Ocean's Tides Explained

[@sherlok1280]

Tides bulge at C point is due to centrifugal force.

[@hanifhanif293]

While explaining the TWO TIDES, How the Point B ( earth ) moves towards the Moon.... ?? This will create Gravitational imbalance and Earth and Moon will merge and Destruction and a second Big Bang!

[@BeedrillYanyan]

For the same reason our space satellites don't fall into the ground.

[@atomf9143]

No? The Earth moves towards the Moon, yes, but it is also rotating and moving in orbit around the sun. It's a bit hard to visualize, but it's like two runners running parallel to each other around a circular track. Because of how they're positioned they'll always be close, and will eventually adopt similar speeds, but due to the shape of the track they can never touch each other.

[@kankudevikothari5522]

If you have taken a science class, you know that the moon attracts earth just as much as the other way around. The force is equal both ways. The reason why the Moon and Earth don't fall into each other is because the Moon is too quick and dodges Earth creating an ellipse, like how ISS orbits Earth. Like how a cannon ball takes longer and longer to fall the faster you launch it, there will be a certain velocity where the cannon ball is constantly falling without hitting the ground. I recommend watching a few videos or reading a nice illustrated book 🙂👍

[@lilydash1864]

no one gonna talk about 1:48

[@kpcgkhn]

that explanation is not true purely. moon and earth are orbitting around each other, and the barycenter is in earth. so earth is swaying around this causing a orbital velocity inequality. the far side from barycenter on the earth is speeding with a high orbiting velocity, so centripedal force causing antigravititational effect and there gravity decreases. so water tides up. same thing is viable on moon, but more antigravititation would become. (this effect is very little on earth, 0.027m/s around barycenter causing 6.91*10^-8 m/s^2 centripetal acceleration, moon's gravity on earth is about 3.32*10^-5 m/s^2)

[@AtomicSchool]

See the beta video version why centrifugal force is a misunderstanding kzbin.info/www/bejne/ap_PgYyZoNCpg9k

[@TileBitan]

antigravity? shame on those poor 42 souls who upvoted you

[@WAGNERMJW]

The assertion of the far side high tide is illogical. It's not as if the ocean is an independent and static object floating inside a water balloon such that the earth can be pulled back and forth while the ocean stays largely static and unaffected by gravitational pull. When the moon pulls at the ocean on its near side it is also pulling the ocean to flow around the earth toward it, thus causing the bulge, the high tide. It's not that the moon is pulling the h2o molecules apart from each other on the near side causing the bulge, at least not measurably. It's because there's more water that's been pulled toward the side facing the moon.

[@Greg042869]

The water is pinched, not pulled.

[@maxwang2537]

2:40 This explanation of two high tides a day is really bizarre - or beyond comprehension to me. The moon is dragging the earth in the middle of a ball of water with a bulge!

[@AtomicSchool]

It's pulling the near-side bulge more than the earth, and pulling the earth more than the far-side bulge

[@maxwang2537]

@@AtomicSchool you must keep in mind this is only what you *wish* it was or it was the reason why. It’s too far off to explain something without science, but wish. Sorry this might be too harsh. This is of course not saying I understand fully the correct and scientific answer, I don’t and am only on my way of finding it out. But this clearly is not a convincing scientific explanation.

[@danneumann3274]

its beyond comprehension because You are using common sense. You are not easlily convinced of nonsense. as if the earth really shifts as the video shows

[@AtomicSchool]

@@danneumann3274 Yes, it's a bit misleading. I'm going to clarify this. The forces are really 3D whereas this is only showing 2D. A discussion of vectors would be useful.

[@gabrielM1111]

Yeah this guy has no idea what he's talking about or how to write. 01:10 In the beginning he claims that the Earth and the moon's gravitational strength upon each other are equal which is absolutely wrong. The more mass an object have the greater the gravitational strength is so the Earth's gravitational strength is greater than the moon's.

[@davidsmith9756]

Only ever proven in a cartoon or CGI. I don't like to say things like this but it's true and undeniable. They never use real genuine unequivocal evidence, never ever, and it frustrates me.

[@user-ei6ti4db5t]

A perfect description of tides. Thank you. Well actually, it’s only correct, if you’re name is Aristotle (circa. 384-322 BCE) and you still believe in a flat earth and the geocentric model? Fortunately, science has come on a long way since then? Sorry nice try? Please refrain from using outdated information and keep up with the recognised curriculum taught in present day schools. Thank you.

[@AtomicSchool]

Thanks back :)

[@jayjeetnandi2024]

But at the point C there is centrifugal force which balance it I think 🤔

[@AtomicSchool]

Centrifugal force is only valid in a non-inertial frame of reference. There's a video coming on that, Jayjeet.

[@jayjeetnandi2024]

@@AtomicSchool yeah but The water on the far side is “unbalanced” because the moon’s attraction there is weaker than it is at the center of the earth, where it just balances the “centrifugal force".

[@AtomicSchool]

@@jayjeetnandi2024 This requires an entire video to tackle the concept. Coming

[@jayjeetnandi2024]

@@AtomicSchool But when you will make this video and this video is not enough for others

[@AtomicSchool]

@@jayjeetnandi2024 Making videos is time-consuming. At the moment I am making a video about Newton's Laws to cover the basics. Also inertial and non-inertial frames of reference, which is needed to understand centrifugal force. If you can atriculate the centrifugal force explanation maybe you can make a video yourself?

[@thegoodthebadandtheugly579]

By the sounds of it, the earth and the moon should get a room.. 🙄

[@ourtruth216]

🤣😂😂😂😂

[@laundrymat2963]

I get it haha

[@michaelmorrison4201]

Na, they got plenty of space...

[@iamsophie3885]

@@michaelmorrison4201 Ha, I see what you did there.

[@pebbleinthecorner]

damn ok... 0.0

[@darrenralph5845]

it push not pull

[@AtomicSchool]

See the beta version kzbin.info/www/bejne/ap_PgYyZoNCpg9k

[@CaptainLSB]

Thats what your mother said. Shut the fuck up.

[@DaumbFauckProduction]

it's both

[@tempus.narrabo]

So many unanswered questions, right off the bat, what's stop the pull of the earth and moon? Same pull? The moon is supposed to be 1/3 size of the earth, earth surely would have more pulling power, mass?

[@mikev4621]

the Moon is 1/80 th the mass of the Earth

[@Doruc7158]

Gravitational force of The Earth and the Moon are opposed. Moon's gravitational force is 6 time weaker than Earth"s. So, should be no gravitational attraction from the Moon. The Moon is not "falling" on Earth because of the rotational speed of Moon, "Centrifuge force" that nullifies the Earth's pulling force of the Moon towards the Earth. Comments?

[@AtomicSchool]

It's hard to respond with a short reply. The moon's gravity is actually much weaker than 1/6th of the earth's gravity on, say, a kilogram of seawater on the surface of the earth as it is much further away. However, this tiny difference applies to the entire diameter of the earth/ocean, distorting the ocean's shape over its 12,000 km diameter. So the tidal variation of about 1 metre is a tiny effect, and not unreasonable.

[@suharins]

If moon pulls ocean toward itself. Then why it can pull rain droplets toward the moon. And if fill balloon with water and place it on flat surface it will roll toward moon?

[@bespokepenguin103]

Because water on Earth experiences higher gravity than from the moon the moon. Inverse square law. An increase in distance causes less gravity.

[@kankudevikothari5522]

Basically, the tides in the illustrations are very exaggerated. In reality the change is only a foot in most places. A similar change is also seen with the atmosphere and the raindrops. I hope it's clear now :)

[@michaelsrowland]

Does the moon stay still?

[@AtomicSchool]

The earth and moon revolve around a common centre of mass (barycentre) which is slightly inside the earth surface. However, this revolving is not the cause of tides and is not covered in this video. Future videos will cover it.

[@vikrantsharma3946]

But this is due to centripetal and centrifugal force

[@mitchbollo1100]

Yes indeed, you are correct on both counts. Here try this-(kzbin.info/www/bejne/i6TJcqSbmK-hgdU and offer them your support instead, because the teacher here doesn’t understand the barycentre.

[@vikrantsharma3946]

Yeah absolutely wrong

[@ronsemneck162]

If the forces shown in the cartoon were the only forces acting, then the points A, B, and C would have different accelerations, and the earth would soon be torn apart. So, this is not a good explanation for tides. You Failed the test.

[@AtomicSchool]

These are net centripetal forces that result in orbital motion about the barycentre, rather than tearing the earth apart.

[@KathleenGuillot]

Please make a video and explain it better. 2years later...still waiting.

[@mreatboom1314]

The earth is torn apart by the tide force that make it a bit non spherical

[@jebythomas9840]

Sir I have doubt. Is there any effect of centrifugal force at the POINT C

[@pennywise5670]

You are actually not correct here, because this common explanation is completely wrong. If it were the moon pulling water towards it, tides would only go up on the side facing the moon accumulating at the point nearest it. It’s more like the Earth’s centre of mass is squeezing all the water towards its centre. Thus, the pressure of this squeeze is least at the line made by the Earth’s centre of mass and the moon’s centre of mass. That is why it bulges on both sides. Imagine your hands on either side of a balloon pressing together.

[@dominicsierakowski6755]

Great explanation. Love it

[@KHJ4D]

How about the time the tides happen?

[@RareVBlue]

Well who ever made this video definitely misunderstood the assignment. This is incorrect in many ways.

