Observables of Physical Geodesy and Their Analytical Representation

Abstract

As we have shown in Chap. 1, the gravity potential W can be split into a known normal potential U plus the anomalous potential T; thus knowing T means knowing W.

Appendix

1.1 A.1

We want to find a manageable expression for the sum of leveling increments along a line, proving (2.58).

To this aim we go back to (2.55) and substitute

$$\begin{array}{rcl} \mathit{dh} = \mathit{dH} + dN = \mathit{dH} -{\epsilon }_{0} \cdot d{\mathbf{r}}_{0}& & \\ \end{array}$$

in it. We receive (see Fig. 2.2 for the notation)

$$\begin{array}{rcl} \delta L& =& (\epsilon -{\epsilon }_{0}) \cdot d\mathbf{r} +{ \epsilon }_{0} \cdot (d\mathbf{r} - d{\mathbf{r}}_{0}) + \mathit{dH} \\ & =& \left [(\mathbf{n} -{\mathbf{n}}_{0}) - (\nu -{\nu }_{0})\right ] \cdot d{\mathbf{r}}_{h} + \frac{H} {R}{\epsilon }_{0} \cdot d{\mathbf{r}}_{0} + \mathit{dH}, \\ \end{array}$$

because, with a good approximation, $d\mathbf{r} - d{\mathbf{r}}_{0} = \frac{H} {R}d{\mathbf{r}}_{0} + \mathit{dh}\nu$ and ${\epsilon }_{0}$ is orthogonal to $\nu$. Since $\int\limits_{{ \frown \atop AB} }{\epsilon }_{0} \cdot d{\mathbf{r}}_{0}$ is the variation of N, which is at most a few meters, even for points A, B far away dozens of kilometers, and $\frac{H} {R} < 1{0}^{-3}$, we can drop the term $\frac{H} {R}{\epsilon }_{0} \cdot d{\mathbf{r}}_{0}$; in other words we can take $d{\mathbf{r}}_{h} \sim d{\mathbf{r}}_{0}$ in this computation. Now, recalling (1.75), we can write

$$\begin{array}{rcl} \mathbf{n} -{\mathbf{n}}_{0} =\int\limits_{{P}_{0}}^{P}{\nabla }_{ h}\log g\mathit{dH}& & \\ \end{array}$$

and similarly

$$\begin{array}{rcl} \nu -{\nu }_{0} = 0 =\int\limits_{{P}_{0}}^{P}{\nabla }_{ h}\log {\gamma }_{0}\mathit{dh};& & \\ \end{array}$$

where γ₀ is constant, along the vertical line, so that the latter identity reduces to 0 = 0, because $\nu$ is indeed constant along the normal to the ellipsoid. So we have

$$\begin{array}{rcl} \epsilon -{\epsilon }_{0}& =& \int\limits_{{P}_{0}}^{P}{\nabla }_{ h}\log \frac{g} {{\gamma }_{0}}\mathit{dh} =\int\limits_{{P}_{0}}^{P}{\nabla }_{ h}\log \left (1 + \frac{g - {\gamma }_{0}} {{\gamma }_{0}} \right )\mathit{dh}\cong \\ & =& \int\limits_{{P}_{0}}^{P}{\nabla }_{ h}\left (\frac{g - {\gamma }_{0}} {{\gamma }_{0}} \right )\mathit{dh} = {\nabla }_{h}\int\limits_{{P}_{0}}^{P}\left (\frac{g - {\gamma }_{0}} {{\gamma }_{0}} \right )\mathit{dh} + \\ & & -\left (\frac{g(P) - {\gamma }_{0}} {{\gamma }_{0}} \right ){\nabla }_{h}{h}_{P}.\end{array}$$

The last step is justified by the well-known differentiation rule

$$\begin{array}{rcl}{ D}_{x}\int\limits_{0}^{g(x)}f(x,t)dt = f[x,g(x)] \cdot g \prime (x) +\int\limits_{0}^{g(x)} \frac{\partial } {\partial x}f(x,t)dt.& & \\ \end{array}$$

