Understanding the Moon's Role in Tides

[pixel01]

In reality, the two tide bulges are symmetrical (or nearly). If proved by centrifugal force and center of mass, the two bulges are not.

 

[D H]

But the explanation given in thet book is correct, the only thing is that he calls the fictitious force the "centrifugal force."

From the book,

Consider now the revolving frame where both the Earth and the Moon state at rest. Inasmuch as the reference frame is noninertial every mass element experiences not only the force of gravity but a centrifugal force as well. ...​

That is definitely a centrifugal force. Unfortunately, the authors (I gather this is a physics for Russian poets text) did not do any math. If they had, it would have been wrong because the explanation in the book is, simply put, wrong.

This physics for Russian poets text is wrong. The teaching of how tides arise in American oceanography classes also invokes centrifugal force and in doing so is wrong.

Here is what one of my profs has to say:
http://www.math.nus.edu.sg/aslaksen/teaching/tides.html

This link provides several links to how oceanographers envision the tides as arising. For example, lecture notes from E&ES 106 at Wesleyan University in 1999 (not your average Whatsamatta U.). Starting at slide 20,
http://soconnell.web.wesleyan.edu/ees106/lecture_notes/lecture-tides/HTML%20Presentation%20folder/sld020.htm,

Tides are differential gravitational forces, the difference between the gravitational forces exerted on two neighboring particles by a third more distant body.​

This is exactly correct. Unfortunately, the subsequent slides do not develop this concept. They instead incorrectly invoke the concept of centrifugal force.

The same explanation is http://web.vims.edu/physical/research/TCTutorial/origin.htm" (NOAA).

The problem with invoking the concept of centrifugal force is that doing so will inherently give the wrong answer for the simple reason that the apparent force in a rotating frame is equal to the force in a non-rotating frame at the frame origin only. Asking what the forces are from the perspective of a non-accelerating frame (e.g., the Earth-Moon barycenter, ignoring the Sun) is equally wrong because the ultimate goal is to determine the acceleration with respect to the center of the Earth.

There are at least a couple of ways to arrive at the correct answer when reasoning from the basis of a rotating frame or a non-Earth centered frame. The right way is to properly develop the equations of motion in the selected frame and then reframe the answer in terms of an Earth-centered inertial frame. The equations of motion as properly developed in the selected frame will include a differential gravity term plus a whole lot of other terms such as centrifugal and coriolis acceleration. Reframing the answer in terms of an Earth-centered frame will remove all that extraneous stuff, leaving only differential gravity. The answer will be correct, but why bother.

These centrifugal force explanations did not use this route. The promulgators instead mix frames, ignore coriolis force, mislabel some hand-wavy thing as the centrifugal force, and do not recast the answer in terms of an Earth-centered frame. They do come up with the right answer, but that is only because they know what the right answer is ahead of time.

 

[granpa]

http://www.math.nus.edu.sg/aslaksen/teaching/tides.html

3.
Some people, including Aslamazov and Varlamov in the book “The Wonders of Physics” use a centrifugal force, but they let the force depend on the distance from the center of rotation, which is clearly wrong. If we ignore the rotation of the Earth[/color], then all points on the Earth describes circles with the same radius, but different centers, in the course of the month.

why would we 'ignore the rotation of the earth'?

 

[granpa]

as the Earth spins on its axis it produces an outward centrifugal force but over millions of years the whole Earth itself (not just the water) has bulged at the equator thereby cancelling out this force. most of the rotation can therefore be safely ignored. but not all of it. the fraction of its rotation equal to 1 rotation per month does not produce a regular equatorial bulge. it produces a bulge on the side facing the moon and a bulge on the opposite side away from the moon. in between there is no bulge.

therefore most of the Earth's spin can be ignored but the once per month rotation can not be ignored.

hence to a first approximation when calculating tidal force we can treat the Earth as though it rotated once per month. and the centrifugal force (due to the Earth and moon orbiting each other) is proportional to distance from the earth-moon center of mass (around which they orbit). the centrifugal force at the center of the Earth (and roughly speaking at the poles) is equal to the gravitational force. to a first approximation the gravitational force can be considered to be a constant.

 

[D H]

why would we 'ignore the rotation of the earth'?

Rotation of an object means angular motion about an axis that passes through an axis internal to the object in question. The Earth rotates once per sidereal day. If you want to attack the issue of tides from the perspective of a rotating Earth, have at it. We ignore the daily rotation of the Earth to get a first-order approximation of what makes the tides arise.

That said an Earth-centered, Earth-fixed frame is the proper reference frame for describing the tides as the oceans are more-or-less rotating with the Earth. A real-world model of the tides is incredibly messy and is also quite ad-hoc (37 frequency components that vary from place to place over the world).

as the Earth spins on its axis it produces an outward centrifugal force but over millions of years the whole Earth itself (not just the water) has bulged at the equator thereby cancelling out this force.

