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Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: thedoc on 16/11/2016 20:53:01

Title: How does the moon impact the tides?
Post by: thedoc on 16/11/2016 20:53:01
Tom Ettinger asked the Naked Scientists:
   My question is about the super moon. On the national public radio station in United States there was a discussion of the super moon. In this discussion they stated that the biggest tides were during the full moon.  I can understand why tides are stronger when the effects of the sun and moon influence each other. Why would tides be higher when the moon and sun are opposite in the full moon, rather than to acting together When there is a new moon?
What do you think?
Title: Re: How does the moon impact the tides?
Post by: evan_au on 17/11/2016 09:51:30
The relative position of the Sun and Moon is a major force driving tide height, with maximum force around Full Moon and New Moon, when the Sun and Moon are in a straight line (syzygy). This occurs for both Full Moon and New Moon.

The minimum drive is when the Sun and Moon are at right angles (Half Moon).

The moon's orbit is a bit eccentric, so the tidal range is slightly greater on a "Super Moon", when the Moon is closest in its orbit (perigee). But this also applies when there is an (invisible) New Moon and the Moon is at perigee.

But the tide is a result of water sloshing around in ocean basins (resonance), and so the actual tide height at a particular port depends on many local factors such as the shape of the ocean basin, depth of the water, and the rate of sloshing in adjacent ocean basins, etc.

In many locations, the highest tidal range occurs 1-2 days after syzygy. This delay is called the "age of the tide" at a particular port.

See https://en.wikipedia.org/wiki/Tide#Timing

The highest astronomical tide would occur when the Earth is at it's closest point to the Sun, the Moon is at its closest point to the Earth, and the Earth, Sun and Moon are aligned. The next one is due around the year 6580 AD.
Title: Re: How does the moon impact the tides?
Post by: Janus on 17/11/2016 19:33:10
Tom Ettinger asked the Naked Scientists:
   My question is about the super moon. On the national public radio station in United States there was a discussion of the super moon. In this discussion they stated that the biggest tides were during the full moon.  I can understand why tides are stronger when the effects of the sun and moon influence each other. Why would tides be higher when the moon and sun are opposite in the full moon, rather than to acting together When there is a new moon?
What do you think?

The thing to remember is that the Moon and Sun both produce two tidal bulges each on opposite sides of the Earth.  The tidal bulges are caused by the difference in gravitational force caused by each body across the diameter of the Earth.  If we consider the Moon, Because gravity falls off by the square of the distance, the near side of the Earth feels a stronger pull from the Moon than the center of the Earth does and the far side less pull than the center. In addition, the gravity from the Moon acting on the limbs of the Earth has a slight inward component that squeezes inward.  So just like when you squeeze the middle of a rubber ball the ends tend to bulge out, you get an additional lift to the tidal bulges.   

Now because the Moon is much closer to the Earth, the Earth's diameter is a lot larger fraction of the total distance between them than is the case for the Sun. So even though the net gravitational force of the Sun on the Earth is greater than that of the Moon, the difference in gravitational force across the Earth(or tidal force) is larger for the Moon than for the Sun and the Moon creates the bigger bulges.
Title: Re: How does the moon impact the tides?
Post by: rmolnav on 01/11/2017 09:22:44
Why would tides be higher when the moon and sun are opposite in the full moon, rather than acting together when there is a new moon?
First of all, when full moon sun and moon DO act together, and then we have higher tides too.
We long discussed this issue last year:
https://www.thenakedscientists.com/forum/index.php?topic=68025.msg525412#msg525412
https://www.thenakedscientists.com/forum/index.php?topic=49715.msg504031#msg504031
I showed there what I do consider is correct Physics of this question. Too long to suggest anybody to have a look at.
I´m going to put here only main errors I´m seeing above.

1) "But the tide is a result of water sloshing around in ocean basins (resonance)” (evan_au)
As a hole, water doesn´t actually slosh around … It oscillates vertically, as an answer to changes in all forces acting on each spot (static gravity from Earth, Moon and Sun, and dynamic forces such us what due to inertia.
Sloshing and resonance phenomenon can occur only locally, due to continents/isles solid obstacles for a “free” deformation of ocean surface shape.
That deformation is what kind of “goes” around, not the water itself.