[@AtomicSchool]

See the beta version of why centrifugal force does not explain the C bulge. kzbin.info/www/bejne/ap_PgYyZoNCpg9k

[@pascalstriangle8708]

No. Logic failure. Your example of “moon on the right, creating a high tide on the left” with your rationale being that, the “earth is pulled away from the water and towards the moon” fails both logic and physics. The southern ocean is not anchored to something, so that the earth can be pulled away from it. Whatever gravitational pull the moon has upon the earth, it also has on the water, such that the water should, can, and does, flow in both physical and density terms, cyclically chasing the source(s) of the extraterrestrial gravitational pull, in constant marginal distribution of the majority of the oceans. And volumetrically toward the source(s) of gravitational pull. Therefore, the high tide on the side opposite the moon is due entirely to a gravitational source on that side, or from some as yet unknown, physical law or power offsetting the moon’s weaker gravitational influence. The earth does not move towards the moon in a straight line.

[@somewherenorthofstarbase7056]

The answer is that the center of mass of the earth moon system is 3/4 the way from the center of the earth. Therefore the mutual gravitational effect creates a wobble in the earth that manifests a centrifugal force directly opposite to moon that creates a bulge here.

[@pascalstriangle8708]

@@somewherenorthofstarbase7056 That is absolutely correct. Then why are you watching this? It makes no sense or mention of the barycentre or centrifugal force and instead says the earth moves towards the moon in a straight line and somehow magically stops! Maybe watch - What Atomic School Gets Wrong About Tides - instead first? They will answer any questions for you.

[@xtian9493]

How does the earth turn into the ocean when the ocean is in the earth? The planet is not surrounded by the ocean. Im really confused with this explanation. This explanation only makes sense if the earth is flat.

[@BeedrillYanyan]

He had to exaggerate to better illustrate. Just think of it as one object. The part that's closer to the moon is attracted more than the part at the opposite side.

[@atulnamdev7546]

Assume that sun is not there.Now moon revolves around the earth so the points behind the earth, as seen by moon experiences centrifugal forces, Hence there is bulge of water at opposite direction also .

[@nialloftara5909]

Water flows a lot more readily than rock or dirt, so the oceans have a bulge that flows across the surface pointing towards and away from the moon, the planet still rotates but the surface of the water in the bulge is simply higher, so as the world turns the water flows up that bulge and then back down. Similar to the way low pressure storm systems can actually raise the local sea level because the storm reduces the air pressure resulting in less force pushing down on that area of the water.

[@hansdegroot652]

The earth is a ribbon

[@vincentmcardell8183]

The main problem with this explanation is that we do not know it to be true.

[@SuperStargazer666]

I always thought that the tide swept around the Earth. Now it transpire it’s the the Earth is moving and sweeping past the tides. 😐

[@pillarsofhercules8266]

Scusa, non è così che funziona la gravità e non è così che vengono calcolate le maree e i tempi delle maree.

[@CaptainLSB]

Bro really just said no 💀

[@natecote1971]

Dude the more I look into this the more I feel like it's all bs and makes no sense! If the moons gravity is soooo strong it pulls our entire ocean towards it why doesn't it affect us? Why can't I jump higher at certain times of the day? Why doesn't the tide change with the earth's tilt?

[@AtomicSchool]

Hi Nate. If you were as high as the diameter of the earth, then you would stretch at both ends by a metre or so, as long as you were completely liquid. One meter over 12,000 km is a tiny effect.

[@goodester6924]

you know there is a slight difference between your mass and the oceans'

[@dino-by7vc]

vc e o cara tão errados a gravidade imposta pela lua não é tão forte a ponto de te deixar mais leve porém como a água é fluida ela pode ser distorcida por essa gravidade, mas se fosse por isso toda a água iria ser atraída porém como os oceanos são contínuos como uma malha em volta da terra ela pode ser afetada

[@dino-by7vc]

English You and the guy so wrong the gravity imposed by the moon is not so strong as to make you lighter but as water is fluid it can be distorted by this gravity, but if that were why all water would be attracted but as the oceans are continuous like a mesh around the earth it can be affected

[@goddess2885]

This man still replies to the comments after 3 YEARS What a legend

[@AtomicSchool]

Or you could just say that I'm slow :)

[@leinadpq9527]

@@AtomicSchool heeey... How come the moon and earth doesn't colliede wheb gravity pulling them together?

[@someonewholikesgames4944]

leinaD PQ earth gravity is more powerfull

[@itsawesome7110]

@@leinadpq9527 Because the gravitational pull is equal between the earth and the moon, he literally talked about it in the video

[@deviprasathp8759]

@@leinadpq9527 The Moon doesn't fall into the Earth attracted by the gravity because it moves fast enough to avoid this. So instead of falling, it keeps an orbit around the Earth at constant Tangential velocity. If it moved too slow, it would crash into Earth. If it moved too fast, it would escape into space.

[@hansdegroot652]

Mmm ithe tidal differences are the same here in holland at least more,than 1.93 meter on a calm day but i was in denmark itcwas less than 30 cm difference. Very strange. I checked. It was realy open sea and they did not have dunes or any ithe barriers to block high tide.

[@maryann8794]

what does "bulge" really mean? Is it like... the earth stretches???

[@AtomicSchool]

The ocean's shape becomes like an egg's. The earth also stretches a bit because it's not completely solid, but not by as much.

[@maryann8794]

@@AtomicSchool how can the ocean's shape change!?

[@AtomicSchool]

@@maryann8794 In a uniform gravitational field the ocean is spherical. With a gravity gradient caused by the presence of the moon the shape changes to an oblate spheroid. In outer space without gravity a drop of water would be spherical. If a planet or sun were nearby it would become distorted in shape and accelerate towards it.

[@jockymc5114]

This isn't how I believed the side at the opposite end of the Moon worked. My understanding is that the high tide at point C is caused by a) centrefugal force due to the Earth spinning and b) the Moon's gravity being weakest at this point

[@maxwang2537]

Your understanding is correct and shared in general and in a simplistic sense.

[@logankageorge]

@@AtomicSchool A "fictitious force" does not mean that it is fictional and does not exist. It means that it exists in a rotating reference frame, which is what we have when we stand on the rotating Earth. Your Lockhaven link is broken now, btw.

[@AtomicSchool]

@@logankageorge The centrifugal force only appears when we subtract the earth's real gravitational force from all real forces in the rotating frame of reference to make our framework seem stationary. This subtractive force is equivalent to a repulsive gravitational force even though all gravitational forces are attractive. A fictious force does not arise from any physical interaction between two objects, such as electromagnetism or gravitation.

[@AtomicSchool]

The Lockhaven link worked at the time of writing 2 years ago. Maybe the Professor has retired?

[@DaveGIS123]

With respect, the tide generating force really is stronger on the side of the earth closest to the moon and weakest on the side opposite the moon. The thing that trips people up is that the tide generating force can be broken into horizontal and vertical components. The vertical components are weak and are opposed and overwhelmed by the strength of the earth's own gravity. The horizontal components are also weak, but the horizontal components are not opposed by any other force. Tides go up and down, not because the moon's gravity pulls water upward, but because the horizontal components make water slosh back and forth. Imagine trying to carry a frying pan full of water. As the water sloshes back and forth, the level of the water goes up and down when measured against the sides of the pan. Now imagine a pan the size of an ocean basin, with ocean water sloshing against the shore. Tides are like that.

[@mitchbollo1100]

I didn’t think the earth was wobbling inside a gigantic water bubble! haha

[@cm9748]

as a sailor . I know of several places get get 4 tides a day ... ... ...

[@cm9748]

@@enpassant1119 as a word class symphonist, I know that if the moon had 82x as much mass it would suck up all our water and at the moment (considering it wouldn't be stuck in orbit around the earth/moon, or left behind the moon because of the moons speed of orbit around the sun/milky way. Or just stuck between the earth and the moon) AT the Waters moment of Impact, There Would Be No Sound !

[@cm9748]

@@enpassant1119 as a famous aviator, I know that if the moon weighed 81x the mass of the earth, I wouldn't be needed . All you would have to do is jump when the moon is directly overhead and the moon would gently lift you into and were you could just flap your arms to get around.

[@cm9748]

@@enpassant1119 as the most well know mouse 🐁. If the moon was really made of cheese, I'd want to go there .

[@cm9748]

@@enpassant1119 As THE piece of petrified wood that the US ambassador to the Netherlands was given . I'd go back to the moon because that is obviously my home. Nothing was faked ever . SO , I really miss all my petrified wood friends those astronaust tool me away from

[@cm9748]

@@enpassant1119 But, I am a suface supplied closed bell gas Diver and I know , with no shadow of a doubt , that those stupid new and aold space suits would pool co2 in the helmet and those astronauts would be dead in 2 mins . None the less blind because of condensation on their single pane plastic "bubble"...

[@matthysuys9084]

Unfortunately this is still wrong. This explanation is similar to 90% of the videos or information on the internet all educating the incorrect physics. For one, the earth does not move towards the moon which in this case does not explain the tide on the other side of the earth’s moon face. If this was the case earth and the moon would’ve been merged by now. The moon is actually moving away from the earth in real time. The bulge of water on the “other side” is caused by the centrifugal force that exists between the earth moon (and sun) system. This system revolves around a center point not in the middle of the earth. Very simplified, one can subtract the gravitational force from the centripetal force and that will give you a value explaining the smaller tide.

