A body starts from rest from a point at a dis | vedclass

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) According to the law of conservation of energy,the total mechanical energy at the initial point $(r = R_0)$ is equal to the total mechanical energy at the surface of the Earth $(r = R)$.Initial Energy $(E_i)$ = Initial Kinetic Energy + Initial Potential Energy = $0 + \left( -\frac{GMm}{R_0} \right) = -\frac{GMm}{R_0}$.Final Energy $(E_f)$ = Final Kinetic Energy + Final Potential Energy = $\frac{1}{2}mv^2 + \left( -\frac{GMm}{R} \right)$.Equating $E_i = E_f$:$-\frac{GMm}{R_0} = \frac{1}{2}mv^2 - \frac{GMm}{R}$.Rearranging for $v^2$:$\frac{1}{2}mv^2 = GMm\left( \frac{1}{R} - \frac{1}{R_0} \right)$.$v^2 = 2GM\left( \frac{1}{R} - \frac{1}{R_0} \right)$.$v = \sqrt{2GM\left( \frac{1}{R} - \frac{1}{R_0} \right)}$.

$A$ body starts from rest from a point at 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 ($R$ represents the radius of the Earth).

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