# A.3 **Estimation of the magnitude of the circumferential forces driving tectonic movements**

The mathematical analysis is based on the concept of the outer rim being allowed to slide relative to the main rotating body (Figs 17b, 18a, & 19 ).

In order to determine the forces postulated as being responsible for tectonic movement the model used is one in which the thin crust can slide relative to the solid body at the crust /mantle interface. By way of illustration Fig 18a 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. 18b 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 span, then a simple force diagram (Fig 17b & 19) can be constructed by making the following assumptions: (a) the crust is a thin shell that is able to slide relative to the mantle, (b) the forces owing to eccentricity are superimposed on the stress caused by the general rotation and gravity, and (c) the stress that is of interest for the purposes of tectonic movement is the differential stress owing 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, in Africa, 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 effects of the circumferential tensile forces (F) 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 derived circumferential stress (F) will be dependent on the distance between the geometric centre and the centre of mass, i.e. (E) the ‘radius of eccentricity’. In a limiting case, if the ‘radius of eccentricity’ is zero, the rotating body will be balanced, and the centripetal forces will be zero.

Consider a thin shell cut across the Earth’s diameter at the Mid-Atlantic ridge (Fig 19 below). 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} Figure C in figure 19 shows this concept of ‘vertical force’. As the semi-circle is symmetrical there are two sides resisting the parting force. Thus, only one side needs to be considered for integration of the ‘vertical’ forces from 0 to π/2.

Fig C in Fig 19 shows 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 17A**

**Fig 17B**

**Fig 18A & 18B**

Models used for the calculation of the differential circumferential stress forces required to move the crust to the mantie. The movement to the lighter side is independent of the hard rotation

**Fig 19**

**Notation**

M = Mass per unit length of crust (kg)

R = Radius of Earth. (m)

E = Radius of eccentricity. (m)

ω = Angular velocity. (rad s^{-1})^{ }

θ = Angle. (rad)

δe = Effective eccentricity at angle θ

F = Total force at point X (cf. Fig.11) (N)

F_{1 }= Radial force due to eccentricity at θ

**Value**

2.8 x10^{6} kg

6.4x10^{6} metres

1x 10^{3} metres

7.27x10^{-5} rads.sec^{-1}

Then from the 'force vector diagram' at surface at an angle θ:

Vertical component of F_{1 }: δf = F_{1}Sinθ_{ }

Effective eccentricity at angle θ: δe = Ε sinθ

And Mass of segment: Rδθ = M R δθ.

Thus,

F_{1} = M. Rδθ.ω^{2}. Ε sinθ = M. R. ω^{2}.Ε. Sinθ. δθ

The vertical force component: δf = F_{1}. Sinθ

= M. R.ω^{2}. Ε. Sinθ. Sinθ. δθ

=** M.R.ω2. Ε. Sin**^{2}**θ. δθ (1)**

Thus,

the total vertical force F = Σ_{0}^{π/2 }M.R.ω^{2}E.Sin^{2}θ. δθ^{ }

= M.R.ω^{2}.E (½.θ -¼Sin2θ) ^{π/2} - (½.θ- ¼ Sin2θ) ^{0}

= M.R.ω^{2}.E (π/4-¼.0) - (½.0-¼.0)

= **M.R.ω**^{2}**.Eπ/4.**** ****(2)**

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.

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:

The radius of the Earth (r) = 6400 km

The angular velocity of the Earth^{55} at the equator (ω) = 7.27x10^{-5} rads.sec^{-1}

The radius of eccentricity at the Core (E) = 1 km.

Hence substituting into equation 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 = 6.64 x10^{7} N.

Since, the magnitude of the circumferential stress is Force/Area

this becomes 6.64x10^{7} / 1 x10^{3}: and hence

the circumferential tensile stress is = 6.64x10^{-2} Nmm^{-2}, 0.644 Bar or c. 9.7 lbs.in^{-2 }

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. 35 is used for this analysis. Fig 36 shows the relationship between the Radius of Eccentricity and the circumferential stresses. Fig 37 shows the relationship between F, E and µ.

As above:

the mass of the segment R δθ = M R δθ

the radial force F_{1 }= Mass.R.ω^{2 }

the radial force F_{1 }= (M R δθ). R. ω^{2 }= M ω^{2} R^{2} δθ.

Thus,

δf = M ω^{2} R^{2}sinθδθ.

With reference to Fig. 35: R = R0 + Esinθ.

thus, δf = M ω^{2} (R_{0}+Esinθ)^{2} sinθ δθ

which approximates to

δf = M ω^{2} (R_{0}^{2}+2E R_{0} sinθ) sinθ δθ

= M ω^{2} (R_{0}^{2}+2E R_{0} sinθ) sinθ δθ.

Thus, the increase of

δf = δf - M ω^{2} Ro^{2} sinθ δθ

= M ω^{2} (R_{0}^{2}+2E R_{0} sinθ) sinθ δθ - M ω^{2} Ro^{2} sinθ δθ

= M ω^{2} sinθ δθ (R_{0}^{2}+2E R_{0} sinθ- Ro^{2})

= M ω^{2} sinθ δθ2E R_{0} sinθ

= **MR**_{0}** ω**^{2}** 2 Esin**^{2}**θ δθ**** ****(Eq. 3)**

This equation has the same form as (Eq. 1) above. As E is small in comparison to R, and R_{0} and R have essentially the same values, the factor 2 that appears in (Eq. 3) does not invalidate (Eq. 1). Hence the derivation of (Eq. 1) from the force diagram (Fig 19 above and 20 below) is considered valid for determining (Eq. 2) by integrating between 0 and π/2.

**Fig 20**

**Fig 35**

**Fig 36: Relationship between the radius of eccentricity & the circumferential stresses**

**Fig 37: Relationship between the force (Newtons) needed to move a 1km x 1m x 1m element of crust and the coefficient of friction at the crust/mantle interface**