AIIMS 2017 Physics Escape Velocity MCQ Question

To determine the maximum height attained by a body projected with a velocity equal to one-third of the escape velocity (vₑ) from the Earth's surface, we first calculate the escape velocity, given by $v_{e} = \sqrt{2gR}$, where $g$ is the acceleration due to gravity and $R$ is the Earth's radius. Thus, one-third of the escape velocity is $v = \frac{1}{3}v_{e} = \frac{1}{3}\sqrt{2gR}$. Using the energy conservation principle, the initial kinetic energy at the surface equals the potential energy at maximum height $h$:

$$\frac{1}{2}mv^2 = mg(R + h) \implies h = \frac{v^2}{2g} - R.$$

Substituting $v = \frac{1}{3}\sqrt{2gR}$ results in $h = \frac{(1/3)^2(2gR)}{2g} - R = \frac{R}{9} - R = \frac{R}{9} - \frac{9R}{9} = \frac{-8R}{9}$, which leads to a maximum height of $R/2$ above the Earth's surface when considering the total height from the center of the Earth. Other options (R/3, R/5,