Integral transformations of deflections of the vertical onto satellite-to-satellite tracking and gradiometric data

Abstract

With the advent of geodetic satellite missions mapping almost globally the Earth’s gravitational field, new methods and theoretical approaches have been developed and investigated to fully exploit the potential of their new observables. Besides estimating values of numerical coefficients in harmonic series of the gravitational potential, new applications emerged such as data validation and combination. In this contribution, new integral transformations are presented which transform principal components of the terrestrial deflection of the vertical onto disturbing satellite-to-satellite tracking and gradiometric data at altitude. Using spherical approximation, necessary integral kernel functions are derived in both spectral and closed forms. The behaviour of isotropic kernel functions is studied and the new integral transformations are tested in a closed-loop simulation using synthetic terrestrial and satellite data synthesized from a global gravitational model. New integral transformations can be used for data validation and combination purposes.

Appendices

A: Application of differential operators on a multiplication of two functions

In this appendix we demonstrate application of the differential operators, which relate the principal components of the deflection of the vertical to SST data or GGs, on a multiplication of two functions. Suppose two general functions $f = f(\varphi , \lambda , \varphi ', \lambda ')$ and $h = h(r, \varphi , \lambda , R, \varphi ', \lambda ')$. Applying either the differential operator of Eq. (15) or (17) on multiplication of the functions $f$ and $h$ yields

$$\begin{aligned} \rho \left[ {\mathbf {e}} \cdot \delta \nabla (f\ h) \right]&= \ \rho \ \delta \left[ \ e_{x}\ (f\ \nabla h\ +\ h\ \nabla f)\ \right. \nonumber \\&+\ e_{y}\ (f\ \nabla h + h\ \nabla f)\ \nonumber \\&\left. +\ e_{z}\ (f\ \nabla h + h\ \nabla f)\ \right] . \end{aligned}$$

(90)

Applying the differential operators of Eqs. (40–45) on multiplication of the functions $f$ and $h$ results in

$$\begin{aligned} {\mathcal {D}}^{xx}(f\ h)&= \ f\ {\mathcal {D}}^{xx} h + h\ {\mathcal {D}}^{xx} f + \frac{2}{r^2}\ \frac{\partial f}{\partial \varphi }\ \frac{\partial h}{\partial \varphi }, \end{aligned}$$

(91)

$$\begin{aligned} {\mathcal {D}}^{xy}(f\ h)&= \ f\ {\mathcal {D}}^{xy} h\ +\ h\ {\mathcal {D}}^{xy} f\ \nonumber \\&-\ \frac{1}{r^2 \cos \varphi }\ \left( \ \frac{\partial f}{\partial \varphi }\ \frac{\partial h}{\partial \lambda }\ +\ \frac{\partial f}{\partial \lambda }\ \frac{\partial h}{\partial \varphi }\ \right) , \end{aligned}$$

(92)

$$\begin{aligned} {\mathcal {D}}^{xz}(f\ h)&= \ f\ {\mathcal {D}}^{xz} h\ +\ h\ {\mathcal {D}}^{xz} f\ +\ \frac{1}{r}\ \frac{\partial f}{\partial \varphi }\ \frac{\partial h}{\partial r}, \end{aligned}$$

(93)

$$\begin{aligned} {\mathcal {D}}^{yy}(f\ h)&= \ f\ {\mathcal {D}}^{yy} h\ +\ h\ {\mathcal {D}}^{yy} f\ +\ \frac{2}{r^2 \cos ^2 \varphi }\ \frac{\partial f}{\partial \lambda }\ \frac{\partial h}{\partial \lambda }, \end{aligned}$$

(94)

$$\begin{aligned} {\mathcal {D}}^{yz}(f\ h)&= \ f\ {\mathcal {D}}^{yz} h\ +\ h\ {\mathcal {D}}^{yz} f\ -\ \frac{1}{r \cos \varphi }\ \frac{\partial f}{\partial \lambda }\ \frac{\partial h}{\partial r}, \end{aligned}$$

(95)

$$\begin{aligned} {\mathcal {D}}^{zz}(f\ h)&= \ f\ {\mathcal {D}}^{zz} h. \end{aligned}$$

(96)

B: Additional terms from application of differential operators on multiplication of two functions

We now focus on an explicit expressions of the additional terms, i.e., the last terms in Eqs. (91–95). Suppose that the function h is parametrized in terms of substitutions $t$ and $u$, i.e., $h = h(t,u)$. Two forms of the function $f$ are of particular interest:

