Analytic Expressions for the Gravity Gradient Tensor of 3D Prisms with Depth-Dependent Density

Abstract

Variable-density sources have been paid more attention in gravity modeling. We conduct the computation of gravity gradient tensor of given mass sources with variable density in this paper. 3D rectangular prisms, as simple building blocks, can be used to approximate well 3D irregular-shaped sources. A polynomial function of depth can represent flexibly the complicated density variations in each prism. Hence, we derive the analytic expressions in closed form for computing all components of the gravity gradient tensor due to a 3D right rectangular prism with an arbitrary-order polynomial density function of depth. The singularity of the expressions is analyzed. The singular points distribute at the corners of the prism or on some of the lines through the edges of the prism in the lower semi-space containing the prism. The expressions are validated, and their numerical stability is also evaluated through numerical tests. The numerical examples with variable-density prism and basin models show that the expressions within their range of numerical stability are superior in computational accuracy and efficiency to the common solution that sums up the effects of a collection of uniform subprisms, and provide an effective method for computing gravity gradient tensor of 3D irregular-shaped sources with complicated density variation. In addition, the tensor computed with variable density is different in magnitude from that with constant density. It demonstrates the importance of the gravity gradient tensor modeling with variable density.

Appendix: Solutions for Some Indefinite Integrals

We provide the solutions of some indefinite integrals, which are necessary for the derivation of the analytic expressions referred to above. In the solutions, we omit the integral constants. The notations here are consistent with those in the text body.

According to Eqs. (4) and (5) in García-Abdeslem (1992), we get

$$\int {\frac{1}{{r^{3} }}{\text{d}}x} = \frac{x}{{(y^{2} + z^{2} )r}}.$$

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According to Eqs. (12) and (13) in Guo et al. (2004), we get

$$\int {\frac{1}{{(x^{2} + z^{2} )r}}{\text{d}}z} = - \frac{1}{xy}\arctan \frac{xy}{{x^{2} + z^{2} + rz}},$$

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$$\int {\frac{1}{{(y^{2} + z^{2} )r}}{\text{d}}z} = - \,\frac{1}{xy}\arctan \frac{xy}{{y^{2} + z^{2} + rz}},$$

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$$\int {\frac{1}{{(y^{2} + z^{2} )r}}dy} = - \frac{1}{xz}\arctan \frac{xz}{{y^{2} + z^{2} + ry}}.$$

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According to Eqs. 20, 131, 141, 183 and 184 in the table of indefinite integrals from Zwillinger (2011), we get

$$\int {\frac{1}{r}{\text{d}}z} = \ln \left( {z + r} \right),$$

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$$\int {\frac{{(x^{2} + z^{2} )^{k - 1} }}{r}{\text{d}}z^{2} } { = 2(} - 1 )^{k - 1} y^{2(k - 1)} r\sum\limits_{l = 0}^{k - 1} {\frac{{( - 1)^{l} (k - 1)!r^{2l} }}{{(2l + 1)l!(k - l - 1)!y^{2l} }}} ,$$

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$$\int {\frac{1}{{(x^{2} + z^{2} )r}}{\text{d}}z^{2} } = \frac{2}{y}\left[ {\ln \sqrt {x^{2} + z^{2} } - \ln (y + r)} \right],$$

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$$\int {\frac{{z^{2l} }}{r}dz} = \frac{{( - 1)^{l} (2l)!(x^{2} + y^{2} )^{l} }}{{2^{2l} (l!)^{2} }}\left[ {ln(z + r) + r\sum\limits_{q = 1}^{l} {\frac{{( - 1)^{q} q!(q - 1)!(2z)^{2q - 1} }}{{(2q)!(x^{2} + y^{2} )^{q} }}} } \right],$$

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$$\int {\frac{{z^{2d + 1} }}{r}dz} = r\sum\limits_{q = 0}^{d} {\frac{{(2q)!(d!)^{2} }}{{(2d + 1)!(q!)^{2} }}} ( - 4x^{2} - 4y^{2} )^{d - q} z^{2q} .$$

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