A particle is projected vertically upwards from the surface of earth (radius Re) 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
Answer & explanation
Correct answer: option 1
(K.E.)escape = $\frac{1}{2} m\left(\sqrt{\frac{2 G M}{R_\theta}}\right)^2=\frac{G M m}{R_e}$
(K.E.)body initially = $\frac{1}{2} \frac{\mathrm{GMm}}{\mathrm{R}_{\mathrm{e}}}$
By Law of Conservation of Energy
$\left(\begin{array}{c}\text { Total Initial } \\ \text { Mechanical } \\ \text { Energy }\end{array}\right)=\left(\begin{array}{c}\text { Total Final } \\ \text { Mechanical } \\ \text { Energy }\end{array}\right)$
$\text { (K. E. +P. E. })_{\text {surface }}=(\text { K. E. +P. E. })_{\text {at height h }}$
$\Rightarrow \frac{1}{2} \frac{\mathrm{GMm}}{\mathrm{R}_{\mathrm{e}}}-\frac{\mathrm{GMm}}{\mathrm{R}_{\mathrm{e}}}=0-\frac{\mathrm{GMm}}{\mathrm{R}_{\mathrm{e}}+\mathrm{h}}$ {∵ velocity at maximum height is zero}
⇒ h = Re