[@AtomicSchool]

I can see how the video seems to make an assumption that the earth and moon are on a collision course. That's because it's the simplest case of explaining tidal forces, and why 90% of physicists explain it this way. But of course, the earth and moon are orbiting around each other, explaining why they don't collide. However, the forces involved in straight-line motion and circular motion are the same. I will make a video to explain this. In either scenario, the moon and earth are in "free fall" towards each other. By subtracting the moon-earth gravitational force from the moon-ocean forces, you invent a non-inertial reference system that gives false signals regarding what is and what is not a real force. Centrifugal force is not real, and it called a fictitious force. en.wikipedia.org/wiki/Centrifugal_force These physics professors also explain why centrifugal force is irrelevant: www.lockhaven.edu/~dsimanek/scenario/tides.htm www.lockhaven.edu/~dsimanek/scenario/centrifugal.htm butikov.faculty.ifmo.ru/TidesOD.pdf

[@AtomicSchool]

@@lorenat4129 Kevin, I agree with you that Matthys should make a video explaining tides. I look forward to it.

[@jayjeetnandi2024]

Matthys uys you are correct because I read it in the book written by Richard Feynman ( six easy pieces)

[@modernaashole5083]

Eventually the moon's gravity will suck out all of the earth's ocean into space , after an asteriod hits the moon and brings it closer to earth. Big brain time

[@JLMetaMix]

The chat lol. We can all agree the moon is part of the creation of High tides and Low tides.

[@oneeleven7897]

‘This is not only not right, it’s not even wrong’ Wolfgang Pauli’s quote fits perfectly here.

[@hunati31]

I still haven't learned when is the high tide related to moon rise for example.... Since I've been living in Florida East coast for seven years I can tell you this and most explanations seem completely wrong here. We have high tides here at the end of February 2023 1h 30' to 2h 10' after the moon rises and low tides after the Moon is directly above us or on the opposite side. I am a frequent beach goer looking at tides and the Moon. In Honolulu the Low Tide is 2h 48' to 3h 47' after the Moon rise... Vastly different from here... Interesting...

[@adarshkumar7238]

Misleading

[@sailorman8668]

In what way is the explanation in this video 'misleading'? If you wanted your opinion to have some credibility, then you would have explained what it is that is 'misleading' in the video. Currently, your 'misleading' comment is completely useless to the discussion.

[@ericformgren8676]

The Atomic School does not understand the mechanism why the opposite tidal wave occur. That's quite sad.. The real reason is simple and easy to explain.. The Atomic School should do some research before presenting a silly video like this.

[@AtomicSchool]

Hi Eric. I'm just now making a video on the far-side ocean bulge and why it is also caused by the moon's gravity rather than centrifugal forces. If that's what you are referring to? Maybe you could post one too, as you seem clear as to the mechanism?

[@ericformgren8676]

@@AtomicSchool Hi! First of all, thanks for the answer! Your far-side ocean bulge has kept my interest for years, especially when it looked impossible to read, or get an understanding of the phenomenon. It took long time for me to understand the mix of gravety, centrifugal force, inertia, friction, corioli effect, hydraulics and other parameters to make the opposite bulge "work"... Finally I made an animated video -questionable technical quality- to my grandkids in January 2020.. Good you are making a video about the secondary bulge.. I really looking forward for that one.. You are writing "caused by the moon's gravity rather than centrifugal forces" when I have the completely opposite understanding, therefore I will wait for your video, before I send you my own :-) Have Fun!

[@keziagreste]

.... I lost it when u said moon's gravity pulls the earth

[@MrNeilo911]

It does

[@sarangpahwa619]

It's just Newton's third law of motion....

[@truthwarrior7826]

@@MrNeilo911 Lol!!

[@truthwarrior7826]

@@sarangpahwa619 suure...

[@tempus.narrabo]

Wakey wakey, when the penny drops, you're whole reality will disintegrate or stay in the bubble, enjoy 😉

[@jacobnutter1639]

The moon pulls the earth away from the far side ocean?? Who tf made this

[@AtomicSchool]

The earth is centripetally accelerating towards the moon because there is a net force acting on the earth towards the moon. This means it's constantly displacing towards the moon from a vectorial point of view. However, the earth-moon distance is constant from a scalar point of view as the path is circular. But the vectorial point of view is the relevant one. The centipetal force toward the moon in a circular path would be the same as the force in a straight-line path.

[@Bonzi94]

@@AtomicSchool no

[@Starkman2367]

@@AtomicSchool no

[@physics.school]

no it does not

[@blueskythinking9987]

@@AtomicSchool no

[@blueskythinking9987]

Very poor demonstration. The simple answer is "No". On all counts. The earth never moves on a straight line. It never suddenly stops. It never "moves directly" toward the moon. "Leaving one tidal bulge behind". This is totally misleading and unhelpful for understanding what's going on with the tides. Please remove your cartoon, it’s misinformation and then issue an apology to the science community and your students. Thank you.

[@oneeleven7897]

Well said. If only he would do as you ask, but the creator of this rubbish is too arrogant to respond to his many critics and seems unable to admit he’s in error.

[@strangecat7147]

youre correct about this but the demonstration shown isint the earth moving its the gravity pulling the water to the moon

[@blueskythinking9987]

@@strangecat7147 Sorry, but I really don’t want to discuss this here, as this demonstration is an absolute debacle, and the complete defamation of actual physical law, laid down by the likes of Newton and Laplace. In the demonstration the teacher clearly says, and I quote, _“At the centre of the Earth point B, also moves towards the Moon.”_ So, if you’d like to continue this discussion please search, and I quote again, _“What atomic school gets wrong about tides!”_ Thank you.

[@SOS_SOS_PLEASE]

THANK YOU SO MUCH, I THOUGHT THIS WAS TRUE AND I HAVE A SCIENCE EXAM TOMORROW, THANK YOU!

[@AtomicSchool]

True. The earth accelerates towards the moon but doesn't get closer to it, because it's centripetal motion. However the forces are as if it is accelerating linearly. See the beta version why centrifugal force is a misunderstanding. kzbin.info/www/bejne/ap_PgYyZoNCpg9k

[@Doruc7158]

Similar thing happens with Sun. Sun attraction is nullified by Earth's revolution around the Sun.

[@Doruc7158]

Think about why the cosmonauts experience zero weight in an orbit around the Earth. Force and counter force nullifying each other. Common sense. No degree required.

[@AtomicSchool]

The cosmonaut is not experiencing zero weight and she has almost the same weight as on the surface of the earth. This is what causes her to centripetally accelerate towards the earth and orbit around it. Without an unbalanced force she would zap off into outer space.

[@kylemcmullen6381]

@@AtomicSchool the cosmonaut would maintain the same mass but they would be weightless since they are in free fall around the earth though the gravational force of the earth is only about 10% weakear aboard the iss

[@AtomicSchool]

@@kylemcmullen6381 The gravitational force of earth acting on the cosmonaut (and the low orbit iss) is the same as their weight. The cosmonaut's weight is about 89% that which she would have on earth, slightly reduced due to being further from the earth.

[@thephaneron3676]

0:11 *Dumbledore:* (After consuming the Potion of Despair) _"Harry... Water."_ *Potter:* (Points Wand) _"Aguamenti!"_ 😂

[@mikewho9964]

Highly unlikely that the earth is deviated by the lesser mass of the moon - anyway water is fluid thats why it is drawn more easily to the moon side - the explanation i guess is far more complex i think - and much more complex than some of the other comments too

[@CameronShirts]

Newton's laws of physics hold that all actions come with an equal and opposite reaction. You might be smaller than the earth, but your body's mass pulls on the earth just as the Earth's mass pulls on you. The effect of the wobbling earth could be compared to someone walking a dog on a leash. If the dog frantically runs in circles around its owner, it will be "orbiting" them, but the person will also be pulled slightly in the direction the dog is currently pulling from. The person pulls equally as hard on the leash as the dog does. Otherwise, the dog would either escape or be yanked inward, depending on who is pulling harder. It is the same with the earth and the moon. This behavior is used to detect planets in other solar systems. Planets are too small and dim to see from astronomical distances, but if we can see a star wobbling back and forth, we can assume an orbiting planet is tugging on it as it makes its way around its orbit.

[@RoninSan7]

Moon's pull on Earth in not negligible... The Earth actually "orbits" the moon in a point around 4671 km from the Earth's center. en.wikipedia.org/wiki/Barycenter

[@eggbeaters]

@@RoninSan7 the earth doesn't orbit anything .

[@RoninSan7]

​@@eggbeaters Geocentrism in 2023?

[@jameswilliams1740]

@@eggbeatersthat’s right the sun, all the planets and all the stars orbit the earth. 😂😂

[@nickb5254]

That is the worst example of 2 high tides, has to be another theory.... The earth has a gravity pull of its own as you explain, so why would ocean C even be raised or elevated at all... Earth's Gravity + the weak pull of the moons gravity would leave Ocean C at low tide 24/7, There is no example in this video that shows any reason Ocean C should be raised in anyway... If the earth pulled slightly to the right placing it in the center of your Ocean diagram, than so should the Ocean C example be pulled with it... that being earths gravity and the moon pulling it to the right.... The only way you will get that second high tide Hump on the other side is if another great massive object was orbiting the other side of the earth, or somehow the earth is losing a gravitational pull because it's fighting the moon, like when magnets lose force when you add more objects... thats my theory

[@arpansingh3959]

I think your second hypothesis (the one about the earth losing gravitational pull due to the moon) is what the video is alluding to. This video explains high tide at point C is due to the reduced gravitational force of the moon at point C compared to A/B. While you're right in that the gravity of the earth pulls the ocean at C with it, the influence of the moon's distributed gravitational force is enough to change the sea level at C by about one meter. Here's a video by astronomers at U of Nottingham, whom give a similar explanation: kzbin.info/www/bejne/j4CWlnePr6ujbrs

[@aaishahthaani7929]

How come the moon pulls the earth to itself Does the moon also has force of gravity

[@tims5268]

All mass has gravity, even you.

