A body starts from rest from a distance R_0 f | vedclass

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(B) According to the law of conservation of energy,the total mechanical energy at the initial position is equal to the total mechanical energy at the

Vedclass · Accepted Answer

$\sqrt{2 GM\left(\frac{1}{R}-\frac{1}{R_0}\right)}$

Vedclass · Answer

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

$A$ body starts from rest from a distance $R_0$ from the centre of the earth. The velocity acquired by the body when it reaches the surface of the earth will be ($R=$ radius of earth,$M=$ mass of earth).

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