A particle is projected vertically upwards 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

Question

Dr. Abubacker · Accepted Answer

(a) $(KE)_{escape} = \frac{1}{2}m\left( \sqrt{\frac{2GM}{R_{e}}} ight)^{2} = \frac{GMm}{R_{e}}$
$(KE)_{\begin{array}{r}
body\ \ \ \ \ \ \  \
initially
\end{array}} = \frac{1}{2}\frac{GMm}{R_{e}}$
By law of conservation of energy
$\left( \begin{array}{r}
Total \
mechanical \
energy
\end{array} ight) = \left( \begin{array}{r}
Total\ final \
mechanical \
energy
\end{array} ight)$
$(KE\  + \ PE)_{surface} = (KE\  + \ PE)_{at\ height\ h}$
$\Rightarrow \ \ \ \frac{1}{2}\frac{GMM}{R_{e}} - \frac{GMm}{R_{e}} = 0 - \frac{GMm}{R_{e} + h}$
($	herefore$ velocity at maximum height is zero)
$\Rightarrow \ v = R_{e}$

Q. A particle is projected vertically upwards 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

Explanation

Reviewed by Dr. Abubacker

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