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JEE Main 2009 (AIEEE) · Acceleration Due to Gravity

Acceleration Due to Gravity question from JEE Main 2009 (AIEEE)

Q123JEE Main 2009 · AIEEEEasy

Acceleration Due to Gravity

The height at which the acceleration due to gravity becomes $\frac{g}{9}$ (where $g =$ the acceleration due to gravity on the surface of the earth) in terms of $R,$ the radius of the earth, is:

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Answer: (d)

$\left(\frac{R}{R+h}\right)^2 = \frac{1}{9}$
$R + h = 3R \Rightarrow h = 2R$
So the answer is (d) $2R$ .

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