Do centrifugal forces contribute to the Earth's tidal bulges?

Centrifugal force.

Centrifugal forces are classed as "fictitious" forces because they do not arise from material objects such as those with mass. They are a mathematical trick to simplify calculations carried out in non-intertial (accelerating) frames of reference. In a reference frame rotating with angular velocity ω this requires modifying Newton's law F = ma by adding fictitious force terms to ma:

F_eff ≡ ma_rot = ma_fixed − m ω × (ω × r_rot ) − 2mω × v_rot

The term − mω × (ω × r) is the centrifugal force where ω is the angular velocity. Angular velocity is a vector along the rotation axis of the reference frame. This reduces to −mω²r when all the motion lies in a common plane. The last term is called the Coriolis force.

The subscript "eff" labels effective force as observed in the rotating reference frame. The label "rot" refers to a measurement made in this rotating reference frame. The label "fixed" refers to a measurement made in the fixed (inertial) reference frame.

Students should be a bit disturbed by this conceptual trick. How does it square with Newton's third law? That law is, "If body A exerts a force on body B, then B exerts an equal and oppositely directed force on A." If a centrifugal force acts on a body A in a system, what does that body exert an equal and oppositely directed force upon? Where is body B? Real forces always have a material body source. What physical body is the source of a centrifugal force? Where is it located?

One "solution" to this dilemma is to note that centrifugal forces are always directed perpendicular to and away from the reference frame's rotation axis. Suppose we invent a "fictitious mass" along that axis. That mass would have to be negative, repulsing other masssive objects in the physical system with a force that gets stronger linearly with distance from the axis. Newton's third law is now satisfied. We thus create a fictitious "centrifugal force field" with the force law F_c = − mω²r. The minus sign indicates that it is directed outward from the axis of the reference frame's rotation.

Can such a force field deform an elastic body? As with gravitational fields, deformation results from the elastic properties of the body, and the gradient of the force field. In the gravitational field we found this derivative to be proportional to 1/r³ and we called it the "tide raising force".

What is the gradient of a field whose strength is linearly dependent on distance from a fixed axis? It is constant! It alone cannot deform an elastic body.

This tells us that it is not centrifugal force that causes Earth's tidal bulges, though many textbooks would have you believe that, and students uncritically accept it as "common sense". Fictitious forces are never the cause of anything.

Earth's equatorial bulge.

How do we account for the Earth's equatorial bulge without invoking centrifugal force? Rotation and elasticity of materials are responsible. We have shown that a body of mass m moving in a circle must be acted on by a radial force mω²r where ω is the angular velocity of the mass. For a chunk of matter at the equator that r is the radius of the Earth, R. At Earth's poles it is zero. At other latitudes it is between R and 0. In all cases, it acts perpendicular to the Earth's rotation axis. The Earth's own gravity is nearly constant anywhere on Earth's surface and in a radial direction toward Earth's center. Of course similar considerations apply throughout Earth's volume.

Now imagine a spherical rotating sphere made of Jello™ with gravity negligible. Suppose the tensile strength of Jello is sufficient to maintain all of it rotating at the same angular velocity ω about a fixed axis. It is kept from flying apart by its own tensile force. (Think of this sphere in the language we use for the Earth, with equator, latitude and poles.) Material at the equator has the greatest orbital speed around the axis. At higher latitudes, the material has a smaller orbital radius and smaller speed. Remember F_centripetal = mω²r_axial. At a latitude where r_axial = R/2 the axial force required to keep a chunk of Jello in circular orbit is half that at the equator. The tensile stress on Jello there is also smaller. The original spherical shape is distorted into an oblate spheroid.

This is not the whole story. To stress a chunk of matter to change its shape we must apply differential force across its volume. If forces were constant over its volume there would be no stress to stretch or compress or in any way distort its shape. The answer lies in the previous paragraph. The radial tensile force at a given latitude varies proportional to distance from the rotation axis. So for two chunks of matter of the same size, the stress at distance R/2 is half that at R.

