Rotational Flattening of Earth

The Earth rotates diurnally with an angular velocity vector, $\Omega$, that is directed from the center of the Earth toward its north geographic pole, and is of magnitude

$${\mit\Omega} = \frac{2\pi}{23^{\rm h}\,56^{\rm m}\,04^{\rm s}}= 7.292\times 10^{-5}\,{\rm rad\,s^{-1}}.$$ (1.351)

Here, $23^{\rm h}\,56^{\rm m}\,04^{\rm s}$ is the length of a so-called sidereal day, and is the period of the Earth's diurnal rotation relative to the distant stars (as opposed to the Sun).

Figure 1.19: Centripetal acceleration.

Let the Earth's axis of rotation correspond to the $z$-axis, and let us set up a conventional spherical coordinate system whose origin is the Earth's center, and whose symmetry axis corresponds to the $z$-axis. (See Section A.23.) A general point in the Earth whose spherical coordinates are $r$, $\theta$, $\phi$ rotates at angular velocity ${\mit\Omega}\,{\bf e}_z$ in a circle of radius $r\,\sin\theta$. See Figure 1.19. Thus, according to elementary physics, the point accelerates toward the $z$-axis with an acceleration $a={\mit\Omega}^2\,r\,\sin\theta$. As is clear from Figure 1.19, the point's vector acceleration is

$${\bf a} = - {\mit\Omega}^2\,r\,\sin\theta\,(\sin\theta\,{\bf e}_r +\cos\theta\,{\bf e}_\theta),$$ (1.352)

where ${\bf e}_r=\nabla r/\vert\nabla r\vert$ and ${\bf e}_\theta=\nabla\theta/\vert\nabla\theta\vert$ are unit vectors in the spherical coordinate system. However, it is easily demonstrated that

$${\bf a} = \nabla\chi$$ (1.353)

(see Section A.23), where

$$\chi(r,\theta)= -\frac{{\mit\Omega}^2\,r^2}{2}\,\sin^2\theta = \frac{{\mit\Omega}^2\,r^2}{3}\left[P_2(\cos\theta)-1\right]$$ (1.354)

can be thought of as a kind of centrifugal potential. (Because in the non-inertial frame of reference that co-rotates with the Earth the point in question would appear to be subject to a fictitious centrifugal force $-\nabla\chi$.)

Let us model the interior of the Earth as a fluid of uniform mass density $\gamma$. [It turns out that the centrifugal potential, (1.354), is sufficiently large that the rigidity of the rock that makes up the Earth is insufficient to prevent the Earth from responding to the potential in a fluid-like manner.] Now, if $p({\bf r})$ is the pressure distribution in the interior of the Earth then a small cuboid volume of the Earth lying between $x$ and $x+dx$, $y$ and $y+dy$, and $z$ and $z+dz$, experiences a net pressure force

$${\bf F} = \left[p(x,y,z)-p(x+dx,y,z)\right]dy\,dz\,{\bf e}_x + \left[p(x,y,z)-p(x,y+dy,z)\right]dx\,dz\,{\bf e}_y$$

$$\phantom{=}+ \left[p(x,y,z)-p(x,y,z+dz)\right]dx\,dy\,{\bf e}_z$$

$$=- \frac{\partial p}{\partial x}\,dV\,{\bf e}_x- \frac{\partial p}{\partial y}\,dV\,{\bf e}_y- \frac{\partial p}{\partial z}\,dV\,{\bf e}_z$$

$$= -\nabla p\,dV,$$ (1.355)

where $dV=dx\,dy\,dz$ is the volume of the cuboid. (See Section A.19). Thus, the force per unit mass due to the pressure inside the Earth is

$${\bf f} =\frac{{\bf F}}{\gamma\,dV}= -\frac{\nabla p}{\gamma}.$$ (1.356)

The equation of motion of a general point inside the Earth is

$${\bf a} = -\nabla{\mit\Phi} -\frac{\nabla p}{\gamma},$$ (1.357)

