I assume that the mantle and crust heat flux is proportional to the average surface heat flux. [11] argued that mantle heat flux represents 40% of the regional average surface heat flux. Despite the fact that their measures were based on a small dataset, I consider here that they are valid up to a minimum scale of about 300 km [12] . Average heat flux data suggest an empirical relationship of the form $Q=Q_o+b\cdot \bar{H}$.

For the purposes of this study, I redefine the relationship as follows:

$\bar{Q}=\overline{Q_o}+b\cdot \bar{H}+q_o$ (14)

where $\bar{Q}$ represents average heat flux across the designated area and $\bar{H}$ is heat production, $\overline{Q_o}$ is average reduced heat flux, $b$ represents the thickness of a shallow layer enriched by radiogenic elements, while $q_o$ depends on latitude and its value of around 12 mW/m² is in direct relation with $Ct\left(\varphi \right)$, which is defined above. Equation (14) reflects changes in average heat flux on a larger scale (>200 km) and is based on a very large dataset. It implies that reduced heat flux $\overline{Q_o}$ is the same at a certain depth and latitude of the crust in all areas. The assumed value $\overline{Q_o}$ is clearly shown in (9), which expresses the gravitational action of the Sun on the moving masses of the Earth. This data can be checked against the $Q_o\text{data}$ provided in Table 1 (below), from the study by [13] .

To compare equation (9) (heat production $Q_o$ due to the gravitational action of the Sun on the Earth, $W$) with $Q_o\text{data}$ (reduced heat flux, Table 1 ), I must extend (9) in order to calculate heat flux up to depths of the order of 660 km for the whole of lithosphere and asthenosphere, and the upper mantel. The calculation of heat flow and heat production is based on a 1m² column that is approximately 660 km deep. In this area, most heat transfer occurs through thermal conduction, and I can assume that the heat produced as a result of the gravitational effect of the Sun is equal across wide areas and therefore comparable to the reduced heat flow $Q_o$ shown in (14). Heat propagation is weakest in the lower mantle; it is no longer completely the result of thermal conduction but varies according to the geography of the area and is comparable to heat production $\bar{H}$ shown in (14).

Reduced heat flow ( $Q_o$) in the 660 km-deep lithosphere asthenosphere and the upper mantel is calculated by applying (9) ( $W$). However, this relation represents only one 24-hour rotation of the Earth and relies on a defined value for the position of the Earth ( $\lambda$) around the Sun. It is not, therefore, a weighted average over an entire year. To find the weighted yearly average, I must integrate (9) at angle $\lambda$ from 0 to 2π as follows:

$W_t=Q_o=\frac{1}{T_{earthsun}}\cdot \frac{365}{2\pi }\cdot {\displaystyle {\int }_0^{2\pi }W\cdot \text{d}\lambda }$ (15)

In (15), $W_t$ represents the heat energy produced all the year in the elementary volume measured in watts per square meter. $T_{earthsun}$ represents the time for one revolution of the Earth around the Sun.

There are several rheological models of the continental lithosphere [24] [25] . In broad terms, I consider the lithosphere (including the crust) as a rigid material about 100 km deep that responds to tectonic forces. It sits on top of the ductile upper mantle and of the transition zone that are approximately 560 km deep, where heat is exchanged with the solar gravitational reaction. Consequently, the limits of the integration of (9) for the calculation of reduced heat flow $Q_o$ are as follows: $R_1=R_{earth}-660\text{km}$ and $R_2=R_{earth}-100\text{km}$ and the limits of integration of (12) for the calculation of tectonic forces and the strain rate are: $R_2=R_{earth}$ and $R_1=R_{earth}-100\text{km}$.

The following calculates the heat flow for $\Delta \varphi$ and $\Delta \theta$ (of a length of 1 m):

$\Delta \theta =\frac{1\text{m}}{r_{earth}\cdot \sin \left(\varphi \left(latitude\right)\right)}\text{,}\Delta \phi =0.00000899\cdot \frac{\pi }{180}$ or $\varphi \left(latitude\right)=\pi \cdot \frac{90-latitude}{180}$

The application of (15) leads to the following reduced heat flow ( $Q_o=\text{Calculus}$ and $Q_o\text{data}=\text{data}$) Graph.

It is interesting to note that Figure 9 shows that reduced heat flow $Q_o=\text{Calculus}$ decreases at higher latitudes. This is not unusual, and several other authors have noted the phenomenon. For example, Figure 10 shows the results from a study of the north-northwest of Western Australia (Perth Basin) carried out by the company [26] . For further discussion of the reduced heat flow problem see [10] .

