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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 ______.

  1. A \(\sqrt{3}R\)
  2. B \(2\sqrt{2}R\)
  3. C \(2R\)
  4. D \(\dfrac{4}{9}R\)
Verified Solution

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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