Classical Physical Geodesy

Abstract

This chapter presents the basics of classical physical geodesy , starting from the definitions of the gravity potential and gravity. The normal gravity potential is derived as the potential of a level ellipsoid plus the rotational potential of the ellipsoid. Normal gravity , defined as the gradient of the normal gravity potential, is presented on and above the level ellipsoid. The basic concepts of the geoid, reference ellipsoid, disturbing potential and geoid height are defined, as well as the classical definitions of gravity anomaly and disturbing potential. After derivation of the fundamental equation of physical geodesy , the gravity field components of the disturbing potential, gravity anomaly and its radial derivative are presented in spherical harmonics, followed by Kaula ’s power rule of the geopotential harmonics. The classical integral formulas of Poisson , Stokes, Hotine , Vening Meinesz and the vertical gradient of gravity anomaly are derived by spherical harmonics. Other spherical integral formulas are derived for determining the gravity anomaly and/or disturbing potential from deflections of the vertical (inverse Vening Meinesz formula ) and gravity gradient components. The classical procedures in geoid determination, including direct and secondary indirect topographic effects and downward continuation of gravity and primary indirect topographic effect on the disturbing potential, are described. Finally, the chapter deals with common height systems, such as geopotential numbers, dynamic, orthometric and normal heights, as well as normal-orthometric heights. Some approximate formulas to correct normal-orthometric heights to orthometric or normal heights are also presented.

Appendices

Appendix 1: Closed-Form Kernels

Referring to Bois (1961), the following formulas can be derived for $0 \le s \le 1$ when using the notations $L(x) = \sqrt {\left( {1 - 2xt + x^{2} } \right)}$ and $L(s) = L$:

$$ S_{1} = \sum\limits_{n = 0}^{\infty } {s^{n} P_{n} \left( t \right)} = 1/L $$

(3.106)

$$ \begin{aligned} S_{2} & = \sum\limits_{n = 1}^{\infty } {\frac{{s^{n} }}{n}} P_{n} \left( t \right) = \int\limits_{0}^{s} {\sum\limits_{n = 1}^{\infty } {x^{n - 1} } P_{n} \left( t \right)} ds = \int\limits_{0}^{s} {\left( {\frac{1}{xL(x)} - \frac{1}{x}} \right)dx} = - \left[ {\ln 2\left( {1 - xt + L} \right)} \right]_{x = 0}^{s} \\ & = - \ln \left( {1 - st + L} \right) + \ln2 \\ \end{aligned} $$

(3.107)

$$ S_{3} = \sum\limits_{n = 0}^{\infty } {\frac{{s^{n + 1} }}{n + 1}} P_{n} \left( t \right) = \int\limits_{0}^{s} {\sum\limits_{n = 0}^{\infty } {x^{n} } P_{n} \left( t \right)} dx = \int\limits_{0}^{s} {\frac{dx}{L\left( x \right)}} = \left[ {\ln 2\left( {x - t + L\left( x \right)} \right)} \right]_{s = 0}^{s} = \ln \frac{s - t + L}{1 - t} $$

(3.108)

and

$$ \begin{aligned} S_{4} & = \sum\limits_{n = 0}^{\infty } {\frac{{s^{n + 2} }}{n + 2}} P_{n} \left( t \right) = \int\limits_{0}^{s} {\sum\limits_{n = 0}^{\infty } {x^{n + 1} } P_{n} \left( t \right)} dx = \int\limits_{0}^{s} {\frac{x}{L(x)}} ds = \left[ {{\text{L}}({\text{x}}) + t\,\ln 2\left( {x - t + L\left( x \right)} \right)} \right]_{x = 0}^{s} \\ & = L - 1 + t\,\ln \frac{s - t + L}{1 - t} \\ \end{aligned} $$

(3.109)

Appendix 2: Solutions to Exercises

Solution to Exercise 3.0:

$\Delta W = \Delta V + \Delta\overline{\Omega } = 2 {\omega} + \left\{ {\begin{array}{*{20}c} {0} \\ { - 4}{\pi}{\rho} \\ \end{array} } \right\} \left\{ {\begin{array}{*{20}l} \rm{exterior\; of\; the\; Earth} \\ \rm{inside\; the\; Earth } \\ \end{array} } \right.$

Solution to Exercise 3.1:

