Question

If the escape speed of a projectile on Earth's surface is $11.2km{s}^{-2}$ and a body is projected out with thrice this speed, then determine the speed of the body far away from the Earth

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

Answer 1

Answer: (D)

According to the principal of conservation of energy,
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)
As consider, $v_e=$ escape velocity
$\frac{1}{2}m{v}_e^2=\frac{GMm}{R}$ ........ (ii)
$\therefore$ From equation (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)
Now, $\begin{array}{c}{v}_e=11.2km{s}^{-1}\\ v=3v_e\end{array}\}\dots .$ (iv) .... (iv)
From equation (iii) and (iv), we get
$v^{\prime 2}={\left(3v_e\right)}^2-v_e^2$
$v^{\prime 2}=9v_e^2-v_e^2=8v_e^2=8\times {\left(11.2\right)}^2$
$=8\times {\left(11.2\right)}^2$
$\Rightarrow v^{\prime }=\sqrt{8\times {\left(11.2\right)}^2}=\sqrt{8}\times 11.2$
$v^{\prime }=2\times 1.414\times 11.2=31.68km{s}^{-1}$
$\therefore$ Speed of the body far away from the Earth,
$v^{\prime }=31.68km{s}^{-1}=31.7km{s}^{-1}$