A body is projected vertically upwards from earth's surface. If its K.E. of projection is equal to half of…
A body is projected vertically upwards from earth's surface. If its K.E. of projection is
equal to half of its minimum value required to escape from the gravitational
influence, then the height upto which it rises is $(\mathrm{R}=$ radius of the earth $)$
$4 R$
$\mathrm{R}$
$2 \mathrm{R}$
$3 \mathrm{R}$
Solution
Minimum value of kinetic energy required to escape from the gravitational influence of the earth is given by
$\mathrm{k}_{\mathrm{e}}=\frac{\mathrm{GMm}}{\mathrm{R}}$
$\therefore$ The kinetic energy of projection is
$\mathrm{K}=\frac{\mathrm{GMm}}{2 \mathrm{R}}$
At the highest point the kinetic energy become zero.
Loss of kinetic energy $=$ Gain in potential energy
$\begin{aligned}
& \frac{\mathrm{GMm}}{2 \mathrm{R}}=\frac{\mathrm{GMm}}{\mathrm{R}}-\frac{\mathrm{GMm}}{\mathrm{R}+\mathrm{h}} \\
\therefore & \frac{1}{2 \mathrm{R}}=\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}+\mathrm{h}} \\
\therefore & \frac{1}{\mathrm{R}+\mathrm{h}}=\frac{1}{\mathrm{R}}-\frac{1}{2 \mathrm{R}}=\frac{1}{2 \mathrm{R}} \\
\therefore & \mathrm{R}+\mathrm{h}=2 \mathrm{R} \\
\therefore & \mathrm{h}=2 \mathrm{R}-\mathrm{R}=\mathrm{R}
\end{aligned}$