Q147 / 180 · AIIMS 2012 · Physics
A particle is thrown vertically upwards with velocity $11.2 \mathrm{~km} \mathrm{~s}^{-1}$ from the surface of earth. Calculate its velocity at height $3 R$. Where $R$ is the radius of earth.
- A$\simeq 9.25 \mathrm{~km} \mathrm{~s}^{-1}$
- Correct answerB$\simeq 5.6 \mathrm{~km} \mathrm{~s}^{-1}$
- C$\simeq 11.2 \mathrm{~km} \mathrm{~s}^{-1}$
- D$\simeq 4.3 \mathrm{~km} \mathrm{~s}^{-1}$
Solution
Let a particle of mass $m$ is thrown vertically upward with velocity $v$. Let its velocity be $v^{\prime}$ at a height $3 R$ from the surface of the earth. According to the conservation of energy,
$\frac{1}{2} m v^2-\frac{G M m}{R}=\frac{1}{2} m v^{\prime 2}-\frac{G M m}{(R+3 R)}$
$v^2-\frac{2 G M}{R}=v^{\prime 2}-\frac{2 G M}{4 R}$
$v^2-v^{\prime 2}=\frac{-2 G M}{4 R}+\frac{2 G M}{R}=\frac{6}{4} \frac{G M}{R}$
$v^{\prime 2}=v^2-\frac{6}{4} \frac{G M}{R}$
$v^{\prime 2}=\sqrt{(11.2)^2-\frac{3}{2} g R} \quad\left(\because g=\frac{G M}{R^2}\right)$
$=\sqrt{(11.2)^2-\frac{3}{2} \times \frac{9.8 \times 6400}{1000}}$
$=\sqrt{(11.2)^2-(94.08)}$
$=5.6 \mathrm{~km} \mathrm{~s}^{-1}$
Hence, the correct answer is option: 'B'.