2) "If we consider the Moon, because gravity falls off by the square of the distance, the near side of the Earth feels a stronger pull from the Moon than the center of the Earth does and the far side less pull than the center" (Janus)
That is correct, but not sufficient to explain far from Moon high tides … Near side, center of Earth and far side feel decreasing pull from the moon, all of them “pulls”. Why all those masses don´t move towards the Moon? …
The answer is that Earth, apart from its rotations around Sun and its own N-S axis, is rotating (together with Moon) around Moon/Earth barycenter. That originates centrifugal forces which, at side far from Moon (and from the barycenter), and together with rest of acting forces, originate the far side high tides ...
   
Title: Re: How does the moon impact the tides?
Post by: evan_au on 01/11/2017 10:17:52
Quote from: rmolnav
As a hole, water doesn´t actually slosh around … It oscillates vertically
Since water is fairly incompressible, and doesn't expand much under tension, the only way water could oscillate purely vertically is if the whole water column lifted off the bottom of the ocean, leaving a vacuum.

The reality is that there are both horizontal and vertical components to the movement of water forming tides,
....just as there are horizontal and vertical components to the movement of water forming waves

The horizontal component of the tide (the part that moves water up and down the beach) reflects off landforms, forming tides of high amplitude in areas where the resonant frequency of a bay or inlet is close to the driving frequency of the Sun and Moon.
Title: Re: How does the moon impact the tides?
Post by: rmolnav on 02/11/2017 08:17:21
Why would tides be higher when the moon and sun are opposite in the full moon, rather than to acting together When there is a new moon?
Firstly I´ve seen that in my last post this quote was said to be Janus´s, but it was thedocs´s.
Secondly, I mixed words when saying "when full moon sun and moon DO act together, and then we have higher tides too" ... That, bar the word "too", is what said by thedoc. I meant when "new" Moon.
Without any further discussion about centrifugal forces, now I´ll just say that the kind of surprising higher tides when full Moon (Moon and Sun pull in opposite senses) is due to the fact that Earth and Moon rotate together around their barycenter. Centrifugal forces occur, bigger at Sun´s side (opposite to moon), and those forces and Sun pull add up. When not in line (30, 60, 90º ...) the addition of those force vectors gives a smaller force, and not directly opposite to own Earth´s pull on oceans waters.
Logically, when new Moon both Sun and Moon pulls are the forces which act together and add up directly, at noon side: higher tide there. At mid night side, what add up are centrifugal forces: the Moon related above mentioned, and the one due to Earth´s rotation around the Sun: also a higher tide there.
Title: Re: How does the moon impact the tides?
Post by: rmolnav on 02/11/2017 12:06:19
Quote from: rmolnav
As a hole, water doesn´t actually slosh around … It oscillates vertically
Since water is fairly incompressible, and doesn't expand much under tension, the only way water could oscillate purely vertically is if the whole water column lifted off the bottom of the ocean, leaving a vacuum.
The reality is that there are both horizontal and vertical components to the movement of water forming tides,
....just as there are horizontal and vertical components to the movement of water forming waves

Firstly, I have to say that, as English is not my mother tongue, I misunderstood the meaning of “sloshing around” … That “around” made me think you meant a huge amount of water were going around the Earth, trying to be the nearer to Moon, the better …
That´s an idea many people out there have. They think the “excess” of water we have if high tide comes from where low tide (some 10,000 km journey, at the equator !!). And I wanted to warn them about that error.
It was a surprise to me you could think so, and I should have checked its actual meaning before replying … I beg your pardon.
Secondly, I find correct what you say about costal effects of tides. I had already said something about it.
But those are local secondary consequences of tides, not the essence of the phenomenon.
Estuary “branches” can have any shape (also their bottom) and direction, and there are places where currents (when sea level is rising) can even have sense W—>E, opposite to high tide bulge movement (E—>W).
That can´t be caused directly by Moon´s pull, main reason of tides. But if at “next door” offshore area sea level is rising, water will enter the estuary branches, whatever their direction ...
It can be even more complex in cases of some type of resonance.
Thirdly, I´m sure you know that at open ocean the necessary horizontal water movement, certainly real, can be relatively small. Water pressure distribution varies, and with small movements of each “drop” of the water of a bulge, and rest of below and westward water, we can fill the bulge to be, thousands of km away (if we are comparing a current position of a bulge with where it will be some six hours later, now with low tide) . Water from higher latitude areas also helps, similarly.   