[@karhukivi]

The main problem with this explanation is that the Moon does not orbit the centre of the Earth; rather they both orbit the common centre of mass which is between the centre of the earth and the surface on the side facing the Moon. This point is called the "barycentre" of the two bodies which orbit about it, the Moon making the largest orbit while the Earth just wobbles about this point. Once you realise that, then the second tide on the opposite side becomes obvious.

[@tribusmontibus6436]

Hello George. That's one explanation. But that means that you are regarding the Earth-Moon system in a rotating non-inertial reference frame. That's a correct approach. But that doesn't necessarily make this approach incorrect. When regarding the Earth-Moon system in an inertial reference frame, centrifugal force disappears and you are left with the 'naked' accelerations presented in this video. Such an approach yields the exact same results. What makes this video confusing is that the earth is depicted 'center screen' and stays there for the duration. This means that the screen's reference frame is essentially non inertial, as Earth really does move, with respect to the distant stars. Yet the argumentation is based on an inertial approach. Besides, it really shouldn't show an earth moving towards the moon. Acceleration towards the moon only makes sense as a centripetal acceleration, where it doesn't close in on the moon. But I guess that was 'animator's choice'.

[@tribusmontibus6436]

By the way, did you ever notice, during your tidal calculations that the highest high tides are typically a day or two after the New Moon or the Full Moon? (assuming UK or US sailing) There is a very good reason for that and, no, it isn't because it 'needs to get going'. skipper of a Contessa 32

[@tribusmontibus6436]

You see, as a sailor you are more likely to notice certain things about the tides which most people miss: - The high tides typically do not coincide with the passage of the moon overhead (not even in mid-ocean). If they do coincide; just look fifty miles either way, on the same meridian, to find a place where they do not (even though the moon was ‘overhead’ at the same time). - The moon passes overhead from east to west, yet the rising tides come in from all directions (even in mid-ocean). - With the moon on the horizon, water should be ‘pulling away from you’ if the ‘pulling on the oceans’ is true. But no correlation can be seen when you check the tidal rises and falls. - For most locations in the northern hemisphere the highest high tides and lowest low tides occur one or two days after the New Moon and the Full Moon. Most sailors, in my experience, accept vague speculative explanations for these apparent discrepancies. Some are more inquisitive and endeavor to find out what’s really going on.

[@Cg23sailor]

@@tribusmontibus6436 that's because earth rotates once in 24 hours whereas the moon goes around the earth once every 28 days. The earth's faster rotation drags the bulge forward off the Earth moon line. That's why high tide doesn't happen right when the moon is overhead. It's also why the moon is slowly moving away from earth and why the earth is slowing down. A few billion years ago Earth's "day" was only 6 hours long, not 24.

[@mizuqatsi]

Yes George, the moon and the earth rotate together around the barycentre, it is often forgotten.

[@Jetexjet]

So water isn’t touching the ground all the way in the bottom of the ocean? It’s like the earth produces more water. I sound stupid but does anyone know what I mean? If you pull something, the end of it is away from the other side which is the ground

[@AtomicSchool]

The shape of the ocean changes to become a spheroid, but the volume remains the same.

[@raulgalets]

you know, this is how i learnt at college too

[@raulgalets]

@@enpassant1119 thank you

[@lazyupload]

One thing I don't get is how the Earth can get stretched like water and being pulled towards the moon. What on earth?!

[@AtomicSchool]

Yes, the diagram is a bit confusing. I am making another video to explain this point. In short, the earth and its ocean are revolving around each other (missing in this video) so although the earth indeed is being pulled by the moon it isn't getting closer to it.

[@chishtiswaleha9828]

My question is mass of the earth is more as compare to moon so resultant force will be by earth,, effect should be bcz of earth's gravitational pull then how come moon is attracting the water on earth😳

[@AtomicSchool]

@@chishtiswaleha9828 Hi Chrishti. Yes, the earth is more massive and also closer, so its gravity is about 300 000 times stronger than the moon's. So the moon cannot make a piece of the ocean to "levitate" because this would require a force greater than earth's gravity. However, the moon's gravity has the effect of warping the ocean's shape over a distance of the earth's radius, causing a higher level on the near side. I'm making a video on this now.

[@shanpadiyoor]

It's not correct

[@ilyassboutjim572]

Gigachad

[@marksimon8393]

🤯finally, it’s all coming together now🤯

[@AtomicSchool]

Good to hear Mark. I will make some more videos on tides to clarify some of the tricky concepts.

[@austridge31]

@@AtomicSchool Great video. One thing that still doesnt make sense to me though. And I'm probably gonna sound stupid here.. The second high tide (at the 'back'), is caused by the earth being pulled slightly towards the moon right? But surely that would only make sense, if there was a constant movement of the earth towards the moon? Which would mean they would eventually collide.

[@AtomicSchool]

@@austridge31 A great question and this is one of those tricky concepts that the video doesn't cover. If the gravity between earth and moon accelerated towards each other in a straight line they would eventually crash. However, the gravitational forces instead bend the motion of earth and moon into mutual orbits, keeping their distance more or less constant. The centripetal force required to do this is exactly the gravitational forces of attraction. Either way (straight line or centripetal) the differential forces of gravity create bulges on either side of the earth. A video on this is coming :)

[@janj9650]

@@AtomicSchool this sounds so romantic like the moon and earth are joined in a perfect alliance together to never destroy or cause combat with the other, in loyal harmony forever and ever in mutual “orbitry” by the power vested by the universe or greatest almighty being 🌎🌘🌏🌗🌍🌒

[@sleepysheep1181]

The Earth may only have two days left if the predictions in this video play out! Please remove this misinformation from YT?

[@twixxtro]

2 month

[@sleepysheep1181]

@@twixxtro That looks like a cat, not a holy cow? I don’t think it would take them two months to collide. It only takes one month for them to both orbit around their common centre of gravity and inertia needs time to act on the far-side ocean. Please do your math again?

[@BeReal918]

What is this comment about? What predictions did the video make? I saw nothing related to _the earth having 2 days left_ in the video. Put down the pipe! You're obviously tripping (dreaming?) sleepy sheep.

[@sleepysheep1181]

@@BeReal918 You’re obviously the one who’s asleep. It would take approximately _two days_ for the earth and moon to collide under their mutual gravitational attraction, without orbital motion around the barycentre of the earth-moon system. Please try and pay attention in class and wake up!

[@BeReal918]

@@sleepysheep1181 I rewatched the video and he fails to mention the Solar vs lunar tides, near vs spring tides, and tried to give a simple explanation but wasn't 100% accurate. Is that what you're on about?

[@adamsmall6419]

DO NOT PLAY FOR Middle SCHOOL kids, this video is completely wrong and NOT COOL for kids

[@Charleslihan0211]

How?

[@SalsaKingoftheApes]

So why not explain it then

[@percival1137]

If you really want the truth, it will find you. If not, it could never be explained to you.

[@physics.school]

I mistakenly showed this to my students in my Lab and now I’ve been bombarded with questions as to how the tides actually work! I wish people would refrain from posting such simplistic nonsense. It’s not helping and now I’ll have to try and undo the damage your miseducation has done, all under the guise of calling yourself a KZbin school. Please Stop! Regards, Mr Jones.

[@Verradonairun]

Teachers are solely responsible for the material they bring into the classroom. It's immature to blame someone else for that. My guess is you didn't "accidentally" show them a video, you simply took the easy/lazy way out and had KZbin do your teaching for you.

[@Lele_0319]

@@Verradonairun he's js expressing his opinion bro it doesn't mean much and he's not the only one complaining that this video is showing false facts.

[@AtomicSchool]

See the beta version of why centrifugal force is not the explanation of tides kzbin.info/www/bejne/ap_PgYyZoNCpg9k

[@CanaldoFakeflip]

If you want to know how tides work, listen to my explanation in Channel No Fanfare.

[@stewiesaidthat]

​@@AtomicSchoolyou need to research Galileo and his theory on tides. Mass is not an actionable force. F=ma. Force comes from Acceleration of the mass, not the mass itself. Otherwise it would just be m=a or mass equals Acceleration. Where do you get more acceleration by adding more mass. Adding salt (mass) to water decreases it's temperature (acceleration) and increases it's absolute acceleration (c) in the equation E=mc. F=G(m1m2)/R2? That shows the universe being gravitationally bound to the earth. That's flat earth/stationary plane physics. Centrifugal force? The earth is rotating on its axis accelerating its mass outward and forward. Centripetal force? The outer radius portion of the mass has more acceleration than the inner radius. The mass is essentially falling in on itself creating curved space. Gravitational attraction is flat earth science. Relativity is flat earth science. Acceleration is the actionable force. Not mass.

[@andybrownen2226]

The information you give about the cause of the ocean's tides is incorrect. The two ocean bulges are of equal size (equal resultant forces on both sides of the earth) and your explanation of their cause, while convincing, is wrong. I suggest you research 'Gravity and Tides'. The cause of the second bulge can be explained by the Earth-Moon system and the resultant shift in the centrifugal force acting on the oceans. Best wishes, Andy Brownen

[@AtomicSchool]

Thanks Andy. I am aware of the centrifugal force explanation for the far-side bulge, but don't agree with it.

