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

 

26) The Gravity anomaly recorded at a point A on the edge of a faulted block of material of limited throw t(as shown in figure) is 2 milligals

What would be the maximum gravity anomaly that can be produced by a 2d slab of width 4 km , thickness t having the same density contrast as the faulted block and located at a depth of 2 km (assume the faulted block and 2d slab to be horizontal semi-infinite)

(Special thanks to Y.Suresh Sir, GSI)

(Thanks to Chandrasekhar, ANU)

 

Solution:

 The maximum Gravity anomaly of a 2d slab having the

                                              Width(x) = 4 km

                                  Depth (z) =2 km

 The gravity anomaly of a semi-infinite horizontal slab is

                   $\triangledown g_{z}=2 G\triangledown \rho t\left[ \frac{\pi}{2} +tan^{-1}(\frac{x}{z}) \right]$

 

where x is the width of the slab

z is the depth of the slab

 Initially for a Vertical fault the Gravity anomaly is 2 mill gal i.e.  at a x=o

therefore,

                                     

$\triangledown g_{z}=2 G\triangledown \rho t\left[ \frac{\pi}{2} +tan^{-1}(\frac{x}{z}) \right]$

 

$2=2 G\triangledown \rho t\left[ \frac{\pi}{2} +tan^{-1}(\frac{0}{z}) \right]$

 

$2=2 G\triangledown \rho t\left[ \frac{\pi}{2} +0\right]$

 

$2=2 G\triangledown \rho t\left[ \frac{\pi}{2} \right]$

 

$2= G\triangledown \rho t \pi$-------------------------------(1)

 

 

 Now the gravity anomaly for a 2d slab

$\triangledown g_{z}=2 G\triangledown \rho t\left[ \frac{\pi}{2} +tan^{-1}(\frac{x}{z}) \right]$

 

$\triangledown g_{z}=2 G\triangledown \rho t\left[ \frac{\pi}{2} +tan^{-1}(\frac{4}{2}) \right]$

 

$\triangledown g_{z}=2 G\triangledown \rho t\left[ \frac{\pi}{2} +tan^{-1}(2) \right]$

 

$\triangledown g_{z}=2 G\triangledown \rho t\left[ \frac{\pi}{2} + \frac{\pi}{3} \right]$

 

$\triangledown g_{z}=2 G\triangledown \rho t\left[ \frac{ 5\pi}{6} \right]$

 

$\triangledown g_{z}=(2) (G\triangledown \rho t \pi)(\frac{5}{6})$

 

$\triangledown g_{z}=(2 )(2)(\frac{5}{6})$from equation (1)

 

$\triangledown g_{z}=\frac{10}{3}$

 

$\triangledown g_{z}=3.33 milli gals$

Note: we are not getting the answer, if any one know the solution please share you knowledge.  

Reference : Fundamentals of Geophysics by William Lowrie.

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One more solution from Chadrashekar Mishra:

(Thank you so much)