Question: What is the accuracy required in elevation to distinguish the maximal gravity anomaly of a spherical 50% Haematite (Iron mineral) ore with a radius of 2m at a depth of 10m within a shale section.

I'm trying to solve this problem, first we know that the gravity anomaly of a sphere is that of a point mass at the sphere center equal to the product of the density and volume of the sphere, meaning that only the ρ * V product governs the anomaly and neither the size or density can be determined individually.

Directly above the sphere the maximum anomaly is: enter image description here

We know that in order to correct for gravity's elevation we need to use free-air formula which decreases the gravity by 0.3086 mGal for every meter above sea-level.

Haematite density is: 4.9 - 5.3 (Mg/m^3), let's assume we take the lower bound 4.9 for this case.

But we'll be taking only 50% of 4.9 because only 50% of the ore has haematite.

Also we know that R=2m and h = 10m.

I'm still not sure how to continue from here to find the required accuracy in elevation to distinguish the maximal gravity anomaly because I've some questions like what's the other 50% percent of the iron mineral? should we take in consideration that it's the shale section and above it we have the sandstone? and should we just sum the free air/water correction with the Bouguer correction to get the required accuracy? Any help is highly appreciated.

My approach would be to calculate the free air/water correction first: Depth is 10m which means it's 0.3086 * 10 = 3.086mGal

And Bouguer correction is: δgB = 2πGρh = 2π(6.67e-11)(2.45)(-10) and I took h=-10m because it's under the surface.

Should I combine it somehow to gz_max I mentioned above?

  • $\begingroup$ @Fred Any ideas how to solve this? :) $\endgroup$
    – AnaRhisT
    Dec 4, 2021 at 19:30

1 Answer 1



If gz_max is bigger than the air correction then we're done. because we can measure the signal. and we're done. So gz_max > e_h is the requirement.

(If the error is bigger than the signal anomaly then we can't measure the signal.)


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