# Is effective gravity constant over the surface of the ocean?

According to Wikipedia:

The Earth is not spherically symmetric, but is slightly flatter at the poles while bulging at the Equator: an oblate spheroid. There are consequently slight deviations in both the magnitude and direction of gravity across its surface. The net force (or corresponding net acceleration) as measured by a scale and plumb bob is called "effective gravity" or "apparent gravity".

and

In combination, the equatorial bulge and the effects of the surface centrifugal force due to rotation mean that sea-level effective gravity increases from about $9.780\,{m/s^2}$ at the Equator to about $9.832\,{m/s^2}$ at the poles, so an object will weigh about $0.5\%$ more at the poles than at the Equator.

But I don't see how this is physically possible.

If the force on a unit of water at the poles was greater than the force at the equator, shouldn't the ocean fall at the poles and rise at the equator? For the same reason that when I step into the bath the water falls where my foot is pressing down on it and rises elsewhere. Surely the force at the surface of a body of water in equilibrium has to be constant.

So is my physical argument wrong, or is the net force on a body actually the same at the poles and equator?

You suggest that for a body of water to be in stable equilibrium, i.e. to have minimum energy, it should have the same gravitational field everywhere at its surface, as otherwise we would be able to move the water around to lower the energy. This sounds intuitive, but it's just not true.

As an extreme example, consider a puddle of water on Pluto. The gravitational field is weaker there, so by your argument we should be able to harvest energy to sending water to Pluto. But that's completely wrong: it would actually cost an enormous amount of energy to do this.

The right statement is that the gravitational potential is the same everywhere on the surface of water. That has nothing to do with whether the field is the same.

Note well: In this answer, I am including the fictitious centrifugal acceleration due to the Earth's rotation as a part of the gravitational acceleration vector. This is quite standard in geophysics.

Suppose at some point on the surface of the ocean that the local gravitational acceleration vector is not normal to the surface. This means the local gravitational acceleration vector has a non-zero component parallel to the surface at that point. That direction is "downhill": That is the direction in which water will flow. That the gravitational acceleration vector is not normal to the surface at this point means that this point is not in equilibrium. The equilibrium state is a surface such that the local surface downward normal and the local gravitational vector are parallel to one another at every point on the surface.

Both Newtonian gravitation and the fictitious centrifugal acceleration can be described in terms of potential functions. The sum of those two potential functions is the Earth's gravitational potential function. The local gravitational acceleration vector is the gradient of this potential. The gradient of an equipotential surface is everywhere normal to that surface. The reverse also applies: A surface for which the gradient of the potential function is everywhere normal to the surface is an equipotential surface.

The equipotential surface defined by mean sea level (aka the geoid) represents a minimum in the potential energy of the Earth's oceans. Deviations from this will result in the oceans having greater potential energy than that of the geoid and will have points at which the waters will flow downhill to realize the geoid.

good explanation here, image source

Youtube video, explaining it with math

To quickly recap, as the earth spins there is an outward centrifugal force, the further away from the axis of rotation the greater the force/acceleration. Since the earth spins arounds it's pole, in other words the axis of rotation goes through the poles, they don't really experience any centrifugal force. The equator experiences the most centrifugal force. Because of this centrifugal force the earth and oceans bulge around the center. Gravity decreases by $1/R^2$ and water is much less dense than the iron and molten rock below the very thin crust of the earth. also see the image below:

Keep in mind I might have simplified a few things for clarities sake.

• Okay, so why is my argument wrong? – Oscar Cunningham May 30 '18 at 14:31
• let's see what's the easiest way to put it. Let's say the earth stops spinning but stays bulged, then the water would flow towards the poles, because there's a lower potential, basically flowing down a hill of sorts. Now since it's spinning it can nicely stay where it is as the system is in equilibrium. Also be careful because you're comparing a system that's in equilibrium with an example that far from is, better if you would try imaging your foot moving infinitely slow. – Bob van de Voort May 30 '18 at 14:43

As you decrease the moment of inertia of an object, for angular momentum to be conserved, its rotational velocity must increase, and this causes an increase in its rotational energy.

[Math]
Angular momentum is $I\omega$. This means that for angular momentum to be conserved, $d(I\omega)$ must be zero, and thus $Id\omega =-\omega dI$, and $d\omega =-\frac{\omega dI}{I}$

Rotational energy is $\frac12 I\omega^2$, so $dE_r = \frac12\omega^2dI+I\omega d\omega$. Substituting in for $d\omega$, we get $dE_r = \frac12\omega^2dI-\omega^2 dI = -\frac12\omega^2dI$.
[/Math]

Moving water from the equator to the poles decreases the moment of inertia, increasing the rotational energy. The surface of the water is such that these balance out, and moving water between different latitudes does not change the total energy.

This is reflected on the forces on water: it has gravitational force that has a component towards the poles, but it has a centrifugal "force" that reflects the energy gradient in a rotational frame that acts outward and with a component towards the equator.