Q55 / 75 · JEE Main 2026 (04 April Shift 2) · Physics
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$2\sqrt{2}R$
- Correct answerC$2R$
- D$\dfrac{4}{9}R$
Solution
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: $2R$
Hence, the correct answer is option: 'C'.