Question

IL. Following is the field gravity anomaly due to a spherical structure. Estimate the depth to the centre of the body by means of (a) half width method depth Z -1.3 Xu2) and (b) amplitude gradient ratio method treating the target as a 3-D body. Assuming the density contrast as 1 gm/ee calculate the radius of the sphere. Gravity anomaly due to a spherical structure is given as: Az where 'z' is the depth to the centre of sphere and A-(4/3jtRG? 0 microgal dgfx) -50-45-40-35-30-25 -15-10-5 0 5101520253035404550 50 m 50 m Find the horizontal & vertical derivatives of glx) [dgfx)ydx and dg(x)/dz]. Compute these derivatives for the depth that you estimate from the above anomaly by half width method for x 50 to 50 assuming A-1. Plot the derivatives on the same graph and verify whether the point of intersection of the derivatives correspond to the depth assumed or not.

Answer 1

Half width method max .gravity = 35 X_1/2 = 7.5 (since max from the graph is 35 ) Therefore Depth Depth to the centre of an amount z = 1.32 times (x/2)z = 9.9 m Amplitude gradient ratio method Z < 0.86 (a_max/A'_max) where from the graph = A_max = 35 A_max' = 12.5 therefore z = 2.408 \r\n Radius of the sphere g(x) = Az/(x^2 + z^2) ^3/2 where A = (4/3) pi r^2 G sigma Here sigma = 1000 kg/m^3 z = 9.9 m g(x) = 3.5 times 10^-7 m/s^2 therefore 4 pi R^2 G sigma times Z/3(x^2 + z^2) ^3/2 substituting above above values we get Radius r = 128.46