Question

The escape velocity of a body from the surface of earth is $11.2 \,km / s$. It is thrown up with a velocity $4$ times this velocity of escape. The velocity of the body when it has escaped the gravitational pull of earth (neglecting presence of all other heavenly bodies) is

• A. $4 \times 11.2\, km / s$
• B. $\sqrt{15} \times 11.2\, km / s$
• C. zero
• D. $ 3\times 11.2\,km/s $

Answer 1

Answer: (B)

For a given gravitation field and a given position, the escape velocity is the minimum speed on object without propulsion, at that position, needs to have to move away indefinitely from the source field.
From law of conservation -of energy
$\frac{1}{2} m u^{2}=\frac{1}{2} m v_{e}^{2}+\frac{1}{2} m v^{2}$
$\Rightarrow v^{2}=u^{2}-v_{e}^{2}$
$v^{2}=\left(4 v_{e}\right)^{2}-v_{e}^{2}=15 v_{e}^{2}$
$\Rightarrow v=\sqrt{15} V_{e}$
Given, $v_{e}=11.2\, km / s$
$\therefore v=\sqrt{15} \times 11.2 \,km / s$