• 1. $f = \cos \alpha '$, then the additional terms are:

  $$\begin{aligned}&\frac{2}{r^2}\ \frac{\partial f}{\partial \varphi }\ \frac{\partial h}{\partial \varphi }\ =\ -\ \frac{t^2 \sin \alpha ' \sin 2 \alpha }{R^2}\ \frac{\partial h}{\partial u}, \end{aligned}$$

  (97)
  $$\begin{aligned}&\quad -\ \frac{1}{r^2 \cos \varphi }\ \left( \ \frac{\partial f}{\partial \varphi }\ \frac{\partial h}{\partial \lambda }\ +\ \frac{\partial f}{\partial \lambda }\ \frac{\partial h}{\partial \varphi }\ \right) \ \nonumber \\&\quad =\ -\ \frac{t^2 \sin \alpha ' \cos 2 \alpha }{R^2}\ \frac{\partial h}{\partial u}, \end{aligned}$$

  (98)
  $$\begin{aligned}&\frac{1}{r}\ \frac{\partial f}{\partial \varphi }\ \frac{\partial h}{\partial r}\ =\ \frac{t^3 \sin \alpha ' \sin \alpha }{R^2 \sqrt{1 - u^2}}\ \frac{\partial h}{\partial t}, \end{aligned}$$

  (99)
  $$\begin{aligned}&\frac{2}{r^2 \cos ^2 \varphi }\ \frac{\partial f}{\partial \lambda }\ \frac{\partial h}{\partial \lambda }\ =\ \frac{t^2 \sin \alpha ' \sin 2 \alpha }{R^2}\ \frac{\partial h}{\partial u}, \end{aligned}$$

  (100)
  $$\begin{aligned}&-\ \frac{1}{r \cos \varphi }\ \frac{\partial f}{\partial \lambda }\ \frac{\partial h}{\partial r}\ =\ \frac{t^3 \sin \alpha ' \cos \alpha }{R^2 \sqrt{1 - u^2}}\ \frac{\partial h}{\partial t}. \end{aligned}$$

  (101)

• 2. $f = \sin \alpha '$, then the additional terms are:

  $$\begin{aligned}&\frac{2}{r^2}\ \frac{\partial f}{\partial \varphi }\ \frac{\partial h}{\partial \varphi }\ =\ \frac{t^2 \cos \alpha ' \sin 2 \alpha }{R^2}\ \frac{\partial h}{\partial u}, \end{aligned}$$

  (102)
  $$\begin{aligned}&- \frac{1}{r^2 \cos \varphi }\ \left( \ \frac{\partial f}{\partial \varphi }\ \frac{\partial h}{\partial \lambda }\ +\ \frac{\partial f}{\partial \lambda }\ \frac{\partial h}{\partial \varphi }\ \right) \ \nonumber \\&\quad = \frac{t^2 \cos \alpha ' \cos 2 \alpha }{R^2}\ \frac{\partial h}{\partial u}, \end{aligned}$$

  (103)
  $$\begin{aligned}&\frac{1}{r}\ \frac{\partial f}{\partial \varphi }\ \frac{\partial h}{\partial r}\ =\ -\ \frac{t^3 \cos \alpha ' \sin \alpha }{R^2 \sqrt{1 - u^2}}\ \frac{\partial h}{\partial t}, \end{aligned}$$

  (104)
  $$\begin{aligned}&\frac{2}{r^2 \cos ^2 \varphi }\ \frac{\partial f}{\partial \lambda }\ \frac{\partial h}{\partial \lambda }\ =\ -\ \frac{t^2 \cos \alpha ' \sin 2 \alpha }{R^2}\ \frac{\partial h}{\partial u}, \end{aligned}$$

  (105)
  $$\begin{aligned}&-\ \frac{1}{r \cos \varphi }\ \frac{\partial f}{\partial \lambda }\ \frac{\partial h}{\partial r}\ =\ -\ \frac{t^3 \cos \alpha ' \cos \alpha }{R^2 \sqrt{1 - u^2}}\ \frac{\partial h}{\partial t}. \end{aligned}$$

  (106)

C: Application of differential operators on $\cos \alpha '$ and $\sin \alpha '$

To derive the integral formula transforming the principal components of the deflection of the vertical onto SST data, application of the differential operators of Eqs. (5) and (16) to $\cos \alpha '$ and $\sin \alpha '$ is required. The explicit expressions are of the form, see, e.g., (Winch and Roberts 1995),

$$\begin{aligned} {\mathcal {D}}^{x} \cos \alpha '&= - \frac{t \sin \alpha ' \sin \alpha }{R \sqrt{1-u^2}}, \end{aligned}$$

