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Q.
Acceleration due to gravity is maximum at $(R$ is the radius of earth)
Gravitation
Solution:
Acceleration due to gravity at a height $h$ above the earth's surface is
$g_{h}=\frac{g}{\left(1+\frac{h}{R}\right)^{2}}$
where $g$ is the acceleration due to gravity on the earth's surface.
At $h=\frac{R}{2}, g_{h}=\frac{g}{\left(1+\frac{R}{2 R}\right)^{2}}=\frac{4 g}{9}$
Acceleration due to gravity at a depth $d$ below the earth's surface is
$g_{d}=g\left(1-\frac{d}{R}\right)$
At $d=\frac{R}{2}, g_{d}=g\left(1-\frac{R}{2 R}\right)=\frac{g}{2}$
At the centre of earth, $d=R, g_{d}=g\left(1-\frac{R}{R}\right)=0$ Thus, the acceleration due to gravity is maximum on the earth's surface.