JEE Mains · Physics · STD 11 - 7. gravitation
The height in terms of radius of the earth \((R)\), at which the acceleration due to gravity becomes \(\dfrac{g}{9}\), where \(g\) is acceleration due to gravity on earth's surface, is ______.
- A \(\sqrt{3}R\)
- B \(2\sqrt{2}R\)
- C \(2R\)
- D \(\dfrac{4}{9}R\)
Answer & Solution
Correct Answer
(C) \(2R\)
Step-by-step Solution
Detailed explanation
The acceleration due to gravity at a height \(h\) above the surface of the earth is given by \(g' = \dfrac{g}{\left(1 + \dfrac{h}{R}\right)^2}\) Given that \(g' = \dfrac{g}{9}\), we get \(\dfrac{g}{9} = \dfrac{g}{\left(1 + \dfrac{h}{R}\right)^2}\)…
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