As stated above, the rigid body approach while clearly demonstrating the differential stress due to eccentricity is not considered as the model for tectonic movement. The model for tectonic movement as defined in Section 8 is based on having relative movement between the outer rim or crust and the main body or mantle.

The mathematical analysis is based on the concept of the outer rim being allowed to slide relative to the main rotating body. (Figs, 10 &11).

In order to determine the inertial forces postulated as being responsible for tectonic movement the model used in one in which the thin crust is able to slide relative to the solid body at the crust /mantle interface. By way of illustration Fig.10 shows that if an unbalanced disc with an outer annular ring containing fluid is rotated about its principal axis, the liquid will move to the ‘lighter’ side. Fig. 11 shows an analogous situation with the sliding continental plates.

If we consider the crust as being able to move relative to the mantle, albeit it over a long geological time scale, then a simple force diagram ( Fig.12 a &b) can be constructed by making the following assumptions: the crust is a thin shell that is able to slide relative to the mantle, the forces due to eccentricity are superimposed on the stress caused by the general rotation and gravity, and the stress that is of interest for the purposes of tectonic movement is the differential stress due to this eccentricity.

By approaching the problem in terms of a thin shell moving relative to the mantle, it is possible to consider what increments of the tensile force are responsible for putting the Pacific basin under compression and the African Plate under tension. The Rift Valley would be a case in point. The calculations, which follow, are based on the consideration of the eccentrically induced loads on the thin crust. In calculating the inertial effects at the surface of the earth due to the centre of mass being offset from the principal axis of rotation, the term ‘ radius of eccentricity’ (E) is introduced to denote the magnitude of the offset.

The magnitude of the inertial forces or in this case the derived circumferential stress will be dependent on the distance between the geometric centre and the centre of mass, i.e. the ‘radius of eccentricity’. In a limiting case, if the ‘radius of eccentricity’ is zero, the rotating body will be balanced and the inertial forces will be zero.

Consider a thin shell cut across the Earth’s diameter at the mid- Atlantic ridge (Fig.12). The force tending to cause this half of the shell to part is the ‘vertical’ component of the centripetal forces generated by the eccentricity. This is similar in concept to that in thin shell circular vessels subjected to an internal pressure^{7} (Fig 12 c) shows the concept of ‘vertical force’). As the semi-circle is symmetrical there are two sides resisting the parting force. Thus only one side need be considered for integration of the ‘vertical’ forces from 0 to π/2.

Fig. 12 c show the force and vector diagrams used to determine the magnitude of the circumferential stress in the direction of the maximum effective radius. For ease of understanding the force diagram is superimposed on the major geological features on the equatorial belt.

Fig. 12Principal forces superimposed across (A) a section of the equator and (B) the force diagram used to determine the total force 'F' acting in the direction of the maximum effective radius. (C) illustrates the 'vertical' force component.

The derivation of the equation of the total force at the maximum effective radius allows for the determination of the circumferential tensile stress on the crust. The approach given above considers the forces developed as a direct function of the radius of eccentricity. It is also possible to look at the addition of the vertical component of E to the radius of the Earth to determine the expression of the forces in the direction of the maximum effective radius. Fig. 17 is used for this analysis.

This equation has the same form as (1) above. As E is small in comparison to R, R_{0} and R have essentially the same value, and the factor 2 that appears in (3) does not invalidate eq. (1). Hence the derivation of eq. (1) from the force diagram (Fig 12) is considered valid for determining eq.(2) by integrating between 0 and π/2.

If we take into eq. 2 the crust to be 1000 meters thick with an average density of 2.8x10^{3}kgm^{-3} then for a 1metre x 1 metre wide strip, the mass per unit area of crust (m) is = 1000 x 1 x1 x 2.8x10^{3} = 2.8x10^{6} kg: and radius of Earth (r) = 6400 km angular velocity of the Earth at the equator (ω) = 7.27x10^{-5} rads.sec^{-1}and the Radius of eccentricity at the Core (E) = 1 km.

Hence substituting into eq.2 we have

F= 2.8 x 10^{6} x 6.4 x 10^{6}x (7.27x10^{-5})^{2} x 10^{3 }x π /4 = 7.44 x10^{7} N.

Since the magnitude of the circumferential stress is Force/Area this becomes 7.44x10^{7} / 1 x10^{3}: and hence the circumferential tensile stress is = 7.44x10^{-2} Nmm^{-2}, 0.744 Bar or c. 10.9 lbs.in^{-2}.

Thus for every 1 Tonne of Crust, the Outward Force at the Equator due to the rotational speed = 9.65/ 2.8 = c.3.4 kg

This is equivalent to a 0.034% reduction of weight as compared to the zero velocity at the poles. This is sufficient to cause the crustal plates to move around the Earth surface on a frictionless mantle.