The action of Sun on the production of heat in the Earth is directly related to the movement of the continental and oceanic plates. I use the heat flow relationships defined above in my new concept of the movement of the tectonics plates. Here I only discuss the action of the Sun on the moving masses of the Earth and neglect the relatively small contribution of the planets and the Moon. To apply (12) to plate tectonics, I must find the relation between the strain rate and tensile energy of the rigid lithosphere. The laws of viscosity allow us to write the following equation:

given the constraint (Pa) ≈ viscosity (Pa∙s) × strain rate (s⁻¹)

for example, when the pressure from the plate subduction process gives rise to a viscosity. The viscous resistance of the mantle due to plate movements is given by:

$\sigma \approx \eta \cdot \frac{\text{d}{\epsilon }_v}{\text{d}t}$ (16)

In this equation, $\sigma$ represents the constraint in Pa, $\eta$ is viscosity measured

in Pa∙s and $\frac{\text{d}{\epsilon }_v}{\text{d}t}$ is the viscous strain rate measured in s⁻¹. The constraint $\sigma$ is

similar to the pressure $P$ of plate subduction over the mantle and I can therefore assimilate $P$ as $\sigma$. For example, viscosity $\eta$ will be in the order of

$\eta =\frac{\alpha \cdot \rho \cdot \Delta T\cdot g\cdot e\cdot d}{v}\approx {10}^{22}\text{Pa}\cdot \text{s}$, given a subducting plate with thickness e = 100

km, upper mantle thickness d = 700 km, speed $v$ equal to 5cm/year, $\Delta T\approx 20\text{K}$, $\alpha$(coefficient of thermic expansion) equal to 5 × 10⁻⁵ K⁻¹, $\rho =3000\text{kg}/{\text{m}}^{\text{3}}$ and $g=10\text{m}\cdot {\text{s}}^{-2}$.

More accurate calculations lead to depth-dependent values for the viscosity of the upper mantle, in particular the lithosphere (excluding the crust) of 2 × 10²⁰ to 10²² Pa∙s [27] . Here, I take an average value of 1 × 10²² Pa∙s. However, simplification is required, in order to move a plate, the pressure must be at least equal to the resistance due to the viscosity $\eta$ of the upper mantle. To complete (16), which defines the Newtonian viscous behavior, I must also take into account elastic solid behavior together with ${\epsilon }_e$ the elastic deformation, i.e., the following pair of equations:

$\sigma \approx \eta \cdot \frac{\text{d}{\epsilon }_v}{\text{d}t}$

$\sigma \approx E\cdot {\epsilon }_e$ (17)

${\epsilon }_t={\epsilon }_v+{\epsilon }_e$

From (17) I can easily deduce the following relationship:

$\frac{\text{d}{\epsilon }_t}{\text{d}t}=\frac{1}{E}\cdot \left[\frac{\sigma }{{\tau }_m}+\frac{\text{d}\sigma }{\text{d}t}-\frac{\text{d}E}{\text{d}t}\cdot \frac{\sigma }{E}\right]$ (18)

The ratio $\frac{\text{d}{\epsilon }_t}{\text{d}t}$ represents the total strain rate, $E$ is the elastic modulus in Pa (similar to Young’s modulus) and ${\tau }_m$ is Maxwell’s relaxation time, equal to $\frac{\eta }{E}$.

The elastic modulus $E$ and the viscous resistance $\sigma$ are equivalent to pressure forces. In order to apply this to plate movements, I must find a correspondence between these two entities and the dissipated energy responsible for lithospheric movements. One close constant is:

$\sigma \propto W_{\sigma }\text{and}E\propto W_{\eta }$ (19)

In this equation, $W_{\sigma }$ represents the energy necessary to maintain constant motion of the plate, this is a kinetic energy. $W_{\eta }$ represents the energy required to overcome the viscous forces between the lithosphere and the asthenosphere in the upper mantle of the earth. The pressures $P_{\sigma }$ and $P_{\eta }$ in the energy function can be written as follows:

$P_{\sigma }=W_{\sigma }\cdot \frac{1}{d\cdot S}\approx \sigma \text{and}{P}_{\eta }\left(E\right)=W_{\eta }\cdot \frac{1}{d\cdot S}\approx E$.

Without changing the meaning of (18), for a known displacement $d$ and surface $S$, $\sigma$ can be replaced by $W_{\sigma }$, and $E$ can be replaced by $W_{\eta }$, which becomes:

$\frac{\text{d}{\epsilon }_t}{\text{d}t}=\frac{1}{W_{\eta }}\cdot \left[\frac{W_{\sigma }}{{\tau }_m}+\frac{\text{d}{W}_{\sigma }}{\text{d}t}-\frac{\text{d}{W}_{\eta }}{\text{d}t}\cdot \frac{W_{\sigma }}{W_{\eta }}\right]$ (20)

The solution of (20) is relatively easy to find using ${\epsilon }_t$ as the coefficient of deformation:

$W_{\sigma }=W_{\eta }\cdot {\epsilon }_t\cdot \frac{1}{2+\frac{T}{{\tau }_m}}$ (21)

This relation gives the kinetic energy $W_{\sigma }$ needed to maintain the movement of the plate as a function of the total energy $W_{\eta }$ applied to it. Most of the energy $W_{\eta }$ cancels the forces resulting from the viscosity of the asthenosphere.