Summarizing and going back to (2.118), we find

$$\begin{array}{rcl} \delta L = {\nabla }_{h}\left [\int\limits_{{P}_{0}}^{P}\left (\frac{g - {\gamma }_{0}} {{\gamma }_{0}} \right )\mathit{dh}\right ] \cdot d{\mathbf{r}}_{h} -\frac{g(P) - {\gamma }_{0}} {{\gamma }_{0}} \mathit{dh} + \mathit{dH}.& &\end{array}$$

(2.118)

As it is shown in Sect. 2.4,

$$\begin{array}{rcl} \int\limits_{{P}_{0}}^{P}\frac{g - {\gamma }_{0}} {\gamma } \mathit{dh}\cong{N}_{{P}_{0}} - {\zeta }_{P},& &\end{array}$$

(2.119)

for which an explicit formula, as function of H, is given by (2.71). Moreover in (2.118) we can substitute δL back for dh to the effect that one can write

$$\begin{array}{rcl} \delta L = d(N - \zeta ) -\frac{g(P) - {\gamma }_{0}} {{\gamma }_{0}} \delta L + \mathit{dH},& & \\ \end{array}$$

which finally integrated along the line ${\frown \atop AB}$ yields

$$\begin{array}{rcl}{ \Delta }_{AB}L = ({N}_{B} - {\zeta }_{B}) - ({N}_{A} - {\zeta }_{A}) -\int\limits_{{ \frown \atop AB} }\frac{g - {\gamma }_{0}} {{\gamma }_{0}} \delta L + {H}_{B} - {H}_{A},\quad & &\end{array}$$

(2.120)

namely the formula we wanted to prove.

1.2 A.2

We want to prove formula (2.65) for the vertical gradient of Δg as function of the normal height h ^∗. We adopt symbols and notation of Sect. 2.4. To this aim we note first of all that in (2.63) we need Δg, so that we have to convert (2.64) into an equation for the vertical continuation of Δg.

To this aim we write the analogous of (2.64) for the normal field, i.e.

$$\begin{array}{rcl} \frac{\partial \gamma } {\partial h} = -2{\mathcal{C}}_{0}\gamma - 2{\omega }^{2};& &\end{array}$$

(2.121)

note that (2.121) can be written for any point along the ellipsoidal normal, for instance at Q ^∗ instead of Q, but we are not allowed to substitute $\frac{\partial } {\partial {h}^{{_\ast}}}$ for $\frac{\partial } {\partial h}$ in (2.121) because h ^∗ is not a linear function of h. So we must transform $\frac{\partial } {\partial H}$ in (2.64) into $\frac{\partial } {\partial h}$, then we subtract (2.121) computed at Q ^∗ from (2.64) and finally we transform $\frac{\partial } {\partial h}$ into $\frac{\partial } {\partial {h}^{{_\ast}}}$.

As for $\frac{\partial g} {\partial H}$ we can write

$$\begin{array}{rcl} \frac{\partial g} {\partial H} = \mathbf{n} \cdot \nabla g = (\mathbf{n} -\nu ) \cdot \nabla g + \nu \cdot \nabla g\cong\epsilon \cdot \nabla \gamma + \nu \cdot \nabla g.& &\end{array}$$

(2.122)

In (2.122) we evaluate the order of magnitude

$$\begin{array}{rcl} O(\epsilon \cdot \nabla \gamma )& =& O(\epsilon \cdot {\nabla }_{t}\gamma ) = O\left (\vert \epsilon \vert \frac{1} {R} \frac{\partial \gamma } {\partial \varphi }\right ) \\ & =& O\left (\vert \epsilon \vert \frac{5 \cdot 1{0}^{-3}\gamma } {R} \right ) = 5 \cdot 1{0}^{-7} \frac{\gamma } {R}\end{array}$$

(2.123)

where we have used (1.145) and (1.181).

Therefore this term contributes to g, and then to Δg, at height h with an error δΔg of the order of magnitude of $5 \cdot 1{0}^{-7} \frac{\gamma } {R}h$, or, equivalently, of $5 \cdot 1{0}^{-7} \frac{\gamma } {R}H$.

