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)

Answers (1)
M manish

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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