JEE Mains · Physics · STD 11 - 7. gravitation
An object is allowed to fall from a height \(R\) above the earth, where \(R\) is the radius of earth. Its velocity when it strikes the earth's surface, ignoring air resistance, will be :
- A \(2 \sqrt{g R}\)
- B \(\sqrt{g R}\)
- C \(\sqrt{\frac{g R}{2}}\)
- D \(\sqrt{2 g R}\)
Answer & Solution
Correct Answer
(B) \(\sqrt{g R}\)
Step-by-step Solution
Detailed explanation
Loss in \(PE =\) Gain in \(KE\) \(\left(-\frac{ GMm }{2 R }\right)-\left(-\frac{ GMm }{ R }\right)=\frac{1}{2} mv ^2\) \(\Rightarrow v ^2=\frac{ GM }{ R }= gR\) \(\Rightarrow v =\sqrt{ gR }\)
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