 
Title: Re: How does the moon impact the tides?
Post by: rmolnav on 15/11/2017 12:19:08
Looking for something else in Youtube, a couple of days ago I found a VERY interesting video, which thoroughly analyzes both tides basic causes (globally considered: Moon, Sun ... the two "standard" bulges), and local tides (water sloshing commented by evan_au). Don´t miss it!
Said that, its authors are among the certainly big group of scientists that don´t want even mention centrifugal forces ... Short time ago I sent a comment explaining why I consider they are wrong, as far as that concrete issue is concerned:
Title: Re: How does the moon impact the tides?
Post by: rmolnav on 17/11/2017 12:02:46
Said that, its authors are among the certainly big group of scientists that don´t want even mention centrifugal forces ... Short time ago I sent a comment explaining why I consider they are wrong, as far as that concrete issue is concerned
In my last post I intended to include just the link to the youtube page, but the video "play" engine (?) turns up …
But what don´t appear are sent comments …
What follows is what I sent, not literally (I had a lapsus and had to correct some details)
Both Moon and Sun, causes of tides, are similarly considered in the video, without even mentioning centrifugal forces.
As we know, Moon is main cause (particularly in timing), and Sun tides affect total tides only in their cycling intensity.
But Moon/Earth dynamics is rather trickier than Sun/Earth´s. as far as distribution of forces is concerned. That is why I´ll refer to last one before.
Diagram timed in the video 2:23 is about Moon related bulges. Let us imagine a similar diagram, but with the Sun instead of the Moon. Shown "medium"  gravitational pull (average as if acting on Earth´s C.G.) is rested both at nearer and farther sides, trying to explain the bulges ...
The authors don´t consider those vectors any centrifugal force, they´re just comparing to find kind of "relative bulges" caused by "relative forces". In other words, the usual concept of inertial frame of reference, which I consider applies to distances, velocities and accelerations, but that applying it to forces is an erroneous "trick". But I will not discuss that now any further. 
What´s causing "medium" black vector at C.G.? : the centripetal acceleration required for the rotation. But that REQUIRED (to orbit at each distance) acceleration is NOT uniform across the Earth, horizontally in our diagram. Farther parts of the Earth, to rotate at same angular speed (an average of 2π radians a year), NEED higher centripetal acceleration than closer ones: just the OPPOSITE to what happens with gravitational forces.
ONLY internal stresses can compensate that imbalance ... Thanks to them, the excess of pull on closer parts is transmitted, from right to left in the diagram, to farther parts ...
But according to Newton´s 3rd Motion Law, if at any considered Earth´s section, right part is pulling (towards the Sun) contiguous left part, this one is also pulling the former towards the left, same force but opposite direction: OUTWARDS.
All those internal forces are centrifugal REAL forces, which, together with their "mirror" centripetal ones (the fraction not necessary for the rotation), stretch the Earth, solid parts included.
Logically, that deformation is much bigger where liquid parts, our oceans. And the TWO Sun related bulges ARE DUE TO THAT.
In another post, better tomorrow, I´ll put what Moon related …This way could be better for some of possibly interested folks: t could facilitate a posterior understanding of the trickier case of Moon related bulges.


