The value of the acceleration due to gravity is g1 at a height h = ${R \over 2}$ (R = radius of the earth) from the surface of the earth. It is again equal to g1 at a depth d below the surface of the earth. The ratio $\left( {{d \over R}} \right)$ equals :
Solution
Given, ${g_{at\,high}} = {g_{at\,depth}}$<br><br>We know, ${g_{depth}} =$$g\left( {1 - {d \over R}} \right)$<br><br>$\therefore$ $g\left( {1 - {d \over R}} \right) = {{GM_e} \over {{{(R + h)}^2}}}$<br><br>$\Rightarrow$ $g\left( {1 - {d \over R}} \right) =$${{G{M_e}} \over {{{\left( {R + {R \over 2}} \right)}^2}}}$ = ${4 \over 9}{{G{M_e}} \over {{R^2}}}$ = ${4 \over 9}g$<br><br>$\Rightarrow$ ${d \over R} = 1 - {4 \over 9} = {5 \over 9}$
About this question
Subject: Physics · Chapter: Gravitation · Topic: Gravitational Field and Potential
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