Wrong. The shape of the Earth is, ignoring mountains and valleys, an equipotential surface, not a constant force surface. Mean sea level is an equipotential surface.

most of the rotation can therefore be safely ignored. but not all of it. the fraction of its rotation equal to 1 rotation per month does not produce a regular equatorial bulge. it produces a bulge on the side facing the moon and a bulge on the opposite side away from the moon.

Wrong again. This bulge has nothing to do with orbital motion. As Russ mentioned earlier, an asteroid headed straight toward the Earth with zero angular momentum would still raise tides -- and on both sides of the Earth. The tides result from the gradient in the gravitational force, not from centrifugal force.

in between there is no bulge.

In between there is the opposite of a bulge. Tidal forces are directed inward at the points where the Moon is on the horizon.

hence to a first approximation when calculating tidal force we can treat the Earth as though it rotated once per month. and the centrifugal force (due to the Earth and moon orbiting each other) is proportional to distance from the earth-moon center of mass (around which they orbit). the centrifugal force at the center of the Earth (and roughly speaking at the poles) is equal to the gravitational force. to a first approximation the gravitational force can be considered to be a constant.

This is nonsense.

Do the math.

 

[chmate]

Thank you all.

#

The correct explanation was given by Newton in 1687. The Moon's gravity pulls on the Earth and the water on it, but the force of the Moon's gravity varies across of the Earth. The pull is greater on the side facing the Moon, pulling the water there closer to the Moon, while the pull is weaker on the side away from the Moon, making the water there lag behind. This stretches out the Earth and the water on it, creating two bulges. Remember that both the Earth and the Moon are falling towards each other. The reason why they don't collide, is that they already have a motion perpendicular to the direction in which they are falling, so the falling only results in a change in that direction.

http://www.math.nus.edu.sg/aslaksen/teaching/tides.html

Is this explanation correct? My book uses centrifugal force to explain the tides.

Thank you for your contribution.

 

[granpa]

I appear to have made a very basic mistake. the net centrifugal force (for a set of points along the equator due to the Earth's spin and orbital motion around the earth-moon center of gravity) is not the sum of a set of vectors pointing toward the center of the Earth and a set of vectors pointing toward the earth-moon center of mass but appears to be instead the sum of a set of vectors pointing toward the center of the Earth and a set of vectors pointing toward a point at infinity.

the Earth's rotation would then be correctly entirely ignored. all points on the Earth would indeed describe circles with the same radius, but different centers, in the course of the month. the centrifugal force (due to the Earth orbiting the earth-moon center of mass) would be reduced to a constant. and the tides would be due entirely to the difference in the strength of gravity from one side of the Earth to the other.

I stand corrected. my apologies.

DH:your confusing and argumentative comments have not been helpful at all.

and I am very well aware that the surface of the Earth is an equipotential. I am also very well aware that there is an anti-bulge between the moon side bulge and the bulge on the opposite side of the earth. but it was irrelevant to my point.

but its all moot now since my arguments were based on a misunderstanding.

 

[D H]

Thank you all.

http://www.math.nus.edu.sg/aslaksen/teaching/tides.html

Is this explanation correct? My book uses centrifugal force to explain the tides.

Thank you for your contribution.

Yes, although that site is lacking in that it doesn't do the math. There is no need to invoke centrifugal force for the simple reason that centrifugal force does not explain what is going on. Using centrifugal force to explain tides is (pick one or more) the wrong explanation, the wrong math, or the wrong use of the term centrifugal force.

To arrive at the proper explanation all you need to do is compute the apparent gravitational acceleration at the point of interest from the perspective of an Earth-centered reference frame. This is a simple subtraction of the gravitational acceleration of the Earth as a whole toward the Moon from the gravitational acceleration the point of interest (usually a point on the surface of the Earth).

Since the distance between the Earth and Moon is significantly greater than the Earth's radius, first order approximation will yield a very good estimate of this differential gravitational force. Assuming a spherical Earth, the tidal acceleration at some point on the surface is

a_tidal ≅ GM_mr_e/r_m³ (2cosθ xhat - sinθ yhat)

where

• r_m is the distance from the center of the Earth to the center of the Moon
• r_e is the radius of the Earth
• θ is the angle between the vector from the center of the Earth to the center of Moon and the vector from the center of the Earth to the point in question
• xhat is the unit vector along the Earth-to-Moon vector
• yhat is normal to xhat such that the vector from the center of the Earth to the point in question is r_e(cosθ xhat + sinθ yhat).

There is not only a bulge at the sub-Moon point and its antipode, there is also a "squeeze" along the great circle defined by those two points.

 

[DaveC426913]

There is not only a bulge at the sub-Moon point and its antipode, there is also a "squeeze" along the great circle defined by those two points.

There are an infinite number of great circles defined by those two points (the meridia).

I assume you're referring to a great circle that is farthest from both? I'm not sure that two points can define such a locus on anything other than a mathematically perfect sphere. But I am not an expert in this field.

 

[D H]

The simplifying assumption is that the Earth is a perfect sphere, and (oops) the great circle in question is the meridia.