Title: Re: How does the moon impact the tides?
Post by: rmolnav on 18/11/2017 12:22:36
ONLY internal stresses can compensate that imbalance ... Thanks to them, the excess of pull on closer parts is transmitted, from right to left in the diagram, to farther parts ...
AS A CONTINUATION of my post of yesterday, yet another analogy, to help understanding.
Yesterday I happened to have to buy a new tyre for my byke. When I had it hanging from my right hand, I felt how heavy it was … and saw some deformation: its circular shape got slightly oval.
What physically was happening is clear: gravity was acting across the tyre, but my still hand was preventing it from getting a g downwards acceleration. My hand was pulling upwards, and total force acting on the tyre was null (forgetting consequences of Earth´s movements …)
But my hand upward force was neither acting on the tyre´s C.G., nor uniformly distributed across the tyre (as gravity was). Similarly to what I said yesterday, through internal stresses my hand´s pull was being transmitted and distributed along the tyre, acting both upwards and downwards (Newton´s 3rd Law), each part of the tyre was still, but the tyre was being stretched vertically.
Then I tried and spinned myself with my arm almost horizontal, the tyre still gripped by my hand, as if I were going to throw it like a disc by an athlete.
I was feeling I had to pull the tyre inwards, the higher the angular speed, the higher the necessary pull. My hand was making rotate actually only what it was touching, but required force was much bigger than if my hand were gripping only some ten cm piece of tyre ...  Rest of the tyre rotated because, almost identical reason as what quoted, "the excess of pull on closer parts was being transmitted, outwards in this case, to farther parts …"
In Earth rotation around the Sun I was talking about yesterday, centripetal force is not initially exerted through physical contact at a spot as in the tyre case…
Gravitational pull is distributed over Earth sphere. Not having a pull "concentration" at any spot, "excesses of pull" are much smaller, but, as explained yesterday, they also happen. And also centrifugal REAL forces appear, due to mentioned Newton´s 3rd Law.
Because of that, deformations due to all those forces (particularly tidal bulges) have to be relatively much smaller than in the tyre case. And that is what really happens. Sun related bulges (by the way, very much exaggerated term …) are only a few meters from high and low tides … 10,000 km apart !! (at the equator).
 
Title: Re: How does the moon impact the tides?
Post by: rmolnav on 19/11/2017 11:48:33
In another post, better tomorrow, I´ll put what Moon related …This way could be better for some of possibly interested folks: t could facilitate a posterior understanding of the trickier case of Moon related bulges.
Here it is ... though a day later due to additional analogy of yesterday (hope useful).
Here it is ...
Mentioned video diagram timed 2:23 is misleading: it seems like a static situation … Authors not only ignore any possible centrifugal force: they don´t even mention any centripetal force. If only what shown were the physical reality there, why Earth doesn´t moves toward the Moon?. Because of same reason as Moon doesn´t move towards the Earth: both are actually rotating around their barycenter, and gravitational pulls are causing their centripetal accelerations. 
As that barycenter is situated app. 2/3 Earth´s radius from its center (logically, at Moon side), Earth´s rotation in this "couple dance" is a kind of wobbling, similar to when a child plays with a hula-hoop.
Though Earth daily spinning generates much, much stronger internal stresses and centrifugal forces (some 28/29 times higher angular speed …), it acts always same way: trying and increasing Earth´s diameter at the equator and lower latitude parallels.
That Earth´s wobbling generates internal stresses and deformations which we can perceive added to those mentioned stronger, permanent ones.
Let us imagine video diagram timed 2:23 with a vertical line some 2/3 Earth´s radius at right side of its center: the axis of this rotation.
Earth parts at the left of that axis require a total force per unit of mass proportional to its distance to that axis, in order to get required centripetal acceleration at ach considered spot. But gravitational pull from Moon is the farther the smaller (to the square of the distance to the Moon). That imbalance can be compensated only by internal stresses: parts closer to the axis have to add an inward pull on contiguous farther parts … Then we have Newton´s 3rd Law, and each part pulls inner contiguous one with same but opposite (and REAL) force, that is, outwards, "fllying from a center" (axis of rotation), CENTRIFUGAL …
At Earth parts at right side of that axis, required centripetal forces are smaller, also the farther from the axis, the bigger. BUT now gravitational pulls from the Moon are opposite to required centripetal forces. Only own gravitational Earth´s pull can supply required centripetal forces …That´s similar to what happens due to daily Earth´s rotation: it tends to increase sea level at that spherical segment. And that effect adds to what derived from the obvious fact that there Moon´s gravitational pull is bigger, and a bulge also occurs at that side, due to both reasons ...
And due to Earth´s daily spinnig, those two opposite bulges seem to rotate around Earth, trying to keep in line with Moon. Result: the main part (Sun caused tides have to be added) of our "popular" (but not so well known) tides …