The relaxation time of the plate-mantle system is of the order of a million years, and it is important in the calculation of plate speed, where ${\tau }_m$ is much smaller than $T$, $T$ predominates. $T$ represents the orogenic (or tectonic) cycle, and it is the current age of the Earth $t_{earth}=4.5\times {10}^9\text{years}$ (generally written as

450 × 10⁶ years or $\frac{t_{earth}}{10}$). $T$ takes the value $t_{earth}+t$ where $t$ varies (here $t$

is the variable time) from 0 to $T_{earth}$ (24h). Finally, I obtain the simple relationship $W_{\sigma }$:

$W_{\sigma }=W_{\eta }\cdot {\epsilon }_t\cdot \frac{1}{2+\frac{t_{earth}+t}{{\tau }_m}}$ (22)

Obviously, $t$ is much smaller than $t_{earth}$ is almost neglible in (22); however, it is important at the time of the early formation of the Earth and the transformation of Pangea.

As stated above, $W_{\sigma }$ is the kinetic energy of the plate ( $W_{\sigma }\approx \frac{1}{2}\Delta M\cdot v^2$), which

allows us to formulate the lateral and vertical velocities in the three directions $e_r,e_{\theta }$ and $e_{\varphi }$. I can replace $W_{\eta }$ in (22) by $W_r,W_{\theta }$ and $W_{\varphi }$ in (12). Similarly, for speed ${\nu }_{\theta }$ towards the west (or east), ${\nu }_{\varphi }$ to the north (or south), and speed ${\nu }_r$ toward (or away from) the center of the Earth, I can write equivalent equations with the same form as (12). The kinetic energy is given by:

${\nu }_{\varphi }=\sqrt{\frac{\frac{2}{2+\frac{t_{earth}+t}{{\tau }_m}}\cdot {\epsilon }_{\varphi t}\cdot \frac{1}{2\pi }\cdot {\displaystyle {\int }_0^{2\pi }W\varphi \cdot \text{d}\lambda }}{{\displaystyle {\int }_0^{\Delta \theta }{\displaystyle {\int }_{\varphi \left(latitude\right)}^{\varphi \left(latitude\right)+\Delta \varphi }{\displaystyle {\int }_{R_1}^{R_2}{r}^2\cdot \sin \varphi \cdot \mu \left(r\right)\cdot \text{d}r\cdot \text{d}\varphi \cdot \text{d}\theta }}}}}$ (23)

${\nu }_{\theta }=\sqrt{\frac{\frac{2}{2+\frac{t_{earth}+t}{{\tau }_{m\theta }}}\cdot {\epsilon }_{\theta t}\cdot \frac{1}{2\pi }\cdot {\displaystyle {\int }_0^{2\pi }{W}_{\theta }\cdot \text{d}\lambda }}{{\displaystyle {\int }_0^{\Delta \theta }{\displaystyle {\int }_{\varphi \left(latitude\right)}^{\varphi \left(latitude\right)+\Delta \varphi }{\displaystyle {\int }_{R_1}^{R_2}{r}^2\cdot \sin \varphi \cdot \mu \left(r\right)\cdot \text{d}r\cdot \text{d}\varphi \cdot \text{d}\theta }}}}}$ (24)

${\nu }_r=\sqrt{\frac{\frac{2}{2+\frac{t_{earth}+t}{{\tau }_{mr}}}\cdot {\epsilon }_{rt}\cdot \frac{1}{2\pi }\cdot {\displaystyle {\int }_0^{2\pi }{W}_r\cdot \text{d}\lambda }}{{\displaystyle {\int }_0^{\Delta \theta }{\displaystyle {\int }_{\varphi \left(latitude\right)}^{\varphi \left(latitude\right)+\Delta \varphi }{\displaystyle {\int }_{R_1}^{R_2}{r}^2\cdot \sin \varphi \cdot \mu \left(r\right)\cdot \text{d}r\cdot \text{d}\varphi \cdot \text{d}\theta }}}}}$ (25)

(modification for the relationship (23), read τ_mφ instead of τ_m)

In these equations ${\epsilon }_{\varphi t},{\epsilon }_{\theta t},{\epsilon }_{rt}$ represent the strain coefficients for the three directions $e_r,e_{\theta }$ and $e_{\varphi }$ (see Figure 5 ) and ${\tau }_{m\theta },{\tau }_{m\varphi },{\tau }_{mr}$ represent the relaxation time for the same orientations. The denominators in equations 23, 24 and 25 represent the portion of mass $\Delta M$ of the plate in question. In order to simplify these equations, the energies $W_{\sigma }$ and $W_{\eta }$ are applied between the lithosphere, which has viscosity $\eta$ and is 100 km thick (including the rigid crust) and the ductile asthenosphere in the upper mantle that is 550 km thick ( Figure 7 ).

