# A particle is projected upward from the surface of earth(radius R) with a K.E. equal to half the minimum value needed for it to escape from earth's gravity. To what height does it rises above the surface of earth?

$\\ \text{The kinetic energy needed to escape from the surface is given as,}\\ E=\frac{1}{2} m v_{e}^{2} \\ \begin{array}{l} =\frac{1}{2} m\left [ \sqrt{\frac{2 G m}{R_{e}}} \right ]^2 \\ =\frac{G M m}{R_{e}} \end{array} \\ \text{ Now the sum of half kinetic energy and the potential energy at the surface of the earth} \\ \text{will be equal to the potential energy at the maximum height },\\ -\frac{G M m}{R_{e}}+\frac{1}{2} \frac{G M m}{R_{e}}=-\frac{G M m}{R_{e}+h} \\ \frac{1}{2 R_{e}}=\frac{1}{R_{e}+h} \\ h=R_{e}$

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