[@Cg23sailor]

@@AtomicSchool yes it's wrong, but so is your explanation

[@physics.school]

Well, thanks for reposting your valiant attempt to try and defend your chemistry teacher inspector cluedo. I’m positive he’ll reward you with a smilie later? 🙃 However, I still seriously doubt you inspected the video in question closely enough. At precisely 1:48 mr ian says, the gravity moves point B towards the moon and somehow magically stops!! Unfortunately, that means the Earth has now left its joint orbit with the Moon around the barycentre, and reduced the separation distance between the two, and sent it into orbital decay, with the only plausible outcome leading to a collision. That is not how Newton, Euler or Laplace ever described the tides or the motion of Planets! Sorry, my little buddy. 👌 Secondly: I’ll leave your incorrect vectorial analysis part, regarding the apparent levitation of ocean water off the surface, to someone whose more qualified in that area, i.e. an Oceanographer, as I don’t think you quite understand what you’re actually saying? Despite to say, that the ocean is never pulled upwards by such an infinitesimal small force, Sorry, again my little buddy. 👌 So, no thanks. Your teacher is wrong and has not been educating your children correctly over all these years. He should have stuck to atom theory, which is the chosen field of science he’s more familiar with. Regards, Mr Jones to Inspector Clouseau. Who sadly disappeared after commenting?

[@msshobanaapnagalingam1334]

Hi, the bulge on the far side is causes by inertia, right?

[@AtomicSchool]

Surprisingly, no, Shobana. The far side bulge would happen even if the earth and moon were not revolving around their common centre of mass. I am making a video on this. BTW, there is a division of opinion on this amongst scientists.

[@AtomicSchool]

@Prime Thanatos It would be great if you could make a video on this. Your animation is great.

[@randomdude9758]

I would rather believe the earth was flat than this.

[@markwalton7298]

You and the rest of the woke idiots. Flat out Dumbo's.

[@AtomicSchool]

Go ahead! See the beta version of why centrifugal force is a misconception. kzbin.info/www/bejne/ap_PgYyZoNCpg9k

[@michadyrek3607]

Aren't tides caused by the centrifugal forces as well?

[@azap12]

No

[@BallMuncher555]

Centrifugal force is not a real force.

[@AtomicSchool]

I am making a video on this topic.

[@maxwang2537]

@@BallMuncher555 You are right, it's not a real force. But when there need to be a centripetal force in the movement of an object along a curve (to maintain the centripetal acceleration), the problem can be treated, in a static model, as if there was a centrifugal force. And this treatment is not suggesting it is a real force.

[@maxwang2537]

After some pondering on this problem, now I strongly believe the other bulge (at the far side of the Earth) is indeed largely due to the 'centrifugal force', though of course, the problem is much much more complex than this simplification. It's a system of the Sun, Earth and Moon interacting all to one another and the tides contain many harmonics due to the inclination angle of the Earth's rotational axis relative to its ecliptic plane, as well as the angle between this plane, this axis and the Moon's orbital plane (and its axis of rotation).

[@rayagoldendropofsun397]

Thanks for the spelling corrections,

[@Olaf_Schwandt]

That's a wrong explanation

[@ZoeZparkles]

1:02 Moon: I'm attracted to the Earth Me: "Do you think Earth likes you back?" 1:06 Earth: And I'm attracted to the Moon Me: "SHIP!" But really, this threw out my confusion into the trash

[@jameseldridge3445]

This doesn't seem real

[@matzr3000]

Looks like magic, isn't it

[@shafs9027]

You failed to mention the gravitational force of the sun aswell

[@AtomicSchool]

Hi Darth, this is covered in a separate video at kzbin.info/www/bejne/fpWskIOvgquBqNE​.

[@IgnatiusVaz]

The hero of the plot "Gravity". The impossible becomes possible. This is worst than a kid's story telling. A fantasy tale

[@slim5816]

you think gravity doesnt exist?

[@IgnatiusVaz]

Absolutely! over to you. Defend it.

[@slim5816]

@@IgnatiusVaz wait what am I supposed to defend? You think gravity isnt a thing? Then what is keeping you on this planet?

[@IgnatiusVaz]

@@slim5816 Tell me in your words, and your understanding, what is Gravity. Dumb it down for me. Don't quote me what science tells you. DEFINE ! gravity for me.

[@slim5816]

@@IgnatiusVaz well the official definition is 'the force that attracts a body towards the centre of the earth, or towards any other physical body having mass.' according to general relativity its rather an effect of warped spacetime We prove both pretty well. i.e Space travel wouldnt work at all if we didnt understand gravity. Time dilation. Time is not universal and is influenced by the intensity of gravity. The concept of spacetime makes that observation pretty intuitive. GPS satellites wouldnt work without frequent time adjustments as time in earths orbit is different.

[@DellHell1]

I'm afraid this explanation is not correct, if it was then tides would be detected in large lakes like the Great Lakes. For the correct explanation search on PBS Space Time Tides.

[@AtomicSchool]

The Great Lakes have tides: tidesandcurrents.noaa.gov/water_level_info.html. They are smaller (about 5cm) because they are shallower and not as wide as the oceans. This means that the gravity differential is not as large as in oceans. Same with bathtubs. And you are also tidally stretched due to your head's stronger attraction to the moon compared to your feet's attraction if you're standing up. This effect is called "spaghettification" but is too small to notice with weak lunar gravities. But near a black hole, you would be stretched to death! Regarding SpaceTime his main concern is that it's only the horizontal component of the lunar attraction that does all the tidal moving. That is, the so-called "tractive forces". I disagree with that, but that's another story. I also recall that he incorrectly claims that there are no tides in the Great Lakes? Can't remember.

[@AtomicSchool]

@@lorenat4129 Kevin, are you saying there are NO tides in the Great Lakes? Or that "gravity differential" and "tractive forces" are gobbledygook terms? If so, it would be hard to have a conversation with you.

[@AtomicSchool]

​@@lorenat4129 I delete threads if they are abusive or off-topic. Calling an argument "gobbledygook" is off-topic (it's got nothing to do with the physics of tides) and also a tad insulting. I'm very close to deleting your comments again.

[@AtomicSchool]

​@@philipmartin2919 Hi Philip, the so-called centrifugal force is just re-badged version of centripetal force, seen from a non-intertial frame of reference. It doesn't explain the far side bulge. I am making a series of videos on this now.

[@AtomicSchool]

@@philipmartin2919 Good point. Maybe you could also contact the following explainer videos which also uses this explanation: Crash course kzbin.info/www/bejne/gZ26oXmCm7F5eKs Adam Hart-Davis kzbin.info/www/bejne/eYW0Z5affaqBnas Brainstuff: kzbin.info/www/bejne/a6DLdXpth9aEb6s Minute Physics: kzbin.info/www/bejne/o4etdpxjhdCel7s Julien Huguet: kzbin.info/www/bejne/e4q0ZmeYiNenfas The problem is that my video treatment (and the others) have simplified the motion for school kids, by ignoring the orbital motion aspect, and just focusing on the forces. This needs clarifying which I am doing in the next video. You seem to have a strong understanding of the physics of tides, so it would be good to see you publish some explainer videos to communicate your view. Or have you already done that, Philip?

[@oneeleven7897]

Unfortunately most of what you say here is simply wrong. The tides are not raised by the axial/vertical component of the Moon’s gravity which you show, but instead by the tangential components which arise off the Earth-Moon axis and which act as tractive forces in the horizontal plane of the oceans. This certainly isn’t obvious, and could be said to be counter-intuitive, when the Moon is after all, up in the sky so it’s effect could be assumed to be in the vertical. This is a widely held misconception that is all too frequently perpetuated by otherwise reputable scientists, websites and text books. The axial model which you describe here does not and never will explain the second or antipodal tidal bulge, despite all the grotesque distortions of Physics, including the one quoted in this video, that are put about to try to force it to do so. In tidal systems where the axial component does dominate, a Roche Lobe is formed and the body distorts into a teardrop or raindrop shape with a single accentuated peak. This is how accreting binary stars are modelled in astrophysics. Non-scientists often instinctively challenge the existence of the second tidal bulge when faced with the erroneous explanation quoted here. The Moon’s vertical tidal component at the Earth’s surface is about one ten millionth of a ‘g’ and the oceans are simply too shallow for this to accumulate any effect. If the vertical force was locally sufficient to raise a tide then all bodies would show the same tidal range regardless of their mass, as they passed the sub lunar point. The true source of the tides was discovered by Euler and mathematically modelled by Laplace in his eponymous Tidal Equations. The tractive forces are similarly tiny but are, crucially, cumulative across the vast surface area of the oceans, and sea water is of course fluid and incompressible. This combination of tiny but cumulative tractive forces and huge areas is sufficient to raise tides of a few metres in magnitude. This model successfully predicts the second tidal bulge, and also the shape of the bulges which are domed or flat-topped because along the Earth-Moon axis the tractive forces tend to zero. Laplace’s Tidal Equations are a set of differential equations that treat the vertical component as negligible, and they are still used today as the basis for detailed modelling and tidal predictions.

[@oneeleven7897]

@Prime Thanatos The Euler-Laplace model does explain the antipodal bulge because the tractive forces are present and pro-axial on both hemispheres, as you can see in your own animation. Euler and Laplace resolved the question of Earth’s tides by about 1775, so for nearly 250 years we have had a detailed explanation for the observed phenomena which is also capable of making verifiable predictions. The scientific process has been satisfied and there is no deficiency in our understanding of the tides. As I said in my original comment, the truth here is certainly not obvious and is probably counter-intuitive, which may explain why so much misinformation is still put out and why so many people still find the tides hard to comprehend. Videos and would-be explanations that persist with long outdated and disproven ideas, and gross distortions of Physics, are no better than the flat-Earthers.