Here, the first term on the right-hand side of the previous equation is the gravitational force per unit mass acting at the point [see Equation (1.269)], whereas the second term is the force per unit mass due to internal pressure. Making use of Equation (1.353), we deduce that force balance inside the Earth requires that

$$\nabla\! \left({\mit\Phi} +\chi+\frac{p}{\gamma}\right)= 0.$$ (1.358)

The previous equation can be integrated to give

$${\mit\Phi} +\chi +\frac{p}{\gamma} = c,$$ (1.359)

where $c$ is a constant.

Let us model the Earth as a spheroid whose outer radius, $R_\theta(\theta)$, is specified by Equation (1.328). (See Section 1.10.1.) It follows from Equation (1.350) that the gravitational potential at the Earth's surface is

$${\mit\Phi}(R_\theta,\theta)\simeq -\frac{G\,M}{R}\left[1+\frac{4}{15}\,\epsilon\,P_2(\cos\theta)\right],$$ (1.360)

where $M$ is the Earth's mass, $R$ its mean radius, and $\epsilon$ its ellipticity. It is clear from Equation (1.354) that the centrifugal potential at the Earth's surface is

$$\chi(R_\theta,\theta) \simeq \frac{{\mit\Omega}^2\,R^2}{3}\left[P_2(\cos\theta)-1\right].$$ (1.361)

Note that we have neglected the slight difference between $R_\theta$ and $R$, when evaluating the previous expression, because the centrifugal potential is relatively small compared to the gravitational potential [see Equation (1.366)], and $\epsilon$ is also assumed to be small [see Equation (1.367)]. Now, the pressure at the Earth's surface must be zero, otherwise the surface would not be in equilibrium with outer space. (Here, we are neglecting the relatively small pressure due to the atmosphere.) It follows from the previous three equations that, on the surface of the Earth,

$$-\frac{G\,M}{R}\left[1+\frac{4}{15}\,\epsilon\,P_2(\cos\theta)\right] +\frac{{\mit\Omega}^2\,R^2}{3}\left[P_2(\cos\theta)-1\right] = c.$$ (1.362)

We can separately equate the components of the previous equation that are independent of $\theta$, and that vary with $\theta$ as $P_2(\cos\theta)$, to give

$$c = -\frac{G\,M}{R} - \frac{{\mit\Omega}^2\,R^2}{3},$$ (1.363)

and

$$\epsilon = \frac{15}{4}\,\zeta.$$ (1.364)

where

$$\zeta=\frac{{\mit\Omega}^2\,R^3}{3\,G\,M}$$ (1.365)

is the ratio of the typical centrifugal acceleration to the typical gravitational acceleration at $r=R$.

Given that ${\mit\Omega} = 7.292\times 10^{-5}\,{\rm rad\,s^{-1}}$, $R=6.371 \times 10^{6}\,{\rm m}$, and $M=5.972\times 10^{24}\,{\rm kg}$, we deduce that

$$\zeta= 1.15\times 10^{-3},$$ (1.366)

and

$$\epsilon = 4.31\times 10^{-3}.$$ (1.367)

In other words, as consequence of the Earth's rotation, the shape of the Earth is an oblate (because $\epsilon>0$) spheroid. Thus, the Earth is slightly flattened along an axis passing through its geographic poles. This result was first obtained by Newton. The predicted difference between the Earth's equatorial and polar radii is ${\mit\Delta}R = R_e-R_p = \epsilon\,R= 27.5\,{\rm km}$. In fact, the observed ellipticity of the Earth is

$$\epsilon=3.35\times 10^{-3},$$ (1.368)

with an associated difference between the equatorial and polar radii of $21.4\,{\rm km}$. Our analysis has overestimated the Earth's rotational flattening because, for the sake of simplicity, we modeled the Earth as a uniform body. In fact, the interior of the Earth is much denser than its crust.