A particle is projected vertically upward from the surface of earth (radius $R_{e}$ ) with a kinetic energy equal to half of the minimum value needed for it to escape.The height to which it rises above the surface of earth is

h = R_{e}

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h = 2 R_{e}

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h = R_{e} / 2

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h = 3 R_{e}

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Solution

The correct option is A $h = R_{e}$
The kinetic energy needed to escape from the surface is given as,

$E = \frac{1}{2} m {v_{e}}^{2}$

$= \frac{1}{2} m {(\sqrt{\frac{2 G m}{R_{e}}})}^{2}$

$= \frac{G M m}{R_{e}}$

Since, the potential energy at the maximum height will be equal to the sum of half kinetic energy and the potential energy at the surface of the earth.

It can be written as,

$- \frac{G M m}{R_{e}} + \frac{1}{2} \frac{G M m}{R_{e}} = - \frac{G M m}{R_{e} + h}$

$\frac{1}{2 R_{e}} = \frac{1}{R_{e} + h}$

$h = R_{e}$

The maximum height is $R_{e}$ .

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A particle is projected vertically upward from the surface of earth (radiusRe) with a kinetic energy equal to half of the minimum value needed for it to escape.The height to which it rises above the surface of earth is

A particle is projected vertically upward from the surface of earth (radius $R_{e}$ ) with a kinetic energy equal to half of the minimum value needed for it to escape.The height to which it rises above the surface of earth is