Question

If the escape speed of a projectile on Earth's surface is $11.2{kms}^{-1}$ and a body is projected out with thrice this speed, then determine the speed of the body far away from the Earth :

• A. $56.63{kms}^{-1}$
• B. $33{kms}^{-1}$
• C. $39{kms}^{-1}$
• D. $31.7{kms}^{-1}$

Answer 1

Answer: (D)

$31.7{kms}^{-1}$

According to the principle of conservation of energy, we have
Initial kinetic energy $+$ Initial potential energy $=$ Final kinetic energy $+$ Final potential energy
$\Rightarrow \frac{1}{2}m{v}^2-\frac{GMm}{R}=\frac{1}{2}m{v}^{{\prime }^2}+0$
$\Rightarrow \frac{1}{2}m{v}^{{\prime }^2}=\frac{1}{2}m{v}^2-\frac{GMm}{R}$ ....(i)
Also consider, $v_e=$ escape velocity
$\frac{1}{2}m{v}_e^2=\frac{GMm}{R}$ .....(ii)
$\therefore$ From equations (i) and (ii), we get
$\frac{1}{2}m{v}^{{\prime }^2}=\frac{1}{2}m{v}^2-\frac{1}{2}m{v}_e^2$ .....(iii)
$\Rightarrow v^{{\prime }^2}=v^2-v_e^2$
Now, $v_e=11.2{kms}^{-1}$ and $v=3v_e$ .....(iv)
From equations (iii) and (iv), we get
$v^{{\prime }^2}={\left(3v_e\right)}^2-v_e^{2^{\prime }}$
$v^{{\prime }^2}=9v_e^2-v_e^2=8v_e^2=8\times {\left(11.2\right)}^2$
$v^{\prime 2}=8\times {\left(11.2\right)}^2$
$\Rightarrow v^{\prime }=\sqrt{8\times {\left(11.2\right)}^2}=\sqrt{8}\times 11.2$
$v^{\prime }=2\times 1.414\times 11.2=31.68{kms}^{-1}$
$\therefore$ Speed of the body far away from the Earth $v^{\prime }=31.68{kms}^{-1}=31.7{kms}^{-1}$