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

CSIR-NET 2015

98. Two sedimentary  layers of infinite horizontal extension sit one over the other. The layer parameters are

Layer1 : depth to top= 2000m ; thickness=1000m ; density contrast= 0.4 gm/cc

Layer2: depth to top=3000m; thickness= 1000m; density contrast= -0.4 gm/cc

The total gravity anomaly produced by these two layers on the surface is

A) 84 mGal              B) 42 mGal          C) 0 mGal                 D) 21 mGal

Solution:

A thick horizontal sheet of infinite width,

Gravity anomaly (g_z) = 2*ᴨ* G* ρ* t -------------------------(1)

G- gravitational constant = 6.67 * 10 ^-11 N m² /kg²

ρ-density contrast

t- thickness of sedimentary bed

Layer1:

ρ₁= 0.4 gm/cc  

   =400 kg/m³

t= 1000 m

g_z = 2*3.14* 6.67*10^-11 *400 *1000

    = 16755040 * 10^-11 m/s²

   = 16.75 mGal

Layer2:

ρ₂= -0.4 gm/cc

    = 400 kg/ m³

t=1000m

g_z = 2*3.14* 6.67*10^-11 *-400 *1000

  = -16.7 mGal

Therefore, the total gravity anomaly produced by these layers on the surface is= layer1( gravity 

anomaly)- layer2 ( gravity anomaly)

        = 16.7 mGal – 16.7 mGal

        = 0  mGal