A body of mass m is dropped from a height h = | vedclass

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(B) According to the law of conservation of mechanical energy,the total energy at the height $h = \frac{R}{2}$ is equal to the total energy at the Ear

Vedclass · Accepted Answer

$\frac{v_e}{\sqrt{3}}$

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(B) According to the law of conservation of mechanical energy,the total energy at the height $h = \frac{R}{2}$ is equal to the total energy at the Earth's surface.Total energy at height $h = \frac{R}{2}$:$E_i = K_i + U_i = 0 + \left( -\frac{GMm}{R + \frac{R}{2}} \right) = -\frac{GMm}{\frac{3R}{2}} = -\frac{2GMm}{3R}$Total energy at the Earth's surface:$E_f = K_f + U_f = \frac{1}{2}mv^2 + \left( -\frac{GMm}{R} \right)$Equating $E_i = E_f$:$-\frac{2GMm}{3R} = \frac{1}{2}mv^2 - \frac{GMm}{R}$$\frac{1}{2}mv^2 = \frac{GMm}{R} - \frac{2GMm}{3R} = \frac{GMm}{3R}$$v^2 = \frac{2GM}{3R}$Since the escape velocity $v_e = \sqrt{\frac{2GM}{R}}$,we have $v_e^2 = \frac{2GM}{R}$.Substituting this into the expression for $v^2$:$v^2 = \frac{v_e^2}{3}$$v = \frac{v_e}{\sqrt{3}}$

$A$ body of mass $m$ is dropped from a height $h = \frac{R}{2}$ above the surface of the Earth,where $R$ is the radius of the Earth. Find its speed when it hits the Earth's surface. (Given: $v_e$ is the escape velocity from the Earth's surface).

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