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Q.
A body is dropped from a height equal to radius of the earth. The velocity acquired by it before touching the ground is
Gravitation
Solution:
To find the velocity, we conserve the energy between the point of release and at Earth's surface.
Since the distance from the Earth's surface to the body $( R )$ is large, we shouldn't just take $g$ while calculating the potential energy.
Total energy before dropping:
$
\frac{- GMm }{ R + R }
$
Total energy as the body hits Earth's surface:
$
\frac{1}{2} mv ^{2}-\frac{ GMm }{ R }
$
Now these two are equal:
$
\Rightarrow \frac{- GMm }{2 R }=\frac{1}{2} mv ^{2}-\frac{ GMm }{ R } \Rightarrow \frac{1}{2} v ^{2}=\frac{- GM }{2 R }+\frac{ GM }{ R }
$
$
\Rightarrow \frac{1}{2} v ^{2}=\frac{ GM }{2 R } \Rightarrow \frac{1}{2} v ^{2}=\frac{1}{2} gR \Rightarrow v =\sqrt{ g R }
$