 
Title: Re: How does the moon impact the tides?
Post by: rmolnav on 12/12/2017 11:31:33
Let us imagine video diagram timed 2:23 with a vertical line some 2/3 Earth´s radius at right side of its center: the axis of this rotation.
Earth parts at the left of that axis require a total force per unit of mass proportional to its distance to that axis, in order to get required centripetal acceleration at ach considered spot. But gravitational pull from Moon is the farther the smaller (to the square of the distance to the Moon). That imbalance can be compensated only by internal stresses: parts closer to the axis have to add an inward pull on contiguous farther parts … Then we have Newton´s 3rd Law, and each part pulls inner contiguous one with same but opposite (and REAL) force, that is, outwards, "fllying from a center" (axis of rotation), CENTRIFUGAL …
I wish we ALL had a better knowledge of nature, or at least more imagination ...
I´ve long been arguing in line with what above, and neither myself, nor anybody else so far (unless they kept it "in secret") have imagined what below, and realized that what quoted is WRONG !!
Some people actually said it was wrong, but not for the correct reason. Most of them just considered that centrifugal forces are never real, but only apparent of ficticious ... And I gave many examples of REAL centrifugal forces ...
But recently I learnt that the 28/29 days circular movement of the couple Earth/Moon around their barycenter, though for the Moon is a rotation, for the Earth is NOT: Earth only REVOLVES around mentioned barycenter ...
That implies that all Earth points, center of mass included, follow equal circular paths.
Subsequently, my explanation, that would be valid for the Moon (it rotates at 2π radians/some 28 days angular speed), is NOT for the Earth !!
And possible imbalances between Moon´s pull on each point (either where solid or liquid parts), that could cause centrifugal forces, are not so straightforward to tell (?) ...
I´m trying to sort that out, but not easy ... For all my discussions on the subject, I´ve read many, many articles. But I NEVER saw any reference to what above !!
Title: Re: How does the moon impact the tides?
Post by: rmolnav on 15/12/2017 10:54:22
But recently I learnt that the 28/29 days circular movement of the couple Earth/Moon around their barycenter, though for the Moon is a rotation, for the Earth is NOT: Earth only REVOLVES around mentioned barycenter ...
That implies that all Earth points, center of mass included, follow equal circular paths.
Subsequently, my explanation, that would be valid for the Moon (it rotates at 2π radians/some 28 days angular speed), is NOT for the Earth !!
And possible imbalances between Moon´s pull on each point (either where solid or liquid parts), that could cause centrifugal forces, are not so straightforward to tell (?) ...
I´m trying to sort that out, but not easy ...
As a continuation of #11, I can say that, as far as I can understand, I´ve already sorted it out ...
Another day I´ll further elaborate on the issue, but now I´m going to put here main points.
Curiously, as Earth only revolves around the barycenter, all particles of the Earth follow equal circular paths, its center of mass included, logically at same angular speed (2π radians every some 28 days). So, centripetal accelerations required for the circular movement of each particle are the same in value, but the direction of each one is towards the center of its circular path.
As those accelerations can´t match with Moon´s pull on each particle, there is an imbalance, similar (but more complex) to what in #10. That imbalance only can be countered by internal stresses. If we analyze them along line Moon´s C.G./barycenter/Earth´s C.G., and similarly orientated lines, we find (as in #10) that particles closer to the Moon have to add a pull on contiguous farther parts … And due to Newton´s 3rd Law, each part pulls inner contiguous one with same but opposite (and REAL) force, that is, outwards from Moon.
Whether those last forces should be called "centrifugal" is arguable ... But they, in any case, together with all Moon´s pulls on each particle,  stretch considered line of particles, both where solid Earth, and where water.
Other similarly orientated lines of material suffer similar stretches, but smaller ...
Deformations were solid Earth depend on elasticity. But water logically is more free to move, and that´s the cause of high tide so called "bulges".
