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

CSIR_NET_JUNE_2019

91) Compared to the gravity field measured at the surface, the one measured at the bottom of the 200 m deep bore hole, through a medium of density(d) such that πdG = 0.068 m gal/m will be

A) Larger by 62 m gals

B) Larger by 38 m gals

C) Larger by 14 m gals

D) Smaller by 14 m gals .

(Special Thanks to Chandrashekar, ANU)

 

Solution:

Solution:

Given that $\triangledown h $=200 m

    $\pi \rho G = 0.06 mgal/m$

The measured Gravity in a Borehole Gravimetry is 

    $\triangledown g =(0.3086 - 0.0838\times \rho)\triangledown h$  


$\pi \rho G = 0.06 mgal/m$

$\rho = \frac{0.06}{\pi G}\frac{mgal}{m}$

$\rho=\frac{0.06}{(3.14)(6.67\times 10^{-11} m^{3} kg^{-1}s^{-2})}\frac{mgal}{m}$

$\rho=\frac{0.06}{20.9438\times 10^{-11}}\frac{mgal}{m}\frac{kg sec^{2}}{m^{3}}$

$\rho=0.002864 \times 10^{11}\frac{10^{-3} gal}{m}\frac{(10^{3} gm) sec^{2}}{(10^{2} cm)^{3}}$

$\rho = 0.002864 \times 10^{11} \times 10^{-3} \times 10^{-2}\frac{m}{sec^{2}}\frac{sec^{2}}{m}\frac{10^{3}gm}{10^{6}cc}$

$\rho = 0.002864 \times 10^{11}\times 10^{-3} \times 10^{-2}\times 10^{3} \times 10^{-6}\frac{gm}{cc}$

$\rho = 0.002864 \times 10^{3} \frac{gm}{cc}$

$\rho= 2.864 gm/cc$

  The Gravity from Bore hole Gravimetry is 

$\triangledown g =(0.3086 - 0.0838\times \rho)\triangledown h$  

$\triangledown g =(0.3086 - 0.0838\times 2.864)200$  

$\triangledown g =(0.3086 - 0.24)200$  

$\triangledown g =0.0686\times 200$

$\triangledown g = 13.72 $

The positive sign indicates larger at a value 13.72 nearly 14 mgals.

$\triangledown g =14 mgals$  

Note:

Let $g_{1}$,$g_{2}$  be the values of gravity measured in a vertical borehole at heights $h_{1}$ and $h_{2}$ respectively 

     The Values of $g_{2}$ will be larger than $g_{1}$ for two reasons

    1.  The lower measurement level is closer to the earth’s center, $g_{2}$ will be greater than $g_{1}$ by the amount of combined elevation correction.

    2.  At the lower level $h_{2}$ the gravimeter experiences an upward Bouguer attraction due to the material between the two surfaces. This reduces the measured gravity $h_{2}$ and the requires a compensation increase to $g_{2}$  of amount $(0.0419\times\rho)\triangledown h$mgals

_References: 

_{1. Fundamentals of Geophysics by William Lowrie.}

_{2. Whole Earth Geophysics by J. Lilliwe}