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 ______.
Pick an option to check your answer.
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: \(2R\)
\(g' = \dfrac{g}{\left(1 + \dfrac{h}{R}\right)^2}\)
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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