Question

A body is released from height 5R where R is
the radius of the earth. Then that body reaches
the ground with a velocity equal to
EXPLAIN PLZ WITH CLEAR SOLUTION​

Answer 1

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Answer 2

Answer:

The velocity with which the body reaches the ground is $V_{1} =\sqrt{\frac{5GM}{3R} }$

Explanation:

We know that the total energy in both the cases given will be equal.

That is,

$TE_{5R} =TE_{R}$

Here the height given is $5R$.

$\frac{GMm}{R-h} =\frac{1}{2} mv_{1} ^{2} - \frac{GMm}{R} \\\\\frac{GMm}{6R}=\frac{1}{2} mv_{1} ^{2}- \frac{GMm}{R} \\\\\frac{1}{2} mv_{1} ^{2}= \frac{GM}{R} -\frac{GM}{6R}\\\\v_{1} ^{2} =\frac{GM}{R} -\frac{GM}{6R}$

Therefore, $V_{1} =$ $\sqrt{\frac{5GM}{3R} }$