A body is projected vertically from the Earth | vedclass

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(C) According to the law of conservation of energy,the total energy at the surface equals the total energy at the maximum height $h$.At the surface: $

Vedclass · Accepted Answer

$\frac{R}{3}$

Vedclass · Answer

(C) According to the law of conservation of energy,the total energy at the surface equals the total energy at the maximum height $h$.At the surface: $E_i = K + U = \frac{1}{2}mv^2 - \frac{GMm}{R}$.Given $v = \frac{v_e}{2}$,where $v_e = \sqrt{\frac{2GM}{R}}$,so $v^2 = \frac{v_e^2}{4} = \frac{2GM}{4R} = \frac{GM}{2R}$.$E_i = \frac{1}{2}m(\frac{GM}{2R}) - \frac{GMm}{R} = \frac{GMm}{4R} - \frac{GMm}{R} = -\frac{3GMm}{4R}$.At maximum height $h$,velocity is $0$,so $E_f = 0 - \frac{GMm}{R+h}$.Equating $E_i = E_f$: $-\frac{3GMm}{4R} = -\frac{GMm}{R+h}$.$\frac{3}{4R} = \frac{1}{R+h} \implies 3R + 3h = 4R \implies 3h = R \implies h = \frac{R}{3}$.

$A$ body is projected vertically from the Earth's surface with a velocity equal to half the escape velocity. The maximum height reached by the body is ($R =$ radius of the Earth).

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