A body is projected vertically upwards from earth's surface. If velocity of projection is…
A body is projected vertically upwards from earth's surface. If velocity of projection is $\left(\frac{1}{3}\right)^{\text {rd }}$ of escape velocity, then the height upto which a body rises is
$(\mathrm{R}=$ radius of earth $)$
$2 \mathrm{R}$
$\frac{\mathrm{R}}{4}$
$\frac{\mathrm{R}}{8}$
$\mathrm{R}$
Solution
$\mathrm{V}=\frac{1}{3} \sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}$
$\mathrm{TE}_{\mathrm{At} \text { surface }}=\mathrm{TE}_{\text {at height } \mathrm{h}}$
$-\frac{G M m}{R}+\frac{1}{2} m v^{2}=-\frac{G M m}{R+h}+0$
$-\frac{G M m}{R}+\frac{1}{2} m \frac{1}{9} \times \frac{2 G M}{R}=-\frac{G M m}{R+h}$
$\left(\frac{1}{9}-1\right) \frac{G M m}{R}=-\frac{G M m}{R+h}$
$-\frac{8}{9} \frac{G M m}{R}=-\frac{G M m}{R+h}$
$\frac{8}{9} \frac{1}{R}=\frac{1}{R+h}$
$\Rightarrow 8 \mathrm{R}+8 \mathrm{h}=9 \mathrm{R}$
$8 h=R \quad \therefore h=\frac{R}{8}$