In addition to the axial component of stress (acting away from the rotation axis) there's also another component of stress that acts in a direction toward the equator. A rotating globe of water in space would deform to an oblate spheroid by physically relocating some water toward the equator. Water does not stretch or compress much. This is especially important when considering the Earth's ocean waters.

Earth's gravity alone produces a spherical shape. Add rotation to this and you get the modification of that spherical shape to an oblate spheroid. This shape is entirely due to rotation, gravitational and tensile forces. No centrifugal forces are necessary to account for it.

Centrifugal forces due to motion around Earth's barycenter?

Earth orbits around barycenter (O) with rotation about its center (C) suppressed.

To isolate this possible effect, we temporarily suppress Earth's axial rotation, and consider only its monthly motion around the barycenter. This has an unexpected geometric result. Earth's center moves in a circle of radius 4,670 km. and each point on the Earth moves around its own circle of the same radius. This is an uncommon sort of motion, seldom seen in textbook physics problems. But we all have experienced this if we ever took a ride in a Ferris Wheel. The passenger seats or cars on the wheel do not rotate with respect to the ground.

In this motion, adjacent chunks of the Earth experience the same centrifugal force, equal in size and direction. This force field has no gradiant across Earth's volume. Therefore it cannot cause deformations of the Earth, and no tides.

Ferris wheel in Vienna.

Why did the Earth's axial spin cause deformation of the Earth, but this movement did not?

The centrifugal forces acting on each piece of the axially spinning Earth had different sizes (and directions). Adjacent chunks of earth therefore experienced different size forces and this meant the compression or expansion of them was not zero. This difference (the force gradient) deformed the Earth.

Turning to the Earth/Moon system, we suppressed Earth's axial spin and changed our refererence frame to one that rotates around the Earth/Moon barycenter, matching the motion of the moon.

Then we considered whether there might be an additional deformation due to the Earth's movement around the barycenter. As with any rotating reference we must include centrifugal forces. But they are constant in size and direction at any time and have no gradient. So they can not deform the Earth's shape.

This little exercise was futile and redundant. We already established that choice of reference frame and its fictitious forces cannot ever be the cause or reason for any physical effect.

Motion of earth around the earth-moon barycenter has one interesting effect. It causes the earth-barycenter distance to increase by about 1 km. But since it doesn't change the shape of earth or oceans, it has no noticable effect on tides.

Summary.

A look at the formula clarifies this. The centrifugal force perceived on any body in the rotating system is mω²r. This depends on the angular rotation of the chosen reference frame. If we choose a reference frame rotating around the Earth/Moon baycenter, all points on Earth have the same radius of rotation at any instant. So this results in a rotating centrifugal field of constant strength.

The Coriolis force is perceived in our system as 2mωv where v is the velocity of a mass m moving in the rotating system. In this coordinate system at any instant all points on or within the Earth have the same angular velocity, so the Coriolis force on each and every one has the same size, and therefore they also cannot stress or deform the Earth's shape.

In short, the gravity gradient alone causes the tidal bulges. Earth's axial rotation causes the equatorial bulge (and polar flattening), which is not a tide. Earth's wobbling around the Earth-Moon barycenter does not generate additional stress effects on the Earth's shape, and no additional contribution to tides.

The only force field that can stress or deform a material object must have a field gradient. Uniform force fields cannot stress or deform material objects.

With earth's rotation suppressed, using a reference frame rotating at the same rate as the Moon, each chunk of matter on Earth experiences the same centrifugal force (constant in size and direction), therefore this cannot act to deform the shape of the Earth. I.e., this field exerts the same force on adjacent chunks of earth material, so there's no differential force to squeeze, stretch or displace materials. The same is true for Coriolis forces. [1]

If we consider a frame of reference rotating about the Earth-Sun center of mass, we reach the same conclusion.

Tidal bulges are solely due to the gradient of gravitational forces acting aginst tensile forces of earth and water. Centrifugal and Coriolis forces contribute nothing to them.

Endnotes.

Donald E. Simanek, May 24, 2020.

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