Let us introduce the notations g, a and $g_{\Omega }$ for the magnitudes of gravity, gravitation and the centrifugal force. Then the maximum and minimum gravity are given at the poles and the Equator, respectively, with

$$ g_{\hbox{max} } = a - g_{\Omega } (\theta = 0) = a,\quad {\text{where}}\;a\;{\text{is}}\;{\text{constantly}}\; 9 8 1\;{\text{Gal}}. $$

and

$$ g_{\hbox{min} } = a - g_{\Omega } (\theta = \pi /2) $$

As $g_{\Omega } \left( \theta \right) = \left| {\frac{{\partial \overline{\Omega } }}{\partial r}} \right| = R\omega^{2} \sin^{2} \theta$, it follows with R = 6371 km and $\omega$ according to Table 1.1, that:

$$ g_{\Omega } (\theta = \pi /2) = 6.371 \times 10^{5} \times 7.292^{2} \times 10^{ - 10} = 0. 3 3 9 { }\left[ {\text{Gal}} \right]. $$

Hence, gravity at the Equator is 339 mGal less than at the pole for a spherical, homogeneous Earth model. (For this model the difference is only due to the rotation of the model.)

Solution to Exercise 3.2:

The gravity, gravitation and centrifugal force vectors (with notations adopting those in the previous exercise) are illustrated in Fig. 3.4a. If the Earth stops rotating, vector $\bar{g}$ will move to and be equal to vector $\bar{a}$. From Fig. 3.4b, the sine theorem can be applied, yielding the equation

$$ \frac{\sin \alpha }{{g_{\Omega } }} = \frac{\sin \delta }{a} = \frac{{\sin \left( {\pi - \varphi - \alpha } \right)}}{a} = \frac{{\sin \left( {\varphi + \alpha } \right)}}{a} = \frac{\sin \varphi \,\cos \alpha + \cos \varphi \,\sin \alpha }{a}, $$

which can be simplified to:

$$ \tan \alpha = \frac{{g_{\Omega } \sin \varphi }}{{a - g_{\Omega } \cos \varphi }}. $$

As $\varphi = \pi /3$ and $\alpha$ is a small angel, the equation can be rewritten as:

$$ \alpha \approx \tan \alpha = \frac{{\sqrt 3 g_{\Omega } }}{{2a - g_{\Omega } }}. $$

Fig. 3.4

The gravitation, centrifugal force and gravity a vectors and b their magnitudes in a triangle

In this case, $g_{\Omega } = R\omega^{2} /4 = 0.085$ [Gal], so that the numerical solution for the change of the plumb line is about $(\alpha \approx )$ 15″ towards the north.

Solution to Exercise 3.3

Let us start from the expression

$$ r\Delta g = - H - 2T $$

where

$$ H = r\frac{\partial T}{\partial r} = \bar{r} \cdot grad(T) = x\frac{\partial T}{\partial x} + y\frac{\partial T}{\partial y} + z\frac{\partial T}{\partial z} = ()T. $$

with

$$\left( {} \right) = \left( {x{\partial \over {\partial x}} + y{\partial \over {\partial y}} + z{\partial \over {\partial z}}} \right). $$

Then

$$ \frac{\partial H}{\partial u} = \frac{\partial T}{\partial u} + \left( {} \right)\frac{\partial T}{\partial u} $$

and

$$ \frac{{\partial^{2} H}}{{\partial u^{2} }} = 2\frac{{\partial^{2} T}}{{\partial u^{2} }} + \left( {} \right)\frac{{\partial^{2} T}}{{\partial u^{2} }}, $$

where $u = x,y,z.$

Hence

$$ \Delta H = 2\Delta T + \left( {} \right)\Delta T, $$

and therefore:

$$ \Delta \left( {r\Delta g} \right) = - \Delta H - 2\Delta T = 0. $$

Solution to Exercise 3.4

Insert (3.24b) into (3.23b) and apply it for r _p = R:

$$ \Rightarrow \Delta g = \sum\limits_{n = 2}^{\infty } {\frac{n - 1}{{ {R} }}} {T}_{n} . $$

Insert “the hint”:

$$ \Delta g = \sum\limits_{n = 2}^{\infty } \Delta g_{n} = \sum\limits_{n = 2}^{\infty } {\frac{n - 1}{{ {R} }}} {\Omega}_{n} \Delta g_{n} \Rightarrow {\Omega}_{n} = R/(n - 1). $$

Hence, by applying “the hint” and (3.23b) once more and changing the order of summation and integration one obtains:

$$ T = {R}\sum\limits_{n = 2}^{\infty} {\frac{\Delta g_{n}}{{ {n - 1 } }}} = \frac{R}{4\pi }\iint\limits_{\sigma } {S\left( {\psi } \right)}\Delta gd\sigma, \quad\textit{where}\; {S\left( {\psi } \right)} = \sum\limits_{n = 2}^{\infty} {\frac{2n + 1}{{ {n - 1 } }}}P_{n}(\rm{cos}{\psi}).$$