Earth's rotation and the surface of the ocean

Posted by Diego Assencio on 2016.09.16 under Physics (Mechanics)

In this post, we will show that the Earth's rotation alters the surface of its ocean, with "ocean" here meaning the global ocean of the Earth, i.e., the continuous body of water encircling it. Our model will consist of a spherical Earth with radius $R$ rotating at an angular velocity $\omega$ around a fixed axis.

The easiest way to solve this problem is by using a non-inertial frame of reference which rotates with the Earth and with origin at its center. The vertical axis $z'$ of this rotating frame is chosen to be the axis of rotation of the Earth (see figure 1). In this frame, the ocean is at rest since it moves with the Earth.

Fig. 1: Earth and the surface of the ocean. The reference frame $S$ sees the Earth rotating with angular velocity $\omega$ around the $z$ axis, while in the frame $S'$, the Earth is at rest since $S'$ is also rotating with respect to $S$ with angular velocity $\omega$, i.e., $S'$ is "glued" to the Earth. The origins of both $S$ and $S'$ coincide with the center of the Earth. The effect of the Earth's rotation on the surface of its ocean is greatly exaggerated in the figure.

Let $S$ be a reference frame with origin at the center of the Earth and such that the Earth itself is seen as rotating with angular velocity $\omega$, and let $S'$be the frame which rotates with the Earth as shown in figure 1. As proven in a previous post, the following equation relates the force ${\bf F}_S$ experienced by a fluid element at the ocean with mass $m$ in $S$ and the effective force ${\bf F}^{\textrm{eff}}_{S'}$ experienced by this same fluid element in $S'$ (terms which trivially vanish were omitted below): $$ {\bf F}^{\textrm{eff}}_{S'} = m{\bf a}_{S'} = {\bf F}_S - m{\pmb\omega} \times ({\pmb\omega}\times{\bf x}') = {\bf F}_S + m\omega^2 \rho\hat{\pmb\rho} \label{post_f9a58faf18bdbd019f06a0f09c123d60_eq_forces1} $$ where ${\bf a}_{S'}$ and ${\bf x}'$ are the acceleration and position of the fluid element as measured in $S'$ respectively, ${\pmb\omega} = \omega\hat{\bf z}$ is the angular velocity of the Earth as seen in $S$, $\rho$ is the distance between the fluid element and the $z$ (or $z'$) axis and $\hat{\pmb\rho}$ is the unit vector which points radially outwards from the $z$ (or $z'$) axis (see figure 2).

Fig. 2: Forces acting on a fluid element of mass $m$ on the surface of the Earth as seen in $S'$. The figure shows a cross section of the Earth which passes through the $z$ (or $z'$) axis.

The term ${\bf F}_S$ is the sum of the only two forces experienced in the non-rotating frame $S$: the gravitational force $m{\bf g}$ and the force ${\bf F}_{\textrm{fluid}}$ exerted on the fluid element by the surrounding fluid (up to this point, we have not explicitly assumed that the fluid element is on the surface, so the equation below applies to any fluid element on the ocean): $$ {\bf F}_S = m{\bf g} + {\bf F}_{\textrm{fluid}} \label{post_f9a58faf18bdbd019f06a0f09c123d60_eq_forces2} $$ Since in $S'$ the fluid element is at rest, ${\bf a}_{S'} = {\bf 0}$. This fact, together with equations \eqref{post_f9a58faf18bdbd019f06a0f09c123d60_eq_forces1} and \eqref{post_f9a58faf18bdbd019f06a0f09c123d60_eq_forces2}, yields the following: $$ {\bf 0} = {\bf F}_{\textrm{fluid}} + m\left({\bf g} - {\pmb\omega} \times ({\pmb\omega}\times{\bf x}')\right) = {\bf F}_{\textrm{fluid}} + m\left({\bf g} + \omega^2 \rho\hat{\pmb\rho}\right) \label{post_f9a58faf18bdbd019f06a0f09c123d60_eq_acc} $$ Figure 2 shows these three forces. In equation \eqref{post_f9a58faf18bdbd019f06a0f09c123d60_eq_acc}, the centrifugal acceleration term $\omega^2 \rho\hat{\pmb\rho}$ can be interpreted as an additional component which, together with the gravitational acceleration ${\bf g}$, yields an effective gravitational acceleration ${\bf g}^{\textrm{eff}}$ which is no longer homogeneous in space: $$ {\bf g}^{\textrm{eff}} = {\bf g} - {\pmb\omega} \times ({\pmb\omega}\times{\bf x}') = {\bf g} + \omega^2 \rho\hat{\pmb\rho} = {\bf g} + \omega^2 R \sin\theta\hat{\pmb\rho} \label{post_f9a58faf18bdbd019f06a0f09c123d60_effec_gravity} $$ where above we used the fact that $\rho \approx R\sin\theta$ since the surface of the ocean is close enough to the surface of the spherical Earth, and $\theta$ is the angle between the $z'$ axis and a segment connecting the center of the Earth to the fluid element.

