JEE MAIN - Physics (2009 - No. 26)
The height at which the acceleration due to gravity becomes $${g \over 9}$$ (where $$g=$$ the acceleration due to gravity on the surface of the earth) in terms of $$R,$$ the radius of the earth, is:
$${R \over {\sqrt 2 }}$$
$$R/2$$
$$\sqrt 2 \,\,R$$
$$2\,R$$
Explanation
Given that, at height h from ground the acceleration due to gravity becomes $${g \over 9}$$.
We know acceleration at earth surface due to gravity g = $${{GM} \over {{R^2}}}$$
and acceleration at height h due to gravity g' = $${{GM} \over {{{\left( {R + h} \right)}^2}}}$$
So $${{g} \over 9} $$ = $${{GM} \over {{{\left( {R + h} \right)}^2}}}$$
$$ \Rightarrow $$ $${{g} \over 9} $$ = $${{GM} \over {{R^2}}}.{{{R^2}} \over {{{\left( {R + h} \right)}^2}}}$$
= $$g.{\left( {{R \over {R + h}}} \right)^2}$$
$$ \Rightarrow {1 \over 9} = {\left( {{R \over {R + h}}} \right)^2}$$
$$ \Rightarrow {R \over {R + h}} = {1 \over 3}$$
$$ \Rightarrow 3R = R + h$$
$$\therefore$$ $$h = 2R$$
We know acceleration at earth surface due to gravity g = $${{GM} \over {{R^2}}}$$
and acceleration at height h due to gravity g' = $${{GM} \over {{{\left( {R + h} \right)}^2}}}$$
So $${{g} \over 9} $$ = $${{GM} \over {{{\left( {R + h} \right)}^2}}}$$
$$ \Rightarrow $$ $${{g} \over 9} $$ = $${{GM} \over {{R^2}}}.{{{R^2}} \over {{{\left( {R + h} \right)}^2}}}$$
= $$g.{\left( {{R \over {R + h}}} \right)^2}$$
$$ \Rightarrow {1 \over 9} = {\left( {{R \over {R + h}}} \right)^2}$$
$$ \Rightarrow {R \over {R + h}} = {1 \over 3}$$
$$ \Rightarrow 3R = R + h$$
$$\therefore$$ $$h = 2R$$
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