[@rafaelmolinanavas8862]

​@Tides I´m glad you say: " ... in this part of the planet inertia prevails" but watch out, your expressions are rather confusing. E.g.: 1)"The center of the earth and anything that is in the line perpendicular to it": perpendicular to a point ?? 2) "... away from the line passing through of the center of the earth": which line ?? (through C.E. infinite lines pass !!) 3) "speed of rotation around the moon ... orbital speed ..." Our planet neither "orbits" nor "rotates" around the moon. It actually revolves around the barycenter of earth-moon "couple" (trajectories of all points, where no movable liquids, are exactly equal to earth's center one around the barycenter, but around different points ..." Should you have this not quite clear, don't hesitate to tell me and I´ll send some link and/or an explanation I used years ago that comes in "handy" (it uses one of our hands "revolving" over the surface of a table).

[@rafaelmolinanavas8862]

@Tides Thank you, but you insist on considering earth movement due to mutual gravitational interaction with the moon as an "orbital" motion ... Both earth around the sun, and moon around earth, can be considered orbital motions (they are "free" movements of bodies in fixed gravitational fields), but NOT the movement of our planet in its "dance" with our moon, as I said yesterday: " Our planet neither "orbits" nor "rotates" around the moon (nor around the barycenter). It actually REVOLVES around the barycenter of earth-moon "couple" (trajectories of all points, where no movable liquids, are exactly equal to earth's center one around the barycenter, but around different points ..." Due to that, the "field" of inertial effects is different ... Unless you understand and accept that, your conclusions will be flawed (I´m afraid, I should say ...).

[@rafaelmolinanavas8862]

@Tides Why "on earth" should I ask Tribus? I've long discussed tides with them, and consider they have some good ideas (e.g.: pressure gradient due to "chain" of pulls ...), but also wrong ideas ... Though, as far as I can remember, the issue I referred to yesterday (and the day before) was not among what we discussed ... But, as far as I can understand, they are erroneous about important details regarding tides, and they offer me no guarantee to take them as a "judge", at least regarding tides ... By the way, this very morning I uploaded a post in relation to one of most common, basic details I consider erroneous, in "Tribus" proposal among many others. Where you interested, please have a look: Why do Tides form on the side OPPOSITE the moon? (KZbin)

[@rafaelmolinanavas8862]

​@Inner Worlds On the one hand, I wouldn't "dismiss this as pure nonsense", but really far-fetched", if we had somehow proved real that water charging, and had quantified mathematically its magnitude and distribution ... And on the other, the reason of the existence of both bulges does exist, ALWAYS related to gravity. Some consider only "differential gravity" (to me just a mathematical "trick", but something not physically real), and I personally (and others) consider it is the result of both REAL gravity from the moon (and sun), and inertial effects due to the fact that all earth particles (water included) are being FORCED (by the whole of the earth) to accelerate NOT FREELY, what is exerted by forces exerted between CONTIGUOUS particles (changes in pressure where water) ... Those interactions are ALWAYS in opposite directions (Newton´s laws). Forces which are opposite to moon´s pull (and to the barycenter) could be called "centrifugal" (particles tend to flee from a center), or AT LEAST "outward" forces, as I said here 2 years ago: "Neil F. Comins considers so: "What if the moon did not existed - Neil F. Comins" (youtube video) (see especially 10 or 15 min. starting 06:00)". IN ANY CASE, gravity and Newton´s Motion laws are the root cause of tides !!

[@MichaelRei99]

That does not explain why different areas on the same coast have different high and low time tides.

[@brenthawkins5097]

Yeah and don't expect an answer on that either

[@davidjeffreys9063]

That was well explained in a simple way, yet i still didn't quite understand it. If the moon is strong enough to pull all that heavy weight of the sea, why doesn't it pull the water out me bath when I'm washing me goolies?

[@AtomicSchool]

The difference in distance between the top and bottom of your bath water is small, so the moon's gravitational difference is tiny compared to the ocean, which has an effective distance from the moon difference of an earth's diameter.

[@klolAF]

How can one learn that

[@klolAF]

@@hafsagreer must kill Greeks mythology and

[@shalmali-379]

the water doesnt move along with the earth when it rotates? why is the water stays in the same position while the eart is rotating?

[@AtomicSchool]

Good point. The ocean does get dragged along with the earth as it rotates, but the diagram does not show this. It also doesn't show it NOT being dragged as the ocean colour is plain. This means there is a conflict between the ocean being taken for a ride with the spinning earth, and some being held back by the moon's gravitational attraction causing the tides.

[@billb945]

The Earth is 70% uncarbonated water Therefore the Earth is flat.

[@froilanbas759]

Neil in one of his videos said that the sun is also involve in the earth's tide. Not just the moon.

[@max5250]

Correct, Sun also affects tides by creating spring (during New Moon, and Full Moon phase), and neap (during Frist Quarter and Third Quarter Moon phase) tides.

[@AtomicSchool]

Neil is right (of course). You can see the sun's effect here kzbin.info/www/bejne/fpWskIOvgquBqNE​.

[@max5250]

@hanashite channel It is not "exactly perfect" axis, or design, and it is not engineered (engineering is human activity, and humans haven't designed space/time).

[@danneumann3274]

I have worked around some of the most precision scales in the world. They carry a gram to 4 decimal places. When the sun and moon are directly overhead, there is no affect on the scales. I even talked to a scale calibration technician who works on far more accurate scales. I asked him if he ever considers the sun and moons position when calibrating scales? He said that He had never thought of it. i dont believe any of this

[@AtomicSchool]

Interesting, Dan

[@danneumann3274]

@@AtomicSchool well I guess I'm out of my league here then. Thanks for the info

[@AtomicSchool]

@@danneumann3274 I learnt something too :)

[@danneumann3274]

@@AtomicSchool on a thousand ounce silver bars shaved to an accuracy of .005 gram,, that equates to 1 part in six million accuracy. far more on gold. thats about .0000001 /1 on a thousand ounces. The scales were capable of far more accuracy. about 23 square miles of ocean represents .0000001 of the worlds oceans area at 139million square miles. I am just a regular guy, but it escapes me when I comprehend the north to south bulge of water as a percentage of the total ocean area is undetectable by the instruments that I have used. If the bulge of water was only 1/100th of a mile wide north to south it would represent far more of a percentage of the worlds ocean area than the scales are measuring a percentage of a gold or silver bar. I guess in summary, the amount of water "lifted" is massive as I look at tide charts. trillions of tons perhaps and I cannot detect anything. and to top it off , in one part of canada the tide is 50 foot . thanks for the civil dialog

[@TheShoppingChannelUK]

@@AtomicSchool doesnt sound slight if its moving trillions of tons of water, sounds pretty significant, whats going on ?

[@peachyvxdss7183]

The “I’m attracted to the moon” got me dying😭

[@justaguy4real]

I don't see how the Earth has such stronger gravity yet the moon has gravitational influence. Interesting

[@AtomicSchool]

The moon's gravity is tiny compared to earth's on its surface. But its different gravity strength across the oceans are enough to stretch them into a slightly obloid shape. This causes the tides.

[@aksha_869]

I understand .thank you😊

[@AtomicSchool]

@@aksha_869 Thanks Aksha!

[@guitarlegendizzy]

@@AtomicSchool do object also lose weight when the tide is high?

[@AtomicSchool]

@@guitarlegendizzy Hi Israel. On object's apparent weight is slightly less when the moon is overhead, but this is so small that it is unnoticeable. The high tide doesn't always coincide with an overhead moon because other distorting factors come into play.

[@thadatkins9792]

Yeah, that's wrong.

[@willemjoubert3662]

Wrong reason for the far side high tide. Point B never moves closer to the moon.

[@AtomicSchool]

Hello Willem. If the moon and earth were not orbiting one another, then A, B and C would differentially accelerate towards the moon in straight-line motion, and their distance separation would continually increase, causing tides. Eventually, the earth and moon would crash into each other. However, in orbital motion the different accelerations have a modified effect, in that they don't get closer to the moon (as you said). But the stronger lunar gravity at A will "pull" it into a tighter orbital radius than B, and similar for B compared to C. Because the ocean is fluid, it responds to this differential acceleration by stretching out, with A will have a tighter orbital radius than B, and C will have a larger orbital radius than B. The circular equation of motion for this is F=mv(2)/r. That is, as the stronger lunar gravity at A causes a smaller orbital radius, whereas the weaker lunar gravity at C extends its orbital radius. The combination is a stretching deformation in the fluid ocean.

[@willemjoubert3662]

@@AtomicSchool The centre of gravity for both the moon and earth is far away from the centre of earth. While earth circles around the sun it is also circling around the centre of gravity with the moon. That causes centrifugal forces on the far side that cause a high tide on that side. There is no other reason.