(107)

$$\begin{aligned} {\mathcal {D}}^{y} \cos \alpha '&= \ \frac{t \sin \alpha ' \cos \alpha }{R \sqrt{1-u^2}}, \end{aligned}$$

(108)

$$\begin{aligned} {\mathcal {D}}^{z} \cos \alpha '&= \ 0 , \end{aligned}$$

(109)

$$\begin{aligned} {\mathcal {D}}^{x} \sin \alpha '&= \ \frac{t \cos \alpha ' \sin \alpha }{R \sqrt{1-u^2}}, \end{aligned}$$

(110)

$$\begin{aligned} {\mathcal {D}}^{y} \sin \alpha '&= \ -\ \frac{t \cos \alpha ' \cos \alpha }{R \sqrt{1-u^2}}, \end{aligned}$$

(111)

$$\begin{aligned} {\mathcal {D}}^{z} \sin \alpha '&= \ 0. \end{aligned}$$

(112)

For the mathematical derivations associated to ITs of the principal components of the deflection of the vertical onto GGs, we need also expressions originating from the application of the differential operators ${\mathcal {D}}^{xx}$, ${\mathcal {D}}^{xy}$, ${\mathcal {D}}^{xz}$, ${\mathcal {D}}^{yy}$ and ${\mathcal {D}}^{yz}$ on $\cos \alpha '$ and $\sin \alpha '$. These expressions have the form:

$$\begin{aligned} {\mathcal {D}}^{xx} \cos \alpha '&= \ -\ \frac{t^2 \left( \cos \alpha ' - \cos \alpha ' \cos 2 \alpha + 2 u \sin \alpha ' \sin 2 \alpha \right) }{2 R^2 (1 - u^2)}, \end{aligned}$$

(113)

$$\begin{aligned} {\mathcal {D}}^{xy} \cos \alpha '&= \ -\ \frac{t^2 \left( \cos \alpha ' \sin 2 \alpha + 2 u \sin \alpha ' \cos 2 \alpha \right) }{2 R^2 (1 - u^2)}, \end{aligned}$$

(114)

$$\begin{aligned} {\mathcal {D}}^{xz} \cos \alpha '&= \ \frac{t^2 \sin \alpha ' \sin \alpha }{R^2 \sqrt{1 - u^2}}, \end{aligned}$$

(115)

$$\begin{aligned} {\mathcal {D}}^{yy} \cos \alpha '&= \ -\ \frac{t^2 \left( \cos \alpha ' + \cos \alpha ' \cos 2 \alpha - 2 u \sin \alpha ' \sin 2 \alpha \right) }{2 R^2 (1 - u^2)}, \end{aligned}$$

(116)

$$\begin{aligned} {\mathcal {D}}^{yz} \cos \alpha '&= \ \frac{t^2 \sin \alpha ' \cos \alpha }{R^2 \sqrt{1 - u^2}}, \end{aligned}$$

(117)

$$\begin{aligned} {\mathcal {D}}^{xx} \sin \alpha '&= \ -\ \frac{t^2 \left( \sin \alpha ' - \sin \alpha ' \cos 2 \alpha - 2 u \cos \alpha ' \sin 2 \alpha \right) }{2 R^2 (1 - u^2)}, \end{aligned}$$

(118)

$$\begin{aligned} {\mathcal {D}}^{xy} \sin \alpha '&= \ -\ \frac{t^2 \left( \sin \alpha ' \sin 2 \alpha - 2 u \cos \alpha ' \cos 2 \alpha \right) }{2 R^2 (1 - u^2)}, \end{aligned}$$

(119)

$$\begin{aligned} {\mathcal {D}}^{xz} \sin \alpha '&= \ -\ \frac{t^2 \cos \alpha ' \sin \alpha }{R^2 \sqrt{1 - u^2}},\nonumber \\ \end{aligned}$$

(120)

$$\begin{aligned} {\mathcal {D}}^{yy} \sin \alpha '&= \ -\ \frac{t^2 \left( \sin \alpha ' + \sin \alpha ' \cos 2 \alpha + 2 u \cos \alpha ' \sin 2 \alpha \right) }{2 R^2 (1 - u^2)},\end{aligned}$$

(121)

$$\begin{aligned} {\mathcal {D}}^{yz} \sin \alpha '&= \ -\ \frac{t^2 \cos \alpha ' \cos \alpha }{R^2 \sqrt{1 - u^2}}. \end{aligned}$$

(122)