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Q.
Find ratio of acceleration due to gravity $g$ depth d and at height $h$, where $d = 2h$.
Bihar CECEBihar CECE 2008Gravitation
Solution:
Acceleration due to gravity decreases both at altitude and depth.
If $g '$ is the acceleration due to gravity at a point, at a height h above the surface of earth, and $g$ be acceleration due to gravity on earths surface, then
$g'=g\left(1-\frac{2 h}{R}\right)\,\,\,...(i)$
If $g$'' be the acceleration due to gravity at a point at depth $d$, below the surface of earth, then
$g''=g\left(1-\frac{d}{R}\right)\,\,\,.... (ii)$
But $d=2 h$ (given)
$\therefore g''=g\left(1-\frac{2 h}{R}\right)\,\,\,...(iii) $
$\therefore \frac{g''}{g'}=\frac{g\left(1-\frac{2 h}{R}\right)}{g\left(1-\frac{2 h}{R}\right)}=1$
$ \therefore g'': g'=1: 1$
Note The value of acceleration due to gravity, also decreases due to rotation of earth.