Work Energy Power

JEE Main level, single correct, medium

64 of 183

Question 64 of 183

A body is released from a height equal to the radius (R)\left(R\right) of the earth. The velocity of the body when it strikes the surface of the earth will be: (Given g=g= acceleration due to gravity on the earth.)

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Answer: (B) gR\sqrt{gR}

Using conservation of mechanical energy,
Ui+Ki=Uf+Kf{U}_{i}+{K}_{i}={U}_{f}+{K}_{f}
⇒−GMm2R+0=−GMmR+12mv2\Rightarrow -\dfrac{GMm}{2R}+0=-\dfrac{GMm}{R}+\dfrac{1}{2}m{v}^{2}
⇒12mv2=GMm2R\Rightarrow \dfrac{1}{2}m{v}^{2}=\dfrac{GMm}{2R}
⇒v=GMR=gR\Rightarrow v=\sqrt{\dfrac{GM}{R}}=\sqrt{gR}

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