# The ratio between the values of accelaration due to gravity at a height 1km above and at a depth of 1km below the earth's surface is (radius of earth is R)

We know that,
variation in g with height "h"

$g^{'} = g\left ( \frac{R}{R+h} \right )^2$

$g'\rightarrow$ gravity at height $h$ from surface of earth.

$R\rightarrow$ Radius of earth

$h\rightarrow$ height above surface

Therefore,
$g'\alpha\, \frac{1}{r^{2}}$

Acceleration due to gravity above the earth  $(g_{1})$    $= g_{o}\left ( 1-\frac{2h}{R} \right )$..............(i)

Since h<<R acceleration due to gravity below earth surface $g_{2}=g_{o}\left ( 1-\frac{d}{R} \right )$...........(ii)

Now, put d = h = 1 km
Thus, $\frac{g_1}{g_2}=\frac{R-2}{R-1}$

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