GATE-2018 (24) :Wyllie – Time average equation

28. Given the Bouguer density of 2.8 g/cc , the Bouguer correction for a gravity station at an elevation of 30 m above the datum is _________________mGals.

(Thanks to Saroja)

Solution:

Formula:

Bouguer correction $({\triangle g}_{Bc})=2 \pi G\triangle \rho t$

G= universal gravity constant = $6.67*10^{-11} m^{3}/kg s^{2}$

$\triangle\rho=2.8 gm /cc$

    =$2800 kg/ m^{3}$

elevation t = 30 m

${\triangle g}_{Bc}=2 \pi G\triangle \rho t $

     = $ 2*3.14*6.67*10^{-11}*2800*30 m^{3}/kg.s^{2}*kg/m^{3}*m$

       = $3518554.8*10^{-11}m/s^{2}$

       = $0.352*10^{-11+7}m/s^{2}$

      =$0.352*10^{-4} m/s^{2}$

      =$0.352*10^{-2}cm/s^{2}$

   $(Gal =1cm/s^2) $

      =$0.352*10^{-2}Gal$

      =$ 0.352*10^{-2}*10^{3}mGal$

      =3.52 mGal.