# A.4 Estimation of radial or centripetal forces on the Earth’s crust

From Appendix 2 consider a 1M^{3} of crust with an average density of 2.8 x10^{3} Kg.m^{-3}.

Taking the same values used in Appendix 2

ρ = Average density of the crust: 2.8 kg.m^{-3}

M = Mass of 1M^{3} of element of crust: 2.8 x10^{3}kg

R = Radius of Earth (m): 6.4x10^{6} metres

ω = Angular velocity (rad s^{-1}): 7.27x10^{-5} rads.sec^{-1}

Thus, Fr = Radial Outward Force (N)

= M ω^{2}.R

= 2.8 x 10^{3} x (7.27 x10^{-5})^{2} x 6.4 x10^{6}

= 94.71 N

= c 9.65 kgf

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

This is equivalent to a 0.034% reduction in weight compared with that at the poles, where the rotational velocity is zero. This is enough to cause the crustal plates to move around the Earth surface on a frictionless mantle.