \over 2}m{v^2} - {{GMm} \over R} = - {{GMm} \over {2R}}$$
$${v^2} = {{GM} \over R}$$
$$v = \sqrt {{{GM} \over R}} = \sqrt {gR} $$
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