As Global Positioning System (GPS) measurements of plate speed are relatively accurate, I take them as my reference for calculating the coefficients of deformations ${\epsilon }_{\varphi t},{\epsilon }_{\theta t},{\epsilon }_{rt}$. It is then simple to solve equations 23, 24, and 25 and to calculate these variables ${\epsilon }_{\varphi t},{\epsilon }_{\theta t},{\epsilon }_{rt}$. I have along of $e_{\theta }$ ( Figure 4 ):

$\Delta M={\displaystyle {\int }_0^{\Delta \theta }{\displaystyle {\int }_{\varphi \left(latitude\right)}^{\varphi \left(latitude\right)+\Delta \varphi }{\displaystyle {\int }_{R_1}^{R_2}{r}^2\cdot \sin \varphi \cdot \mu \left(r\right)\cdot \text{d}r\cdot \text{d}\varphi \cdot \text{d}\theta }}}$

$W_{\theta \eta }=\frac{1}{2\pi }\cdot {\displaystyle {\int }_0^{2\pi }{W}_{\theta }\cdot \text{d}\lambda }\text{and}{\tau }_m=\frac{\eta }{W_{\theta }}\cdot \left({\displaystyle {\int }_{\varphi \left(latitude\right)}^{\varphi \left(latitude\right)+\Delta \varphi }{\displaystyle {\int }_{R_1}^{R_2}r\cdot \text{d}r\cdot \text{d}\varphi }}\right)\cdot {\nu }_{\theta }\cdot t$

Note that I use the weighted annual average energy for the period $t=T_{earth}$ or 24 hours (a full rotation of the Earth).

In the expression of ${\tau }_m$ the part $\frac{W_{\theta }}{\left({\displaystyle {\int }_{\varphi \left(latitude\right)}^{\varphi \left(latitude\right)+\Delta \varphi }{\displaystyle {\int }_{R_1}^{R_2}r\cdot \text{d}r\cdot \text{d}\varphi }}\right)\cdot {\nu }_{\theta }\cdot t}$ represents

the lateral pressure exerted on the plate with mass $\Delta M$, moving at speed ${\nu }_{\theta }$ and displacement ${\nu }_{\theta }\cdot t$ where $t$ ranges from 0 to $T_{earth}$ with coordinates $latitude,\Delta \varphi ,\Delta \theta ,R_1,R_2,R_3$(see equation 19 and following), (without the index $t$ for total,) the total deformation coefficients ${\epsilon }_{\theta t}$ become either:

$\begin{array}{c}{\epsilon }_{\theta }=\frac{\frac{1}{2}\cdot {\nu }_{\theta }^2\cdot \Delta M\cdot \left(2+\frac{t_{earth}+t}{{\tau }_m}\right)}{W_{\theta }}\\ =\frac{1}{2}\cdot {\nu }_{\theta }^2\cdot \Delta M\left[\frac{t_{earth}+t}{\eta }\cdot \frac{1}{\left({\displaystyle {\int }_{\varphi \left(latitude\right)}^{\varphi \left(latitude\right)+\Delta \varphi }{\displaystyle {\int }_{R_1}^{R_2}r\cdot \text{d}r\cdot \text{d}\varphi }}\right)\cdot {\nu }_{\theta }\cdot t}+\frac{2}{W_{\theta }}\right]\end{array}$ (26)

If I replace $\theta$ by $\varphi$ or $r$ and the full surface ${\displaystyle {\int }_{\varphi \left(latitude\right)}^{\varphi \left(latitude\right)+\Delta \varphi }{\displaystyle {\int }_{R_1}^{R_2}r\cdot \text{d}r\cdot \text{d}\varphi }}$ in

equations (26) and following by those dependent on $\varphi ,\theta$ or $r$ (corresponding to the direction of movement: westward, northward or vertical), I can calculate the three strain coefficients ${\epsilon }_{\varphi },{\epsilon }_{\theta },{\epsilon }_r$. The calculation of the derivative of ${\epsilon }_{\theta }$

with respect to time $t$ gives us the strain rate $\frac{\text{d}{\epsilon }_{\theta }}{\text{d}t}$ or, if I consider $W_{\theta }$ as

weakly dependent on time $t$:

$\frac{\text{d}{\epsilon }_{\theta }}{\text{d}t}=\frac{1}{2}\cdot {\nu }_{\theta }^2\cdot \Delta M\left[\frac{t_{earth}}{\eta }\cdot \frac{1}{\left({\displaystyle {\int }_{\varphi \left(latitude\right)}^{\varphi \left(latitude\right)+\Delta \varphi }{\displaystyle {\int }_{R_1}^{R_2}r\cdot \text{d}r\cdot \text{d}\varphi }}\right)\cdot {\nu }_{\theta }\cdot t^2}\right]$ (27)

In (26) and (27) the reference speeds ${\nu }_{\varphi },{\nu }_{\theta },{\nu }_r$ come from experimental GPS data.

Given these values for the strain coefficient and strain rate, I can verify the accuracy of my results with experimental GPS velocity measurements (either absolute or with reference to a hot spot) of the Pacific (PA), Eurasia Union (EU), North America (NA), Australian (AU) and South American (SA) plates. Obviously, it is possible to reverse the calculation and calculate strain coefficients, strain rate

${\epsilon }_{\varphi },{\epsilon }_{\theta },{\epsilon }_r$ and $\frac{\text{d}{\epsilon }_{\varphi }}{\text{d}t},\frac{\text{d}{\epsilon }_{\theta }}{\text{d}t},\frac{\text{d}{\epsilon }_r}{\text{d}t}$, knowing the latitude, density and the absolute

GPS speed.