Title: Re: How does the moon impact the tides?
Post by: rmolnav on 20/12/2017 08:01:00
Curiously, as Earth only revolves around the barycenter, all particles of the Earth follow equal circular paths, its center of mass included, logically at same angular speed (2π radians every some 28 days). So, centripetal accelerations required for the circular movement of each particle are the same in value, but the direction of each one is towards the center of its circular path.
As those accelerations can´t match with Moon´s pull on each particle, there is an imbalance, similar (but more complex) to what in #10. That imbalance only can be countered by internal stresses. If we analyze them along line Moon´s C.G./barycenter/Earth´s C.G., and similarly orientated lines, we find (as in #10) that particles closer to the Moon have to add a pull on contiguous farther parts … And due to Newton´s 3rd Law, each part pulls inner contiguous one with same but opposite (and REAL) force, that is, outwards from Moon.
I think I¨ve already managed to grasp the actual mechanism of internal stresses in cases, like Earth attracted by Moon, when the object revolves, but doesn´t rotate.
ALL MATERIAL POINTS not only follow equal circular paths, logically at same speed, but even they ALWAYS are at farthest (from Moon) point of their own path …
Then, ALL centripetal forces required for the circular movement of each particle are equal both in value and in direction: ALWAYS parallel to line Earth´s C.G./barycenter/Moon´s C.G., and towards the Moon.
If we imagine the revolving object cut into slices (all its material equally far from Moon), and analyze in detail all forces (gravitational, interactions, and the ones required for their circular movements, ALL PER UNIT OF MASS), we find that mentioned "differential gravitational forces" are what remain kind of "free" (not necessary for the circular movements), and either stretch solid Earth, or move water (and change pressure) towards the Moon (where excessive Moon´s pull), or in the opposite direction (where excessive centrifugal force).
Being all centripetal forces equal, "excessive" pull at closer slices is transmitted outwards (at same  rate as Moon´s pull decreases) ... Between each pair of contiguous slices opposite pulls occur (Newton´s 3rd Law), that could be arguably called centripetal and centrifugal ...
So, "differential gravitational forces" tool (usually called "tidal forces") kind of hides a "chain of transmission" of forces between contiguous particles, centrifugal forces included, what affect the whole revolving object.
I used to think "how on Earth so many people, even scientists, say those differential gravitational forces are the unique deep cause of tides, without even mentioning any circular movement, let alone centrifugal forces ?? ... Centrifugal forces HAVE TO be taken into consideration !!!"
And THEY WERE, but without most people (including me) noticing ... !!!
Title: Re: How does the moon impact the tides?
Post by: rmolnav on 20/12/2017 20:30:56
As a continuation of #13, just a detail to avoid anybody could get mislead.
When talking about position of material points relative to the Moon, we have to keep in mind that what is being analyzed is exclusively the  Moon /Earth "dance" around  their barycenter... The daily spinning of Earth has to be disregarded.
Title: Re: How does the moon impact the tides?
Post by: rmolnav on 24/12/2017 18:44:45
If we imagine the revolving object cut into slices (all its material equally far from Moon), and analyze in detail all forces (gravitational, interactions, and the ones required for their circular movements, ALL PER UNIT OF MASS), we find that mentioned "differential gravitational forces" are what remain kind of "free" (not necessary for the circular movements), and either stretch solid Earth, or move water (and change pressure) towards the Moon (where excessive Moon´s pull), or in the opposite direction (where excessive centrifugal force).
Being all centripetal forces equal, "excessive" pull at closer slices is transmitted outwards (at same  rate as Moon´s pull decreases) ... Between each pair of contiguous slices opposite pulls occur (Newton´s 3rd Law), that could be arguably called centripetal and centrifugal ...
So, "differential gravitational forces" tool (usually called "tidal forces") kind of hides a "chain of transmission" of forces between contiguous particles, centrifugal forces included, what affect the whole revolving object.
