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

 

54 & 55) The peak gravity anomaly over a 2-D line mass of circular cross-section(horizontal cylinder) of density contrast 500 kg/m³ is 1.674 milli gal. The anomaly decreases to 0.8387 m gal at a distance of 500 m along a principle profile. The universal gravitation constant G= 6.6667*10^-11 m³ sec^-2 kg^-1.

Then calculate the

1.      The depth(m) to the center of line mass and radius(m) to the horizontal cylinder

2.      Compute the access mass per unit length (kg/m) of the line mass (Thanks to Chandrasekhar, ANU)

Solution:

 Given that

      Density contrast (Δρ) = 500 kg/m^3.

  Maximum Gravity anomaly ($\triangledown g_{max}$)= 1.674 milli gal

                                                        = $1.674\times 10^{-5} m/sec^{2}$

 At a distance of 500 m the gravity anomaly = 0.8387 milli gal.

 

 For a horizontal cylinder the anomaly will reduce to its half of its maximum anomaly at the its half width distance.

 According to the above statement the half width ($x_{1/2}$) of the horizontal cylinder is 500m.

 The Depth to the center of the line mass will be defined as Z = $x_{1/2}$

                                                                                                Z= 500m

Similarly the maximum gravity anomaly can be defined using the radius of the cylinder and the depth to the center of line mass as

$\triangledown g_{max} =2\pi G(\frac{\triangledown\rho\times R^{2}}{Z})$

 

$R^{2}= \frac{\triangledown g_{max}Z}{2\pi G \triangledown\rho}$

 

$R^{2}= \frac{(1.674\times 10^{-5})\times (500)}{2\times3.14\times 6.667\times10^{-11}\times500}$

 

$R^{2}= \frac{837 \times 10^{-5}}{20933.438\times 10^{-11}}$

 

$R^{2}= 0.0399838\times 10^{6}$

 

$R = \sqrt{0.0399838\times 10^{6}}$

 

$R= 199.9 m$

 

The Excess mas can be defined as

$M_{exe}= \frac{\triangledown g_{maz}\times Z^{2}}{2G}$

 

$\frac{M_{exe}}{Z}(kg/m)=\frac{\triangledown g_{maz}\times Z}{2G}$

 

$\frac{M_{exe}}{Z}=\frac{1.674\times 10^{-5}\times 500}{2\times 6.667 \times 10^{-11}}$

 

$\frac{M_{exe}}{Z}=\frac{837\times 10^{-5} }{13.334 \times 10^{-11}}$

 

 $\frac{M_{exe}}{Z}=62.77186 \times 10^{6}$

 

  $\frac{M_{exe}}{Z}=6.277186 \times 10^{7}(kg/m)$

 

For the given problem

1.      The depth to the center of the line mass is 500m.

2.      The radius of the horizontal cylinder is 199.9 m

3.      The excess mass for kg/m  is $6.277186 \times 10^{7}$