A body (mass m) starts its motion from rest f | vedclass

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Question

(B) According to the law of conservation of energy,the total mechanical energy at the initial point (at distance $R_0$) is equal to the total mechanic

Vedclass · Accepted Answer

$\left[ 2 GM \left( \frac{1}{R} - \frac{1}{R_0} \right) \right]^{\frac{1}{2}}$

Vedclass · Answer

(B) According to the law of conservation of energy,the total mechanical energy at the initial point (at distance $R_0$) is equal to the total mechanical energy at the surface of the Earth (at distance $R$).Initial energy $E_i = K_i + U_i = 0 + \left( -\frac{GMm}{R_0} \right) = -\frac{GMm}{R_0}$.Final energy $E_f = K_f + U_f = \frac{1}{2} mv^2 + \left( -\frac{GMm}{R} \right)$.By conservation of energy,$E_i = E_f$:$-\frac{GMm}{R_0} = \frac{1}{2} mv^2 - \frac{GMm}{R}$.Rearranging the terms:$\frac{1}{2} mv^2 = \frac{GMm}{R} - \frac{GMm}{R_0} = GMm \left( \frac{1}{R} - \frac{1}{R_0} \right)$.$v^2 = 2 GM \left( \frac{1}{R} - \frac{1}{R_0} \right)$.Therefore,the velocity $v$ is:$v = \left[ 2 GM \left( \frac{1}{R} - \frac{1}{R_0} \right) \right]^{\frac{1}{2}}$.

$A$ body (mass $m$) starts its motion from rest from a point distant $R_0$ $(R_0 > R)$ from the centre of the Earth. The velocity acquired by the body when it reaches the surface of the Earth will be ($G =$ universal constant of gravitation,$M =$ mass of Earth,$R =$ radius of Earth).

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