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 — Using the conservation of mechanical energy, the initial kinetic energy plus potential energy at the surface equals the potential energy at maximum height. Solving -GMm/R + 1/2 m(v_e/2)^2 = -GMm/(R+h) leads to the result h = R/3.

JEE Main · PhysicsHard

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

A. $$R/4$$
B. $$R/3$$
C. $$R/2$$
D. $$R$$

Show correct answer & step-by-step solution

Correct answer: B — $$R/3$$

Solution

Using the conservation of mechanical energy,

- \frac{GM m}{R} + \frac{1}{2} m (\frac{v _{e}}{2})^{2} = - \frac{GM m}{R + h}

. Substituting

v_{e}^{2} = \frac{2 GM}{R}

yields

- \frac{GM m}{R} + \frac{GM m}{4 R} = - \frac{GM m}{R + h}

, which simplifies to

h = R /3

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