Question

A particle is projected vertically up with velocity $v=\sqrt{\frac{4 g R_{e}}{3}}$ from earth surface. The velocity of particle at height equal to half of the maximum height reached by it

• A. $\sqrt{\frac{g R_{e}}{2}}$
• B. $\sqrt{\frac{g R_{e}}{3}}$
• C. $\sqrt{g R_{e}}$
• D. $\sqrt{\frac{2 g R_{e}}{3}}$

Answer 1

Answer: (B)

Conserving mechanical energy at the surface of earth and the maximum height attained,
$ \frac{-G M m}{R_{e}}+\frac{1}{2} m \frac{4 G M}{3 R_{e}^{2}} R_{e}=\frac{-G M m}{r}+0 $
$ P . E_{i}+K . E_{i} =P . E_{f}+K . E_{l} $
$\Rightarrow \frac{-G M m}{R_{a}}+\frac{2 G M m}{3 R^{e}}=\frac{-G M m}{r} $
$=\frac{1}{3} \frac{G M m}{R_{e}}=\frac{-G M m}{r} $
$\Rightarrow r=3 R_{e} $
$\Rightarrow R_{e}+h=3 R_{e} $
$h=2 R_{e}$
Now, let us calculate the velocity of the particle at height equal to half of the maximum height i.e at $h=R_{e}$ Again using mechanical conservation of energy,
$ P . E_{i}+K . E_{i}=P . E_{j}+K \cdot E_{j} $
$ \frac{-G M m}{R_{e}}+\frac{1}{2} m \frac{4}{3} \frac{G M}{R_{e}^{2}} \times R_{e}=\frac{-G M m}{2 R}+\frac{1}{2} m v^{2} $
$\Rightarrow -\frac{1}{3} \frac{G M m}{R_{e}}+\frac{G M m}{2 R_{e}}=\frac{1}{2} m v^{2} $
$\Rightarrow \frac{G M m}{6 R_{e}}=\frac{1}{2} m v^{2}$
$\Rightarrow v=\sqrt{\frac{G M}{3 R_{e}}}=\sqrt{\frac{G M}{R_{e}^{2}} \times \frac{R_{e}}{3}}=\sqrt{\frac{g R_{e}}{3}}$