A particle is projected vertically upwards from the surface of the earth with a velocity equal to half of the escape velocity. The maximum height reached by the particle is measured in terms of the radius of the earth R. What is this height?

B. R/3 — Apply the principle of conservation of mechanical energy to equate the initial energy at the surface to the energy at the maximum height.

JEE Advanced · PhysicsHard

$A particle is projected vertically upwards from the surface of the earth with a velocity equal to half of the escape velocity v_{e} = 2 g R . The maximum height h reached by the particle is measured in terms of the radius of the earth R . What is this height?$

Show correct answer & step-by-step solution

Correct answer: B — $R /3$

$By the conservation of mechanical energy, the initial energy at the surface equals the energy at maximum height h : - \frac{GM m}{R} + \frac{1}{2} m (\frac{v _{e}}{2})^{2} = - \frac{GM m}{R + h} .$
$Substitute the escape velocity v_{e}^{2} = \frac{2 GM}{R} into the equation: - \frac{GM m}{R} + \frac{1}{2} m (\frac{2 GM}{4 R}) = - \frac{GM m}{R + h} .$
$Simplify the expression to - \frac{GM m}{R} + \frac{GM m}{4 R} = - \frac{GM m}{R + h}, which reduces to - \frac{3}{4 R} = - \frac{1}{R + h} .$
$Solving for h gives R + h = \frac{4 R}{3}, which results in h = \frac{R}{3} .$
$Hence the answer is (B).$