Question

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

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

Answer 1

Answer: (D)

According to the principal of conservation of energy,
Initial kinetic energy + initial potential energy,
= final kinetic energy + final potential energy
$\Rightarrow \, \frac{1}{2}mv^{2}-\frac{G M m}{R}=\frac{1}{2}mv^{′ 2}+0$
$\Rightarrow \, \, \, \frac{1}{2}mv^{′ 2}=\frac{1}{2}mv^{2}-\frac{G M m}{R}$ ..... (i)
As consider, $v_{e}=$ escape velocity
$\frac{1}{2}mv_{e}^{2}=\frac{G M m}{R}$ ........ (ii)
$\therefore \, \, \, $ From equation (i) and (ii), we get
$\frac{1}{2}mv^{′ 2}=\frac{1}{2}mv^{2}-\frac{1}{2}mv_{e}^{2}$ ...... (iii)
$\Rightarrow \, \, \, v^{′ 2}=v^{2}-v_{e}^{2}$
NoW, $$ \left.\begin{array}{c} v_{e}=11.2 km \quad s ^{-1} \\ v=3 v_{e} \end{array}\right\} \ldots $$ .... (iv)
From equation (iii) and (iv), we get
$v^{′ 2}=\left(3 v_{e}\right)^{2}-v_{e}^{2}$
$v^{′ 2}=9v_{e}^{2}-v_{e}^{2}=8v_{e}^{2}=8\times \left(11.2\right)^{2}$
$= \, \, 8\times \left(11.2\right)^{2}$
$\Rightarrow \, \, \, v^{′}=\sqrt{8 \times \left(11.2\right)^{2}}=\sqrt{8}\times 11.2$
$v^{′}=2\times 1.414\times 11.2=31.68 \, kms^{- 1}$
$\therefore \, \, $ Speed of the body far away from the Earth,
$v'=31.68kms^{- 1}=31.7kms^{- 1}$