As a consequence of (2.63), to evaluate the error induced by neglecting $\epsilon \cdot \nabla g$ in computing N − ζ one has to assess the order of magnitude of δΔg integrated in H, i.e., observing that in the topographic layer one has $O\left (\frac{H} {R} \right ) \sim 1{0}^{-3}$,

$$\begin{array}{rcl} O(\delta [N - \zeta ]) = O\left (\epsilon \cdot \nabla \gamma \frac{{H}^{2}} {\gamma } \right ) \sim 5 \cdot 1{0}^{-7}\frac{{H}^{2}} {R} \sim 5 \cdot 1{0}^{-10}H;& &\end{array}$$

(2.124)

this shows that the term in question doesn’t matter in our computation. So we can write

$$\begin{array}{rcl} \frac{\partial g(Q)} {\partial H} \cong\nu \cdot \nabla g(Q) = \frac{\partial } {\partial h}g(Q)& & \\ \end{array}$$

in (2.64) and work on the right hand side with an obvious approximation to arrive at the equation

$$\begin{array}{rcl} \frac{\partial g(Q)} {\partial h} = -2[\mathcal{C}(Q) -{\mathcal{C}}_{0}({Q}^{{_\ast}})]\gamma - 2{\mathcal{C}}_{ 0}({Q}^{{_\ast}})g(Q) + 4\pi G\rho - 2{\omega }^{2}.& &\end{array}$$

(2.125)

If we can prove that in (2.125) the term

$$\begin{array}{rcl} [\mathcal{C}(Q) -{\mathcal{C}}_{0}({Q}^{{_\ast}})]\gamma \cong[\mathcal{C}(Q) -{\mathcal{C}}_{ 0}(Q)]\gamma + {\mathcal{C} \prime }_{0}(Q)\gamma \zeta & &\end{array}$$

(2.126)

is negligible, we are left with the equation

$$\begin{array}{rcl} \frac{\partial } {\partial h}g(Q) = -2{\mathcal{C}}_{0}({h}^{{_\ast}})g(Q) + 4\pi G\rho - 2{\omega }^{2}& &\end{array}$$

(2.127)

We evaluate (2.126) in two steps. First we use the following estimate, derived from several numerical experiments,

$$\begin{array}{rcl} O([\mathcal{C}(Q) -{\mathcal{C}}_{0}(Q)])\cong\frac{1{0}^{-3}} {R} \ ;& &\end{array}$$

(2.128)

as always, O( ) means the order of magnitude of the maximum value, as the standard deviation of $\mathcal{C}(Q) -\mathcal{C}({Q}_{0})$ is easily one order of magnitude smaller. Then we evaluate the impact of this term on N − ζ by considering the corresponding error $[\mathcal{C}(Q) -{\mathcal{C}}_{0}(Q)]\gamma$ integrated in H, once to give its impact on g, and then a second time, divided by γ, to give the impact on N − ζ (see (2.63)). The result is

$$\begin{array}{rcl} O(\delta [\mathcal{N}- \zeta ]) = O\left ([\mathcal{C}(Q) -{\mathcal{C}}_{0}(Q)]\gamma \cdot \frac{{H}^{2}} {\gamma } \right )\cong O\left (\frac{1{0}^{-3}H} {R} \cdot H\right ) = 1{0}^{-6}H,& & \\ \end{array}$$

which is negligible because it gives at maximum an error of 1 mm/km of altitude. As for the second addendum in (2.126) we use the rough approximation

$$\begin{array}{rcl} \vert {\mathcal{C} \prime }_{0}\vert \cong \frac{1} {{R}^{2}},& & \\ \end{array}$$

yielding

$$\begin{array}{rcl} O(\delta [N - \zeta ]) = O\left ({\mathcal{C} \prime }_{0}\gamma \zeta \cdot \frac{{H}^{2}} {\gamma } \right ) = O\left (\frac{{H}^{2}} {{R}^{2}} \zeta \right ) = 1{0}^{-6}\zeta ;& & \\ \end{array}$$

this is totally negligible since it is below the millimeter for any height up to 6,000 m.

