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JEE Mains · Physics · STD 11 - Gravitation
  • JEE Mains 2026
  • Medium
  • MCQ

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 ______.

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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

\(\dfrac{g}{9} = \dfrac{g}{\left(1 + \dfrac{h}{R}\right)^2}\)

\(\left(1 + \dfrac{h}{R}\right)^2 = 9\)

\(1 + \dfrac{h}{R} = 3\)

\(\dfrac{h}{R} = 2\)

\(h = 2R\)

Answer: \(2R\)
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