JEE Mains · Physics · STD 11 - 7. gravitation
A body is released from a height equal to the radius ( \(R\) ) of the earth. The velocity of the body when it strikes the surface of the earth will be : (Given \(g\) = acceleration due to gravity on the earth.)
- A \(\sqrt{g R}\)
- B \(\sqrt{4 g R}\)
- C \(\sqrt{2 g R}\)
- D \(\sqrt{\frac{g R}{2}}\)
Answer & Solution
Correct Answer
(A) \(\sqrt{g R}\)
Step-by-step Solution
Detailed explanation
By conservation of mechanical energy \(U _{ i }+ K _{ i }= U _{ f }+ K _{ i }\) \(-\frac{ GMm }{2 R }+0=-\frac{ GMm }{ R }+\frac{1}{2} mv ^2\) \(\frac{ GMm }{2 R }=\frac{1}{2} mv ^2\) \(v =\sqrt{\frac{ GM }{ R }}=\sqrt{ gR }\)
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