[@AtomicSchool]

@@willemjoubert3662 That's a common theory, but imo it's wrong. You can find the explanation for this misconception with these sites: www.lockhaven.edu/~dsimanek/scenario/tides.htm www.lockhaven.edu/~dsimanek/scenario/centrifugal.htm butikov.faculty.ifmo.ru/TidesOD.pdf arxiv.org/ftp/arxiv/papers/1506/1506.04085.pdf en.wikipedia.org/wiki/Tidal_force hyperphysics.phy-astr.gsu.edu/hbase/tide.html Minute Physics: kzbin.info/www/bejne/o4etdpxjhdCel7s&feature=emb_logo www.wired.com/2013/11/how-do-you-explain-the-tides-in-10-seconds/ (this explains why centrifugal force is sometimes used) arxiv.org/ftp/arxiv/papers/1506/1506.04085.pdf butikov.faculty.ifmo.ru/TidesOD.pdf www.badastronomy.com/bad/misc/tides.html

[@AtomicSchool]

@@willemjoubert3662 I'm planning to make my own video on this centrifugal issue to clarify it. But time is my biggest scarcity.

[@CarlosAM1]

For some reason i still find it hard to understand why the smaller tide bulge doesnt just instantly fall to the earth by being so close to it

[@CarlosAM1]

@@tmb9126 but why though, I know I am wrong here but monkey brain tells me it should just fall straight back down like any object instead of hovering slightly higher up

[@CarlosAM1]

@@lorenat4129 thanks.

[@paulm5443]

The moon doesn't 'pull' the sea water into a bulge and it doesn't move the earth as much as you suggest, its more complicated than that. The oceans are squeezed by small amounts all over their surface by the moons gravity. Because the oceans are very large this squeezing effect results in the tidal bulges on both sides of the earth. Very large lakes and even the Mediterranean sea have virtually no tides as there is not enough water (area and depth) to squeeze into a bulge. It's also why you don't get tidal bulges in your swimming pools. The earth's movement caused by the moon's gravity is so small it couldn't account for the tidal bulge on the opposite side of the planet. For a better explanation kzbin.info/www/bejne/pqimmZ5qiJ5sn5Y

[@rafaelmolinanavas8862]

Many experts agree that, instead of direct vertical pull of the moon on the waters at sublunar area (opposed to the much, much greater pull of our planet or weight of the water): “… the tides are produced by that component of the tide-raising force of the moon which acts to draw the waters of the earth horizontally over its surface toward the sublunar and antipodal points. Since the horizontal component is not opposed in any way to gravity and can, therefore, act to draw particles of water FREELY over the earth's surface, it becomes the effective force in generating tides. At any point on the earth's surface … This second (horizontal) component, known as the tractive ("drawing") component of force, is the actual mechanism for producing the tides” (from tidesandcurrents.noaa.gov/restles3.html) This mechanism could act not “alone” (f.e.: others perhaps in line with what exposed at Tribus Montibus Oceanography you tube site could also act, or even something similar to what at What Physics Teachers Get Wrong About Tides! | Space Time ...KZbin · PBS Space Time · Aug 6, 201515:21, I personally find it quite plausible: When more than 90º from sublunar point, net tide-raising force acts towards antipodal point. When around 90º, it is almost completely horizontal, but close to null (any possible “squeezing” downwards force as proposed at last mentioned site quite negligible). The closer to sublunar point, the more increases net rising-force, but the smaller its horizontal component (by the way, much smaller than what appears at Tribus Montibus Oceanography 4:57-5:27 images: moon is much far away, and all its pulls should appear almost parallel). Therefore, especially from intermediate areas, water particles are FREELY (with neither an opposed force nor pressure) drawn towards sublunar point (at closer to moon hemisphere). Horizontal components initially increase starting from null, but later decrease (though net tide-raising force vector increases, its horizontal component decreases and ends being null …). That scenario obliges water surface layers to follow the easiest way: to progressively slide climbing over contiguous inner water (although having to act against own weight, something quite easy because the “slope” is negligible, in the order of less than few meters/10,000,000 m.). At sublunar area those horizontal components are negligible, but logically water level also rises, due to Archimedes principle … Added water own weight, at more distant areas around sublunar point, increases pressure, what especially is radially transmitted towards sublunar point, and water is "pushed" upwards there, similarly to a hydraulic pump.

[@oneeleven7897]

@@rafaelmolinanavas8862 Absolutely correct! You say ‘many experts agree..’ but why don’t all scientists agree on this when it was conclusively modelled by Euler and Laplace way back in the eighteenth century? How do those who disagree think we are able to predict the tides with such extreme accuracy? Some of the attempts to explain the second/antipodal bulge using the erroneous vertical model are painful to see and are an abuse of Physics, yet many otherwise reputable scientists and talking heads perpetuate such rubbish. It seems to be a struggle to spread the truth…

[@black5f]

Correct ... the moon doesn't attract otherwise you would weigh less when it's over head..

[@7F0X7]

@@oneeleven7897 Please explain how the literal water around a ball floating in space with a moon 1% of its mass was modeled to replicate the double bulge tide. Also, why do these writers on government websites feel the compulsion to bamboozle people with made up words just to obfuscate what they are "explaining" instead of just explaining it in reasonably widely understood words? And why does the regularity of an occurrence automatically suggest one contrived explanation is correct? it's not "predicting with extreme accuracy" when some peasant with a sun-dial could mark the time each day the tides came in and easily make the same "prediction"?

[@black5f]

And everything would start to float ha ha. Tidal forces are very complex things. The moon doesn't pull on us and we don't pull on the moon, we are both following a straight path through space that is bent by mass. Like a race car going around a banked bend. The moon always faces us with the same face because its tidally locked, many moons are like this. You young people ... go out and find out why?

[@davidthiel483]

The center of mass in the Earth-Moon system, also known as the barycenter, is located within Earth itself. It lies about 4,671 km (2,902 miles) from Earth's center, which is approximately 75% of Earth's radius on the moon side. While it seems like the Moon orbits Earth, both the Earth and Moon actually orbit this common center of mass. However, because the Earth is much more massive than the Moon, the barycenter is located closer to Earth's center, causing the Earth to wobble slightly as it orbits the barycenter. This means that the elliptical path we generally attribute to Earth's orbit around the Sun is actually the orbit of the barycenter of the Earth-Moon system. So you would observe the earth wobble about the barycenter the side opposite the moon is traveling faster around the barycenter and thus it is centrifugal force that is causing the bulge opposite the moon in exactly the same magnitude as the moon facing side. the earth wobble about the barycenter the side opposite the moon is traveling faster around the barycenter and thus it is centripetal force that is causing the bulge opposite the moon at about the same magnitude as the moon facing side.

[@suleymanbabak1973]

Inaccurate and poorly explained.

[@rubenlee-rolesmusic1692]

Nerd

[@gabrielM1111]

Dear producer and audience, Right off the bat this video is shamefully wrong. 01:10 the Earth and the moon do not have equal attraction to each other regarding gravity. The larger the mass, the stronger the gravitational pull is. Unless hes trying to say that the law of gravity is equal for all objects in the universe. If that's the case he definitely needs to reword it

[@AtomicSchool]

F = GmM/d^2. So if m is increased then F is increased. Or if is increased then F is increased. F depends on BOTH m and M, and is the same size for both objects. That's why it's called a _mutual_ force of attraction. It is an example of Newton's 3rd law of action with an EQUAL and opposite reaction.

[@gabrielM1111]

@@AtomicSchool Newtons 3rd law of motion has nothing to do with gravity related to objects. Watch this simple demo kzbin.info/www/bejne/j56Zn5-tq7aNpKc your wording is wrong and the 2 arrows of equal size you have between the earth and the moon is wrong. At 02:02 you contradict your claim made earlier that the Earth and the moon gravitational strength are equal. Therefore gravitational strength is not fixed but variable but force is constant.

[@tribusmontibus6436]

Hello Gabriel. Yes, the mutual attraction is ALWAYS equal. Easy to explain with a thought experiment. Imagine equal masses which, at a distance of 1 meter apart mutually attract each other with a force of 1 Newton. I like to keep things simple. First of all you suspend mass 1 and mass 2 at 1 meter apart. I hope you will agree with me that mass 1 will experience a force of 1 N towards mass 2 and mass 2 experiences a force of 1 N towards mass 1 Next you suspend another equal mass (mass 3), also at 1 m distance from mass 1, right next to mass 2. Mass 1 now experiences a 2 N attraction towards masses 2 and 3. Mass 2 experiences a 1 N force towards mass 1, and so does mass 3. If you glue both masses together, the combined mass 2/3 experiences a total attraction of 2 N towards mass 1. See, how that works? Always an equal attraction going in both directions, no matter how the masses are divided. Kind regards, Garret

[@AtomicSchool]

@@gabrielM1111 Wow! Thank you for the link as it is a brilliant yet elegant demonstration of Newton's Universal Gravitation. I will try to replicate that in my lab for my students. It is fiddly, but it would be worth it. I thought I would have to use much larger masses to demonstrate this effect. Again, thanks. Newton's Universal Gravitation is not the same as Newton's 3rd Law, but it is an example of it. The two equal and opposite forces of gravity (earth on moon and moon on earth) are 3rd law reaction pairs. You can check this here: kzbin.info/www/bejne/f6OxiItvg9ySfrs where the first example of the 3rd law is the earth-moon pair. I agree with Garret's reasoning about the opposite but equal force between earth and moon, and his reference to different masses.

[@rayagoldendropofsun397]

@@AtomicSchool "Equal force between the Earth and Moon," yet the Moon is the follower, contrary to equal force. Unusually the stronger Energy in Motion overrides the weaker, this is why object's fall to Earth Surface, and why the Moon follows Earth, which follows the Sun. The Webb Telescope liftoff Rocket shows the stronger Energy Flow overrides the weaker, as it break away from Earth 9.8ms downward pull. @AtomicSchool. Gravity is a myth ! Atomic Molecular activitys sets gas in Motion, and when Bonded together it's Motion is lost, causing a downward falling motion to Earth Surface, the stronger Energy in Motion. Newton's equations works perfectly with Energy !