As most likely not everybody buys what quoted, I´m going to put what follows, that I sent last year to a scientist by email, when discussing on the issue. I´m sure this analogy can help understanding:
"Surely you also played what in Spain we call "el látigo", literally in English the whip: a row of several boys, each left hand grasping right hand of contiguous inner boy … "Inner" because they run around one of them at a center, trying to keep the row in a radial line at a certain equal angular speed.
Let us consider five boys, A, B, C, D, and E, A being on a fixed point.
Let us suppose each boy hand-to-hand distance is 1 m (logically, the same between their C.G.), they weight of each 40 kg, and they rotate at 1 radian/sec.
Centripetal acceleration of E has to be 1x1x4 = 4 m/sec2. Centripetal force, 4x40 =160 newtons. It only can by exerted by right hand of D pulling inwards left hand of E.
Centripetal acceleration of D has to be 1x1x3 = 3 m/sec2. Required centripetal force: 3x40 =120 newtons. But he is "suffering" two opposite pulls: one outwards (centrifugal) exerted by E, of 160 newtons (3rd Newtons´Principle), and another inwards, exerted by C. If net force on D has to be 120 newtons inwards, right hand of C has to pull left hand of D with an inward force of 120+160=280 newtons.
Also due to 3rd Newton´s Principle, that means right hand of C is pulled outwards (centrifugal) by left hand of D with a force of 280 newtons ...
… C : 1x1x2 = 2 m/sec2 … 2x40 = 80 newtons … 280+80= 360 newtons.
… B : 1x1x1 = 1 m/sec2 … 1x40 = 40 newtons … 360+40= 400 newtons.
The result is boys, especially their arms, get stretched.
All those hand-to-hand transmitted forces are similar to what happens between contiguous pieces of material of above mentioned "columns" perpendicular to axis of rotation. Though in last case there is another important "source" of centripetal forces, Moon´s attraction, it can´t exert required forces along those columns, because Moon´decreases with distance, and required centripetal force increases with distance from axis of rotation.
There is also the difference that own Earth´s gravity effect, important for tides, does´t  significantly affect rotation of boys (energy wasted in friction could be reduced to a minimum with boys skating on ice). There are also other centrifugal forces due to daily Earth´s own rotation, which produces a permanent bulge around the equator.  But what exposed above is the real scenario in comparison with if Earth were just not rotating around Moon/Earth.barycenter, with the purpose of analyzing forces caused by that rotation alone.
The result of all those forcers is both a stretch of solid Earth´s parts, and the pseudo-equilibrium of oceans with its "theoretical" spherical shape very slightly changed, with two bulges".
By that time I had not yet learnt that Earth doesn´t actually rotate, only revolves. Then, the comments to the analogy should be changed.
On the one hand, in the game required centripetal forces increase proportional to the radius, but all particles of Earth follow identical circles, what requires same centripetal accelerations ...
On the other hand, in the game only "internal" stresses supply force for centripetal accelerations, but in our real case Moon gravitational pulls (on each Earth´s particle) are main cause of centripetal accelerations. But not uniformily ...
BUT, similarly to the game, between contiguous slices of any imaginary "column" of material parallel to line of centers of G. Moon/Earth, opposite pulls have to turn up to compensate the imbalance, in such a way that the net force acting on each slice (three vectors to add: Moon´s gravitational pull, and two opposite pulls from contiguous particles), divided by its mass, gives the required centripetal acceleration, the same across the Earth as previously said.
Then, centrifugal forces resulting distribution is kind of a "mirror" image of gravitational pull distribution. At Earth C.G. there are not "spare" forces (all Moon´s pull is spent in centripetal acceleration), but at closer and farther points that balance does´t occur: at closer points there is an "excess" of gravitational pull, and in farther points a deficit, what results in "spare" outwards forces ("arguably" centrifugal ... perhaps another day I´ll come back to that).
BOTTOM LINE: the well known diagram with average pull at C.G., bigger vector pull at closer side, and smaller one at further side, which after deducing average bigger vector changes direction to side opposite to the Moon ...