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(B) Let the mass of the body be $m$ and the mass of the earth be $M$. The escape velocity is $v_e = \sqrt{\frac{2GM}{R}}$.The initial velocity of the

Vedclass · Accepted Answer

$R/3$

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(B) Let the mass of the body be $m$ and the mass of the earth be $M$. The escape velocity is $v_e = \sqrt{\frac{2GM}{R}}$.The initial velocity of the body is $v = \frac{v_e}{2} = \frac{1}{2} \sqrt{\frac{2GM}{R}}$.Using the law of conservation of mechanical energy between the surface of the earth and the maximum height $h$ (where final velocity is $0$):$K_i + U_i = K_f + U_f$$\frac{1}{2}mv^2 - \frac{GMm}{R} = 0 - \frac{GMm}{R+h}$Substituting $v^2 = \frac{1}{4} \cdot \frac{2GM}{R} = \frac{GM}{2R}$:$\frac{1}{2}m \left( \frac{GM}{2R} \right) - \frac{GMm}{R} = - \frac{GMm}{R+h}$$\frac{GMm}{4R} - \frac{GMm}{R} = - \frac{GMm}{R+h}$$-\frac{3GMm}{4R} = - \frac{GMm}{R+h}$$\frac{3}{4R} = \frac{1}{R+h}$$3(R+h) = 4R$$3R + 3h = 4R$$3h = R$$h = R/3$

$A$ body is projected vertically from the surface of the earth of radius $R$ with a velocity equal to half of the escape velocity. The maximum height reached by the body is

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