Practicing Success
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 |
Re 2Re 3Re 4Re |
Re |
(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 |