Question 230: Gravitation

Question

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:

Accepted Answer

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' = $${{GM} \over {{{\left( {R + h} ight)}^2}}}$$
So $${{g} \over 9} $$ = $${{GM} \over {{{\left( {R + h} ight)}^2}}}$$
$$ \Rightarrow $$ $${{g} \over 9} $$ = $${{GM} \over {{R^2}}}.{{{R^2}} \over {{{\left( {R + h} ight)}^2}}}$$
= $$g.{\left( {{R \over {R + h}}} ight)^2}$$
$$ \Rightarrow {1 \over 9} = {\left( {{R \over {R + h}}} ight)^2}$$
$$ \Rightarrow {R \over {R + h}} = {1 \over 3}$$
$$ \Rightarrow 3R = R + h$$
$$	herefore$$ $$h = 2R$$

Answer

$${R \over {\sqrt 2 }}$$

Answer

$$R/2$$

Answer

$$\sqrt 2 \,\,R$$

Answer

$$2\,R$$

Gravitation

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:

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