The force ${\bf F}_{\textrm{fluid}}$ is a very interesting one since it is always orthogonal to the surface of the ocean for fluid elements on the ocean surface. To understand this point, imagine the case in which there is no rotation, i.e., ${\pmb\omega} = {\bf 0}$. In this case, the frames $S$ and $S'$ coincide and the ocean is at rest on both. From equation \eqref{post_f9a58faf18bdbd019f06a0f09c123d60_eq_acc}, we see that ${\bf F}_{\textrm{fluid}} = - m{\bf g}$, which means ${\bf F}_{\textrm{fluid}}$ is the force on the fluid element exerted by the surrounding fluid against the gravitational force: the fluid element is not moving, so the surrounding fluid must be providing the force ${\bf F}_{\textrm{fluid}} = -m{\bf g}$ to keep it in place. Since the fluid surface is locally horizontal in this case, i.e., orthogonal to ${\bf g}$, then ${\bf F}_{\textrm{fluid}}$ is orthogonal to the fluid surface. When $\omega \neq 0$, ${\bf F}_{\textrm{fluid}}$ plays exactly the same role, but now the effective gravitational force given in equation \eqref{post_f9a58faf18bdbd019f06a0f09c123d60_effec_gravity} is no longer homogeneous in space but depends explicitly on the position ${\bf x}'$: ${\bf F}_{\textrm{fluid}}$ will still be orthogonal to the surface of the ocean for every fluid element on the ocean surface; a tangential force component would set the fluid element in motion since it would not be able to resist this force (but a perpendicular force is resisted by a pressure gradient created inside the water).

Since ${\bf F}_{\textrm{fluid}}$ is always perpendicular to every fluid element on the ocean surface, and since, from equations \eqref{post_f9a58faf18bdbd019f06a0f09c123d60_eq_acc} and \eqref{post_f9a58faf18bdbd019f06a0f09c123d60_effec_gravity}: $$ {\bf F}_{\textrm{fluid}} = -m\left({\bf g} + \omega^2 R\sin\theta\hat{\pmb\rho}\right) = -m{\bf g}^{\textrm{eff}} \label{post_f9a58faf18bdbd019f06a0f09c123d60_eq_acc2} $$ then the ocean surface is always perpendicular to the effective gravitational acceleration ${\bf g}^{\textrm{eff}}$. As shown in figure 3, even though ${\bf g}$ always points to the center of the Earth, ${\bf g}^{\textrm{eff}}$ does not, meaning the surface of the ocean will not be spherical. At the poles, we have $\theta = 0$ and $\theta = \pi$, so ${\bf g}^{\textrm{eff}} = {\bf g}$, meaning the ocean surface is perpendicular to ${\bf g}$ at those points. Along the equator, $\theta = \pi/2$ and therefore ${\bf g}^{\textrm{eff}} = {\bf g} + \omega^2 R \hat{\pmb\rho}$; since $\hat{\pmb\rho}$ is parallel to ${\bf g}$, the resulting ${\bf g}^{\textrm{eff}}$ still points to the center of the Earth and therefore the ocean is perpendicular to ${\bf g}$ there as well, but the fact that $\omega^2 R \hat{\pmb\rho}$ points away from ${\bf g}$ means ${\bf g}^{\textrm{eff}}$ is smaller in magnitude at the equator than at the poles (where it attains its highest magnitude). But at any other point on the surface of the Earth, $\bf g$ and $\omega^2 R\sin\theta\hat{\pmb\rho}$ are not parallel to each other and therefore the surface of the ocean is in general not perpendicular to ${\bf g}$.

Fig. 3: Components of ${\bf g}^{\textrm{eff}}$ for fluid elements on diverse points along the surface of the ocean. Except at the poles and at the equator, the surface of the ocean is in general not orthogonal to ${\bf g}$.

A final comment is necessary here: when studying the surface of the ocean, we only considered the rotation of the Earth. In reality, the centrifugal acceleration $\omega^2 \rho \hat{\pmb\rho}$ causes the Earth itself to be shaped more like an oblate spheroid with an equatorial bulge of $42.77\textrm{km}$, and the ocean is affected as a result. The gravitational forces of the Sun and the Moon also significantly change the shapes of both the Earth and the ocean due to the tidal forces which they generate.

For the curious, $\omega^2 R \approx 0.34\textrm{m/s}^2$, so the rotation of the Earth makes objects on the equator be $\omega^2 R / g \approx 0.35\%$ lighter than at the poles (here we used $g = 9.8\textrm{m/s}^2$). In practice, however, the fact that the Earth is an oblate spheroid means the difference is even higher since objects at the equator are farther from the center of the Earth than objects at the poles.


karl-johan österlund on Oct 28, 2021:
Hello,.. quite great formulas .. As we see it, the result is important... a tilt change of the earth axis with only 0,45 degree, with north pool moving 50 km towards Sibiria (like the magnetic pool) but the earth surface like status quo would give a sea level rise on the new equator & north Atlantic area of ab.50 m ?