GPS measurements are sufficiently reliable ( [28] ; NASA GPS Time Series data in the IGS88 reference frame) to highlight an interesting property for the pairs

${\tau }_{m\varphi },{\tau }_{m\theta },{\tau }_{mr}$ and $\frac{\text{d}{\epsilon }_{\varphi }}{\text{d}t},\frac{\text{d}{\epsilon }_{\theta }}{\text{d}t},\frac{\text{d}{\epsilon }_r}{\text{d}t}$, where:

${}_{}^{}\frac{\text{d}{\epsilon }_{\theta }}{\text{d}t}\cdot {\tau }_{m\theta }\approx constant,\frac{\text{d}{\epsilon }_{\varphi }}{\text{d}t}\cdot {\tau }_{m\varphi }\approx constant\text{and}\frac{\text{d}{\epsilon }_r}{\text{d}t}\cdot {\tau }_{mr}\approx constant$

I will see later how to calculate these constants, which I call $k_{\theta },k_{\varphi }$ and $k_r$ and which are useful for the speed calculation of plates, with the relationships (23 to 25).

First, I define the notion of the tectonic plate that is used in the following calculations. In general, the size of a plate is established by its coordinates $r,\varphi$ and $\theta$. Length is measured by longitude, width depends on $r$ and $\varphi$ measured by latitude and thickness is a function of $R_1$ and $R_2$ measured along the Earth’s radius ( Figure 7 ). Figure 11 is a simple representation of a plate:

Plates move in three directions, along axes $e_{\varphi },e_{\theta },e_r$. Several studies of the asthenosphere have shown that the resultant plate velocity is along an axis from east to west [7] [29] - [36] . Thus, the forces exerted by the action of the Sun on the plate masses $\Delta M$ depend on their position $\Delta \theta$ on the plate and masses $\Delta M$ will be subject to the sum of the pressures exerted by the masses which precedes it from east to west. Thus, points further west and closer to the equator will experience more stress than those lying to the east and further away from the equator. However, as plates are surrounded by other plates, they are likely to be subject to additional stress from both their neighbors and the Earth’s internal activity. The overall direction of movement will be from the west and north (the lateral surface of the Earth’s movement) and vertically along the Earth’s radius. If I apply the results obtained above

by considering the term $\frac{2}{W_{\theta }}$ of (26) to be negligible, and mass $\Delta M$ as equal to: $\Delta M={\displaystyle {\int }_0^{\Delta \theta }{\displaystyle {\int }_{\varphi \left(latitude\right)}^{\varphi \left(latitude\right)+\Delta \varphi }{\displaystyle {\int }_{R_1}^{R_2}{r}^2\cdot \sin \varphi \cdot \mu \left(r\right)\cdot \text{d}r\cdot \text{d}\varphi \cdot \text{d}\theta }}}$ with $t_{orogenic}=t_{earth}$ I can

write the following equation for the strain coefficient and strain rate:

${\epsilon }_{\theta }=\frac{1}{2}\cdot {\nu }_{\theta }^2\cdot \Delta M\left[\frac{t_{earth}+t}{\eta }\cdot \frac{1}{\left({\displaystyle {\int }_{\varphi \left(latitude\right)}^{\varphi \left(latitude\right)+\Delta \varphi }{\displaystyle {\int }_{R_1}^{R_2}r\cdot \text{d}r\cdot \text{d}\varphi }}\right)\cdot {\nu }_{\theta }\cdot t}\right]$ (28)

With $\frac{\text{d}}{\text{d}t}{\epsilon }_{\theta }\approx -\frac{{\epsilon }_{\theta }}{t}$ where $t$ is equal to $T_{earth}$. I note that in (28), only the

value $\theta$ of the longitude (in the expression of mass $\Delta M$) provides a consistent

change in the calculation of ${\epsilon }_{\theta }$ and $\frac{\text{d}{\epsilon }_{\theta }}{\text{d}t}$. Thus, in the equation for the mass

$\Delta M,\Delta \theta$ is calculated using the longitude of the most eastern plate border as follows ( Figure 12 ):

With lithospheric viscosity $\eta$ equal to 10²² Pa∙s along the east-west axis $e_{\theta }$ (see Figure 4 ), I can draw the following graph for the Pacific plate (PA):

I used data from GPS velocities to the east and north [28] to plot this curve using the ITRF2000 terrestrial reference system, which is comparable to the

ITRF2005 NNR-NUVEL-1A system. $\frac{\text{d}{\epsilon }_{\theta }\left(t,i\right)}{\text{d}t}$ represents the strain rate, and

${\epsilon }_{\theta }$ is the strain coefficient. Obviously, I can do the same calculation along the direction of $e_{\varphi }$ and $e_r$ (see Figure 4 ) to obtain the three components of the strain rate and strain coefficient for the point in question. I can also represent the graph shown in Figure 13 for the PA plate as a function of the longitude, and I obtain:

Again, using the PA plate I can verify the validity of the fundamental relationship $\frac{\text{d}{\epsilon }_{\theta }}{\text{d}t}\cdot {\tau }_{m\theta }\approx constant=k_{\theta }$ and, with the expression of ${\tau }_{m\theta }$ I have:

