If g is acceleration due to gravity at earth's surface and r is radius of earth, escape velocity

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7.Gravitation

easy

If g is the acceleration due to gravity at the earth's surface and $r$ is the radius of the earth, the escape velocity for the body to escape out of earth's gravitational field is

A

$gr$

B

$\sqrt {2gr} $

C

$g/r$

D

$r/g$

Solution

(b) $K E+P \cdot E=0$

$\frac{1}{2} m v^{2}-\frac{G M m}{R}=0$

$\frac{1}{2} m v^{2}=\frac{G M m}{R}$

$V=\sqrt{\frac{2 G M \times R}{R \times R}}$

$g=\frac{G M}{R^{2}}$

$v=\sqrt{2 gr}$

Standard 11

Physics

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