What is the value of acceleration due to gravity $(g)$ at the centre of the Earth? What will be the variation of $g$ below and above the surface of the Earth?

(N/A) $1$. At the centre of the Earth,the distance from the centre $r = 0$. The formula for acceleration due to gravity at a depth $d$ is $g_d = g(1 - d/R)$,where $d = R$ at the centre. Thus,$g_{centre} = g(1 - R/R) = 0$. So,the value of $g$ at the centre of the Earth is $0 \ m/s^2$.
$2$. Below the surface: As we go deeper into the Earth (increasing depth $d$),the value of $g$ decreases linearly according to the formula $g_d = g(1 - d/R)$.
$3$. Above the surface: As we go higher above the surface (increasing height $h$),the value of $g$ decreases according to the formula $g_h = g(1 - 2h/R)$ for $h \ll R$,or more generally $g_h = gR^2 / (R + h)^2$. In both cases,$g$ decreases as height increases.