${\tau }_m=\frac{\eta }{W_{\theta }}\cdot \left({\displaystyle {\int }_{\varphi \left(latitude\right)}^{\varphi \left(latitude\right)+\Delta \varphi }{\displaystyle {\int }_{R_1}^{R_2}r\cdot \text{d}r\cdot \text{d}\varphi }}\right)\cdot {\nu }_{\theta }\cdot t$ (29)

This expression, which has been explained above (see also equations 18 and 19). Following the graph of $k_{\theta }$ for the PA plate with the GPS data shown in Figure 13 and Figure 14 :

$k_{\theta }$ (see Figure 15 ) takes the value approximatly equal to: 7.54, and I can simplify the equations for the speed of plates (23, 24, 25) taking into account the longitude of the eastern border (or the latitude of the southern border) of the plate in question with the calculation of $\Delta \theta$ or $\Delta \varphi$ (see Figure 12 ) as follows:

${\nu }_{\varphi }=\sqrt{\frac{k_{\theta }\cdot \frac{2\cdot T_{earth}}{t_{earth}}\cdot \frac{1}{2\pi }\cdot {\displaystyle {\int }_0^{2\pi }W\varphi \cdot \text{d}\lambda }}{{\displaystyle {\int }_0^{\Delta \theta }{\displaystyle {\int }_{\varphi \left(latitude\right)}^{\varphi \left(latitude\right)+\Delta \varphi }{\displaystyle {\int }_{R_1}^{R_2}{r}^2\cdot \sin \varphi \cdot \mu \left(r\right)\cdot \text{d}r\cdot \text{d}\varphi \cdot \text{d}\theta }}}}}+v_{ref}$ (30)

In this relation $t_{earth}$ is 24h (the time for one rotation of the Earth), $t_{earth}$ is the orogenic time defined above and $v_{ref}$ is a dependent additive speed reference from GPS measurements. $k_{\theta }$ can be calculated from a single GPS speed measurement ( $V_{GPS}$), taken in a stable zone of the tectonic plate. Applying (30) I can write:

$k_{\theta }=\frac{-1}{2}\cdot \frac{1}{W_{\theta }}\cdot \left(\frac{t_{earth}}{t}\right)\cdot V_{GPS}^2\cdot {\displaystyle {\int }_0^{\Delta \theta }{\displaystyle {\int }_{\varphi \left(latitude\right)}^{\varphi \left(latitude\right)+\Delta \varphi }{\displaystyle {\int }_{R_1}^{R_2}{r}^2\cdot \sin \varphi \cdot \mu \left(r\right)\cdot \text{d}r\cdot \text{d}\varphi \cdot \text{d}\theta }}}$ (31)

Given GPS speed measurements, I can also calculate the viscosity of the asthenosphere. The principle of laminar flow at low speed gives the graph ( Figure 16 ):

Given a constant force $F$, I can write the following:

$\begin{array}{l}F={\eta }_{asthe}\cdot S\cdot \frac{\text{d}v}{\text{d}h}\\ \text{with}S={\displaystyle {\int }_0^{\Delta \theta }{\displaystyle {\int }_{\varphi \left(latitude\right)}^{\varphi \left(latitude\right)+\Delta \phi }{R}_2^2\cdot \sin \varphi \cdot \text{d}\varphi \cdot \text{d}\theta }}\\ \text{and}\frac{\text{d}v}{\text{d}h}=\frac{P{A}_{i,1}}{T_{Earthsun}}\cdot \frac{1}{\Delta r}\end{array}$ (32)

Finally, given (17) and (29) and with $\eta$ as the viscosity of the lithosphere, I obtain the following:

${\eta }_{asthe}=\eta \cdot \frac{\frac{\text{d}{\epsilon }_{\theta }}{\text{d}t}\cdot {\displaystyle {\int }_{\varphi \left(latitude\right)}^{\varphi \left(latitude\right)+\Delta \varphi }{\displaystyle {\int }_{R_1}^{R_2}r\cdot \text{d}r\cdot \text{d}\varphi }}}{{\displaystyle {\int }_0^{\Delta \theta }{\displaystyle {\int }_{\varphi \left(latitude\right)}^{\varphi \left(latitude\right)+\Delta \phi }{R}_2^2\cdot \sin \varphi \cdot \text{d}\varphi \cdot \text{d}\theta }}\cdot \frac{P{A}_{i,1}}{T_{Earthsun}}\cdot \frac{1}{\Delta R}}$ (33)

In (33) $\Delta R=R_3-R_2$, $R_3$ represents the distance between the boundary of the lithosphere and a given point in the asthenosphere, $R_2$ represents the thickness of the lithosphere. $R_1$ is comparable to the radius of the Earth ( $r_{earth}$, see Figure 7 ). $P{A}_{i,1}$ represents the movement of a GPS position of the PA plate in one year

