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Turns Ratio

A particle is...

Question

A particle is thrown vertically upwards from the surface of earth and it reaches to a maximum height equal to the radius of earth. The ratio of the velocity of projection to the escape velocity on the surface of earth is

A

\frac{1}{\sqrt{2}}

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B

\frac{1}{2}

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C

\frac{1}{4}

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D

\frac{1}{2 \sqrt{2}}

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Solution

The correct option is A $\frac{1}{\sqrt{2}}$
Change in kinetic energy $=$ change in potential energy
$∴ \frac{1}{2} m v^{2} = - \frac{G M m}{R + R} - (- \frac{G M m}{R}) = \frac{G M m}{2 R}$
$v = \sqrt{\frac{G M}{R}} = \frac{\sqrt{(\frac{2 G M}{R})}}{\sqrt{2}} = \frac{v_{e}}{\sqrt{2}}$ (where $v_{e} =$ escape velocity)

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