Question 174: Gravitation

Question

The value of the acceleration due to gravity is
g
1
at a height h = $${R \over 2}$$ (R = radius of the earth) from the surface of the earth. It is again equal
to g
1
at a depth d below the surface of the
earth. The ratio $$\left( {{d \over R}} \right)$$ equals :

Accepted Answer

Given, $${g_{at\,high}} = {g_{at\,depth}}$$
We know, $${g_{depth}} = $$$$g\left( {1 - {d \over R}} ight)$$
$$ 	herefore $$ $$g\left( {1 - {d \over R}} ight) = {{GM_e} \over {{{(R + h)}^2}}}$$
$$ \Rightarrow $$ $$g\left( {1 - {d \over R}} ight) =$$$${{G{M_e}} \over {{{\left( {R + {R \over 2}} ight)}^2}}}$$ = $${4 \over 9}{{G{M_e}} \over {{R^2}}}$$ = $${4 \over 9}g$$
$$ \Rightarrow $$ $${d \over R} = 1 - {4 \over 9} = {5 \over 9}$$

Answer

$${5 \over 9}$$

Answer

$${1 \over 9}$$

Answer

$${7 \over 9}$$

Answer

$${4 \over 9}$$

Gravitation

The value of the acceleration due to gravity is g 1 at a height h = ${R \over 2} (R = radius of the earth) from the surface of the earth. It is again equal to g 1 at a depth d below the surface of the earth. The ratio \left( {{d \over R}} \right)$ equals :

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