How does the gravitational anomaly measured at the Earth's surface and produced by a subsurface body depend on its depth and on the density contrast of the body relative to its surroundings?

  • 2
    $\begingroup$ Maybe do some reading around simple forward modeling of gravity anomalies. I found this and it looks like a good start. $\endgroup$
    – Matt Hall
    Mar 23, 2016 at 11:59
  • $\begingroup$ the subsurface body is too vague/broad $\endgroup$
    – Jan Doggen
    Mar 23, 2016 at 13:44
  • $\begingroup$ A bit more of context in this question would be useful to the reader. What do you need this for? What do you want to understand? What do you know already? $\endgroup$
    – DrGC
    Mar 24, 2016 at 7:54

1 Answer 1


The gravitational anomaly (delta_g) changes linearly with the density contrast, and proportionally to the inverse square of depth. That's simply a version of the 'Universal Law of Gravitation'. Here is an example for the anomaly created by an spherical anomalous spherical body with a density difference of delta_rho relative to the surrounding density. G is the known Gravitational Constant:

From Turcotte and Schubert, 2002, Geodynamics From Turcotte and Schubert, 2002, Geodynamics

  • $\begingroup$ I am not sure you are answering the question: you are referring to the absolute calculation of gravity but not the expected deviation in reference to the geoid (anomaly) $\endgroup$
    – Neo
    Mar 24, 2016 at 5:13
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    $\begingroup$ Also de anomaly (not just the absolute value) of gravity follows the same dependence, since gravity is additive (the field produced at a given location by 2 bodies equals the addition of the two forces exerted by each body). The question does not mention geoid, which is a very different concept than gravity anomaly. $\endgroup$
    – DrGC
    Mar 24, 2016 at 7:48

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