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

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.