[28] based on the parameters of the point in question ( $r,\varphi$ and $\theta$). $\frac{P{A}_{i,1}}{T_{earthsun}}$ is

the speed, where $T_{earthsun}$ is the time for one revolution of the Earth around the Sun. From (40) and assuming that the distance between the bottom of the lithosphere and the asthenosphere ranges from 10 to 2000 km I can draw the following graph with the viscosity of ${\eta }_{asthenosphere}$ in Pa∙s ( Figure 17 ):

I can then develop (30). And by replacing $\frac{\text{d}{\epsilon }_{\theta }}{\text{d}t}$ defined in (28) I obtain the following:

${\eta }_{asthe}=\frac{\frac{1}{2}\cdot {\displaystyle {\int }_{\varphi \left(latitude\right)}^{\varphi \left(latitude\right)+\Delta \varphi }{\displaystyle {\int }_{R_1}^{R_2}{r}^2\cdot \sin \varphi \cdot \mu \left(r\right)\cdot \text{d}r\cdot \text{d}\varphi \cdot \left(\frac{t_{earth}}{t^2}\right)}}}{\frac{1}{\Delta R}\cdot {\displaystyle {\int }_{\varphi \left(latitude\right)}^{\varphi \left(latitude\right)+\Delta \phi }{R}_2^2\cdot \sin \varphi \cdot \text{d}\varphi }}$

This can be simplified as follows, where $t=T_{earth}=24\text{h}$:

${\eta }_{asthe}=\frac{t_{earth}\cdot {\mu }_{earth}}{T_{earth}^2}\cdot \frac{0.415\cdot \left(R_3-R_2\right)\cdot \left(R_2^3-R_1^3\right)}{6\cdot R_2^2}$ (34)

This equation does not depend on latitude and longitude. Therefore, the viscosity of the upper mantle of the Earth should be constant for a given depth and for probably old and stable plates. In these calculations I assume:

$R_1=r_{earth},R_2=r_{earth}-120\text{km}\text{and}{R}_3-R_2=\Delta R$ and ${\mu }_{earth}=5.5\times {10}^3\text{kg}\cdot {\text{m}}^{-3}$

This does not take into account the Earth’s internal forces, or the action of the planets and the Moon, which together make up less than 2% of the total action of the Sun [10] . Nevertheless, together they may make plate movements more complex and further work will be needed to fully understand their contribution.

In the preceding section I examined the known speed of plates determined by GPS measurements on the ground. These sources are fairly reliable and are based on international ITRF 2000 and ITRF 2005 systems and the NASA Time Series GPS data in the IGS88 reference frame. In the following, I apply my results to the calculation of the speed of plates over the asthenosphere and show that my theoretical results correspond well to GPS data.

The Earth is subject to Newton gravitation from the Sun, Moon and other celestial bodies. The Sun and the Moon are the nearest and dominant masses, and they exert a force on all infinitesimal terrestrial masses $\text{d}m$. These forces are largely balanced by centrifugal forces due to rotation around a common center of mass for the Earth-Sun and Earth-Moon pairs. I know that lunar and solar tides (relatively lower than that of the moon), deform the surface of the Earth and they generate around 0.1 TW of energy, which is relatively low compared to the 47 TW of terrestrial heat flow.

In the first part of this article, I showed that there was another force acting on infinitesimal masses $\text{d}m$ and that this force depends on the speed of rotation of the Earth without any centrifugal (or other) compensation. From the calculation of the instantaneous energy in (11) and (12) I see that it is much higher than that of Newton gravitation, and the action of the Sun is greater than that of the Moon and planets combined. Nevertheless, the action of the Moon and other planets on plate movements is not negligible (about 2% of the total energy), and this force could be the subject of further work.

According to (9) and (15), during a complete rotation, the total energy $W$ absorbed by the Earth along the Sun-Earth axis takes the following values (for $\theta$ from 0 to 2π and $\varphi$ from 0 to π):

$W_t=\frac{1}{T_{earth}}\cdot \frac{1}{2\pi }\cdot {\displaystyle {\int }_0^{2\pi }W\cdot \text{d}\lambda }=55.53\text{TW}$

The energy dissipated in the Earth far exceeds that of the lunar and solar tides. As described above, I consider that the lithosphere is a rigid and deformable material while the asthenosphere is a ductile material. Thus, in the $e_r,e_{\theta }$ and $e_{\varphi }$ directions the energy $W$, will behave as the work of a force on a solid foundation for the lithosphere, but be absorbed and converted into heat in the asthenosphere. Although this is a simplified model, it nevertheless allows a first degree of approximation and a credible solution. I applied my results to the PA, NA, AU, SA and EU plates in the international terrestrial reference frames ITRF 2000 and ITRF 2005, based on measurements from GPS stations ( [28] ; NASA GPS Time Series data in the IGS88 reference frame). In my model, the forces applied to a particular lithosphere plate depend on the latitude, longitude and distance $r$ to the center of the Earth from the point in question. The values of $\Delta \theta$ and $\Delta \varphi$ that are used in the calculation of the strain rate and plate speed must be set as accurately as possible.

