Gujarati
Hindi
7.Gravitation
normal

A body of mass $m$ is situated at distance $4R_e$ above the Earth's surface, where $R_e$ is the radius of Earth how much minimum energy be given to the body so that it may escape

A

$mgR_e$

B

$2\,mgR_e$

C

$\frac {mgR_e}{5}$

D

$\frac {mgR_e}{16}$

Solution

Potential energy of the body at a distance $4 R_{e}$ from the surface of earth

$U=-\frac{m g R_{e}}{1+n / R_{e}}$

$=-\frac{m g R_{e}}{1+4}=-\frac{m g R_{e}}{5}$ As $h=4 R_{e}$

So, minimum energy required to escape the body will be $\frac{m g R_{e}}{5}$

Standard 11
Physics

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