So we know that (2.127) is correct and we can subtract (2.121) from it, to get

$$\begin{array}{rcl} \frac{\partial } {\partial h}g(Q) - \frac{\partial } {\partial h}\gamma ({Q}^{{_\ast}}) = -2{\mathcal{C}}_{ 0}({Q}^{{_\ast}})[g(Q) - \gamma ({Q}^{{_\ast}})] + 4\pi G\rho ,& & \\ \end{array}$$

namely

$$\begin{array}{rcl} \frac{\partial } {\partial h}\Delta g = -2{\mathcal{C}}_{0}({h}^{{_\ast}})\Delta g + 4\pi G\rho.& &\end{array}$$

(2.129)

Now from

$$\begin{array}{rcl} h = {h}^{{_\ast}} + \zeta & & \\ \end{array}$$

we see that (cf. (2.62))

$$\begin{array}{rcl} \frac{\partial } {\partial h} = (1 - \zeta \prime ) \frac{\partial } {\partial {h}^{{_\ast}}} = \left (1 + \frac{\Delta g} {\gamma } \right ) \frac{\partial } {\partial {h}^{{_\ast}}}.& & \\ \end{array}$$

So, omitting all second order terms that are easily verified to be negligible, we write (2.129) in the form

$$\begin{array}{rcl} \frac{\partial } {\partial {h}^{{_\ast}}}\Delta g = -2{\mathcal{C}}_{0}({h}^{{_\ast}})\Delta g + 4\pi G\rho.& &\end{array}$$

(2.130)

Finally, we want to show that in (2.130) we can consider ${\mathcal{C}}_{0}$ and ρ as constants.

We reason again in terms of orders of magnitude of maximum errors. So if we use the rough estimate

$$\begin{array}{rcl} O(\vert {\mathcal{C}}_{0}(0) -{\mathcal{C}}_{0}({h}^{{_\ast}})\vert ) = \frac{1} {R} - \frac{1} {R + {h}^{{_\ast}}}\cong\frac{{h}^{{_\ast}}} {{R}^{2}},& & \\ \end{array}$$

we see that one has for the error δ(N − ζ), after the usual double integration on H,

$$\begin{array}{rcl} O(\delta [N - \zeta ]) = O\left ( \frac{{h}^{{_\ast}}} {{R}^{2}}\gamma \frac{{H}^{2}} {\gamma } \right ) = 1{0}^{-6}{h}^{{_\ast}},& & \\ \end{array}$$

namely 1 mm/km of altitude in worst case.

In parallel one can consider that in the crust ρ can vary around its mean value, $\overline{\rho } = 2.67\mathrm{\,g}{\mathrm{\,cm}}^{-3}$, by no more than 10%, but

$$\begin{array}{rcl} 0,1 \cdot 4\pi G\overline{\rho }\cong 0.02\mathrm{\,mGal}{\mathrm{\,m}}^{-1}& & \\ \end{array}$$

so that the corresponding error on δ[N − ζ] is of the order of

$$\begin{array}{rcl} O(\delta [N - \zeta ])& =& O\left (0.1 \cdot 4\pi G\overline{\rho }\frac{{H}^{2}} {\gamma } \right ) \\ & =& 2 \cdot 1{0}^{-8}{H}^{2}(H\mbox{ in meters})\end{array}$$

(2.131)

Therefore, with H = 10³ m, our maximum error becomes 2 cm, which is certainly not too small. Yet the following has to be considered: first of all sometimes we have geological maps that could help us to use a value of ρ good up to 1%, giving in (2.131) an error smaller by one order of magnitude; a variation of 0. 267 g cm^− 3 in the surface density has to be considered very large. Finally, this is certainly the most uncertain information we can have in physical geodesy so that, when we really need N − ζ, we have to live with errors of this magnitude.

So now (2.130) can be written as

$$\begin{array}{rcl} \frac{\partial } {\partial {h}^{{_\ast}}}\Delta g = -2{\mathcal{C}}_{0}\Delta g + 4\pi G\rho & &\end{array}$$

(2.132)

with ${\mathcal{C}}_{0}$ and ρ considered as constants.