As explained above, these forces depend on the eastern (for $\Delta \theta$) and southern (for $\Delta \varphi$) plate boundaries. The following is an example of data for the PA plate:

$\varphi \left(latitude\right)=\pi \cdot \frac{90-latitude}{180}$ and

$R_1=R_{earth}\text{and}{R}_2=R_{earth}-100\text{km}\text{and}\Delta \theta ,\Delta \varphi$

Next, I apply (30) for westward speeds, replace $\theta$ by $\varphi$ and the full surface

${\displaystyle {\int }_{\varphi \left(latitude\right)}^{\varphi \left(latitude\right)+\Delta \varphi }{\displaystyle {\int }_{R_1}^{R_2}r\cdot \text{d}r\cdot \text{d}\varphi }}$ by ${\displaystyle {\int }_0^{\Delta \theta }{\displaystyle {\int }_{R_1}^{R_2}r\cdot \text{d}r\cdot \text{d}\theta }}$, which represents the external sur-

face subject to pressure toward the north in Equation (26) and following, I obtain Figure 18(a) , Figure 18(b) , Figures 19(a)-(c) for the PA plate:

Figure 18(a) and Figure 18(b) show the westward speed of the PA plate. In this case the results do not closely match GPS data. This is because westward speed is not dependent on $\Delta \theta$, unlike, $\Delta \varphi$ which is north-dependent. However, Figure 18(b) shows the results if I hypothesize that orogenic time $t_{earth}$ varies according to longitude and latitude. This graph is very similar to GPS data. Figure 18(b) deserves a thorough analysis as it shows that the age of the place where GPS lithosphere measurements are taken influences the calculation.

Although I could develop graphs based on longitude, it makes no difference to the accuracy of my calculations. In either case the calculations towards the west and north, $k_{\theta }$ and $k_{\varphi }$ are equal to 7.544 (see Equation (41)).

Figure 19(a) shows Calculus of the speed northward, varying with the position of the southern boundary of the PA plate located at an average latitude of 60˚S, which does not take into account the interpenetration of the AU and PA plates. Figure 19(b) shows the influence of the AU plate on the southern border of the PA plate, which is closer to GPS measurements. Figure 19(b) is an adjustment of curve 19A, and Figure 19(c) is a modification of the masses of the moving portion of the plate, which is a function of $\Delta \theta$ and $\Delta \varphi$ and indicates that the action of the AU plate on the PA plate is not uniform along their shared northern and southern borders. It is likely that there is a mini-plate at latitudes of 19 - 25˚S and longitudes of 160˚E - 153˚E. Given that the mass that creates energy $W_{\varphi }=W_{\eta }$ is greater than the movement of the PA plate, it seems that some energy is transferred to another plate or to the formation of volcanoes and mountains. As not all GPS data points are available, it is quite possible that this plate is much larger. For longitudes 183˚E to 160˚W the masses creating energy $W_{\varphi }=W_{\eta }$ are lower than the moving mass, indicating that there is energy input from another plate or sources internal to the asthenosphere. Finally, for longitudes 150˚E - 183˚E small masses create small energies $W_{\varphi }=W_{\eta }$. This data suggest that the AU plate creates a discontinuity in the propagation of pressure from south to north in the PA plate.

A better comparison of the mutual action of the PA and AU plates is found by plotting graphs for the AU plate.

I can compare the Figure 19(c) and Figure 20(c) to the interpretation given above. If I compare the portion of the AU plate at longitude 150˚E - 175˚E with the portion of the PA plate at the same longitude ( Figure 19(c) ), I see that they are complementary; the PA plate has part of the energy of the AU plate and energy is transmitted from the south of the PA plate.

Thus, the position of plate boundaries together with theoretical results (adjusted for orogenic or lithospheric age and mass) provides new information about the movements, stress and other activity of the lithosphere and upper mantle of the Earth. Further, more detailed studies will be needed to fully analyze all the results obtained.

The same calculations could be applied to the NA, EU and SA plates. I expect to find similar, detailed results and rich predictive analyses.

Finally, I calculated the west and north speed, using the same projection of energy $W$ along the vector $r$. Using (30), and replacing $\theta$ by $r$ and the

surface ${\displaystyle {\int }_{\varphi \left(latitude\right)}^{\varphi \left(latitude\right)+\Delta \varphi }{\displaystyle {\int }_{R_1}^{R_2}r\cdot \text{d}r\cdot \text{d}\varphi }}$ by ${\displaystyle {\int }_0^{\Delta \theta }{\displaystyle {\int }_{\varphi \left(latitude\right)}^{\varphi \left(latitude\right)+\Delta \varphi }{r}^2\cdot \sin \varphi \cdot \text{d}\varphi \cdot \text{d}\theta }}$ (this integral

represents the output surface of the vertical pressure; Equation (26) and following), to calculate the vertical velocity of the lithosphere, I have the following graph ( Figure 21 ):

This graph is obtained by considering the general equation (30) to be valid for the calculation of the vertical velocity; I replace $\theta$ by $r$ Although $k_r$ is different to the lateral velocity and equal to 0.002, it can be easily calculated from a single, stable GPS measurement, in a stable place of tectonic plate (see equation 31). In this case, the gravitational pressure on the mass in question opposes the vertical movement. Energy $W_{\eta }$ offsets this inverse gravitational energy and creates slight plate movement.