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7.Gravitation
medium
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
A
$2R$
B
$\frac{R}{{\sqrt 2 }}$
C
$\;\frac{R}{2}$
D
$\;\sqrt {2R} $
(AIEEE-2009)
Solution
We know that $\frac{{g'}}{g} = \frac{{{R^2}}}{{{{\left( {R + h} \right)}^2}}}$
$\therefore \frac{{g/9}}{g} = {\left[ {\frac{R}{{R + h}}} \right]^2}\,\,\,\therefore h = 2R$
Standard 11
Physics