What is the acceleration due to gravity at th | vedclass

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(A) The acceleration due to gravity $(g)$ at a distance $(r)$ from the centre of the Earth is given by the formula $g' = g(1 - d/R)$,where $d$ is the

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$0$

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(A) The acceleration due to gravity $(g)$ at a distance $(r)$ from the centre of the Earth is given by the formula $g' = g(1 - d/R)$,where $d$ is the depth from the surface.At the centre of the Earth,the depth $(d)$ is equal to the radius of the Earth $(R)$.Substituting $d = R$ into the formula,we get $g' = g(1 - R/R) = g(1 - 1) = 0$.Therefore,the acceleration due to gravity at the centre of the Earth is $0 \ m/s^2$.

What is the acceleration due to gravity at the centre of the earth (in $m/s^2$)?

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