Question

A particle is thrown vertically upwards with velocity $11.2km{s}^{-1}$ from the surface of earth. Calculate its velocity at height $3R$. Where $R$ is the radius of earth.

• A. 9.25 $km{s}^{-1}$
• B. 5.6 $km{s}^{-1}$
• C. 11.2 $km{s}^{-1}$
• D. 4.3 $km{s}^{-1}$

Answer 1

Answer: (B)

Let a particle of mass m is thrown vertically upwards with velocity v. Let its velocity be v' at a height 3R from the surface of the earth. According to the conservation of energy,
$\frac{1}{2}m{v}^2-\frac{GMm}{R}=\frac{1}{2}m{v}^{\prime 2}-\frac{GMm}{\left(R+3R\right)}$
$v^2-\frac{2GM}{R}=v^{\prime 2}-\frac{2GM}{4R}$
$v^2-v^{\prime 2}=\frac{2GM}{4R}+\frac{2GM}{R}=\frac{6}{4}\frac{GM}{R}$
$v^{\prime 2}=v^2-\frac{6}{4}\frac{GM}{R}$
$v^{\prime 2}=\sqrt{{\left(11.2\right)}^2-\frac{3}{2}gR}\left(\because g=\frac{GM}{R^2}\right)$
$=\sqrt{{\left(11.2\right)}^2-\frac{3}{2}\times \frac{9.8\times 6400}{1000}}$
$=\sqrt{{\left(11.2\right)}^2-\left(94.08\right)}$
$=5.6km{s}^{-1}$