[@stalinrodrigues6741]

Ok So my question is does sun gravity has any effect on water or is it only moon that effect the water and gives tides

[@tribusmontibus6436]

Yes, the sun’s gravitational attraction on any mass on Earth is more than 175 times stronger than the moon’s (that’s why the earth orbits the sun and not the moon). But tides are all about the differential in gravitational attraction, and the differential of the moon’s weaker gravitational attraction is more than two times stronger than the sun’s. If you’d care to know the specifics, watch OTMS-1

[@stalinrodrigues6741]

@@tribusmontibus6436 thanks a lot 😊 btw what is OTMS-1

[@tribusmontibus6436]

@@stalinrodrigues6741 You’re welcome. OTMS-1 is a video on Differential Gravitational Attraction, in the series Ocean Tides Making Sense.

[@AtomicSchool]

You can see the effect of the sun at kzbin.info/www/bejne/fpWskIOvgquBqNE

[@ThePejsek02468]

Unlike most of the commenters here, you have my support on this. I came to this video from an exercise in a text written by a well-respected physics professor. He mentions no centrifugal force in his explanation of tides. What I would add is that the reason that the tide at C is smaller than at A is because of the hyperbolic relation between gravitational force and distance.

[@AtomicSchool]

Thanks, especially regarding the centrifugal force misconception. Yes, the tidal force is inversely proportional to the CUBE of the distance to the tide generating body, so the near side is affected most. However, I do oversimplify the motion of the earth a bit and am making another video to set the record straight. In particular, the difference between centripetal and straight line motion between earth and moon. Trouble is, making videos is time consuming :(

[@AtomicSchool]

It's refreshing to hear from someone who gets it.

[@stevehammond]

This very interesting but you haven’t mentioned the effect of the sun’s gravity on the oceans?

[@AtomicSchool]

See kzbin.info/www/bejne/fpWskIOvgquBqNE, Steve.

[@AtomicSchool]

See the sun's effect at kzbin.info/www/bejne/fpWskIOvgquBqNE.

[@willemjoubert3662]

The centre of gravity for both the moon and earth is far away from the centre of earth. While earth circles around the sun it is also circling around the centre of gravity with the moon. That causes centrifugal forces on the far side that cause a high tide on that side. There is no other reason.

[@willemjoubert3662]

@@TribusMontibus My reasoning is easy to understand. Why then make it so complicated?

[@Doruc7158]

Sorry to tell you, the scientists, the basic physics of "any force has a counter force" nullifies you illogical explanation of high tide.

[@ralphburnette]

This is totally inaccurate.

[@ralphburnette]

Here’s how tides actually work: kzbin.info/www/bejne/pqimmZ5qiJ5sn5Y

[@hafsagreer]

And here’s why this is totally inaccurate: kzbin.info/www/bejne/i6TJcqSbmK-hgdU

[@Mr.MT3]

Thank you.

[@iancanty9875]

The video is great as far as gravity gradients go but overall is somewhat simplistic. It but doesn't mention the inertia of earths oceans being part of the picture. Also, why doesn't the video also mention centrifugal force due to the barycentre(centre of mass) of both Earth and moon being about 4,600 miles (7,403 Km) towards the moon? The high tides on the away side of the barycentre and to a smaller degree on the moon side of the barycentre must partly be influenced by centrifugal force as well. To be clear, I'm not talking about centrifugal force due to Earth's spin about its own axis, but due to the earth/moon system's rotation about its barycentre.

[@iancanty9875]

@@AtomicSchool Thanks for the links. 👍 I will read all of them. Some of the ones I found online seem somewhat self contradictory, or just plain confusing because they’re poorly explained.

[@AtomicSchool]

Yes, Ian, it's a very tricky topic as I have discovered myself. Over time I have shifted from the centrifugal to the differential gravity explanation myself. Good luck!

[@iancanty9875]

@@TribusMontibus Okay. Thanks. It would seem that I have some careful studying to do. When I looked at several different papers online, I had a feeling that the subject is more complex than any of them led me to believe so I set off cautiously. It looks as if this is the right attitude with which to proceed.

[@robbiehiatt9966]

Well all you people can hate me or whatever from this comment I'm about to make. But I can't see where the moon has anything to do with the tide. I've seen the moon on the horizon when The Tide Is High. I've seen the moon high in the sky When the tide is low. The tide has a schedule, the moon has a schedule. They have absolutely nothing to do with each other. And please explain to me why the tide Rises 15 ft in this video yet out here in North Carolina it only Rises 6 feet. Why does the moon pull the water up further in some places? Three times as much. Here's what I say, the title range I think is the word you call it, is based on temperature and the amount of water in the oceans, expansion and contraction, do you realize how high the water level of the oceans would be if it all froze? This has nothing to do with the Moon

[@robbiehiatt9966]

@@lisaburke1045 every time I make a comment on KZbin someone hassles me about it.

[@tims5268]

Local tides are hugely affected by the geography of the ocean floor and ocean currents. The tides don't follow the moon perfectly. The moon (and the sun) merely provide the force to move the water, water then does its thing and acts like the fluid it is. Temperature does indeed have an affect on tides but it does not provide the force that causes tides.

[@robbiehiatt9966]

@@tims5268 I 100% believe your first sentence saying that tieds are hugely affected by the geography of the ocean floor. But I stand by my belief that there is no correlation between the Moon and the tide, your second sentence says that the tide does not exactly follow the Moon perfectly, where and how does it follow it at all? I can't find any correlation between the tide and the Moon. In fact go back and look when the last I think they call it a super tide but the last supertide was around July 4th and that's the point when the sun is farthest away from the Earth. I believe it's 3.2 million miles closer in January. Wouldn't it make sense that a super tide would come in January when the sun is closest the Earth?

[@robbiehiatt9966]

@@tims5268 I think the earth expands and contracts, the water is effected, I think of the world kinda as like a ball, you press here, it expands there... the ocean is pushing down and then the earth is pushing back in a perfect rhythm/balance. And if I can figure out why it's 50 ft in The Bay of Fundy and find how that correlates to the balance. Recently I looked at where the tides are the highest and I've narrowed that down to 4 vertical locations on the planet equally distances apart. The sun goes by once a day, 4 locations, 4 tides per day. Where's the correlation to the moon?

[@tims5268]

@@robbiehiatt9966 Why don't you try and get in touch with an oceanographer and see if they can help you with your theory?

[@wcottee]

No, this is not right. Richard Feynman explained the tide at "C" is caused by the fact that the Earth and Moon revolve about their mutual barycenter. The centrifugal force at "C" is what causes the high tide at "C".

[@AtomicSchool]

Yes, I've seen his Caltec lecture on this topic, but disagree with it, as do lots of other physicists. But not all. BTW, Feynman omitted this part of his lecture when the book version of his Caltec lectures was published. Maybe he had second thoughts? I don't know.

[@wcottee]

@@AtomicSchool Interesting...didn't notice that. Thanks for the reply. I may have to re-think this. Feynman was right about being careful to not fool yourself because you are the easiest person to fool. :)

[@AtomicSchool]

@@wcottee He was a great man. A hero figure, though he would not be comfortable with that role.

[@o0GzxS41BIbyux3vEoNehl5FH]

I think Faynman see this kzbin.info/www/bejne/pqimmZ5qiJ5sn5Y then decide to forget it.

[@sheetalbhat6664]

Me teacher sent me this-💀

[@mitchbollo1100]

Well, your teacher shouldn’t have. It’s wrong 🎃 try “what atomic school gets wrong about tides!” Instead 🤩

[@AtomicSchool]

🙂

[@sheetalbhat6664]

@@mitchbollo1100 🤣🤣🤣frr tho

[@sheetalbhat6664]

@@AtomicSchool wow this is the first time a youtuber replied to me !

[@mitchbollo1100]

@@sheetalbhat6664 wow 🤣🤣🤣 you are a youtuber Sheetal, you left an interesting comment and I replied with a reply, that might help you understand tides better, because this video isn’t helpful? 💀

[@bunchhoeun2128]

I am in quarantine that why I am wachting

[@AtomicSchool]

We're all in quarantine :(

[@sonyvarghese198730]

Watch such kind of videos even when you are not in quarantine. It will help to increase our knowledge

[@riotsss3473]

Sony Varghese or just causs it’s boring

[@milanchakraborty7564]

Yes ,they are true please watch the vedio

[@oajjajajaj]

My mum made me watch lol

[@tsehaynidaw4863]

when i see your video,it it s better to know oceans tides.but,the reason why to change high tide and low tide?sr

[@StanJan]

The gravitational attraction of the moon must be immense to over come the effects of the angular momentum the earth is subject to by rotating around the sun ! I just tried to quantify this, and the numbers do no, in ant stretch of the imagination provide the slightest clue how this is possible. Any guidance you can offer would be appreciated

[@AtomicSchool]

Good question Stan. Normally people ask about the angular momentum due to its rotation around its own axis, and its revolving around the barycentre with the moon. I would have to think about the momentum of the earth's orbit around the sun. My guess is that it does not matter, but I'm not defending that at the moment.

[@chuckh9168]

Zero mention of the gravitational effect of the Sun. Yes, it makes a difference, and so does the angle it forms with the sun and moon. This is not to say that moon phase determines tide height to an appreciable degree (neap/spring tide excepted), but this explanation completely ignores that part of the overall process. This also ignores that the moon is actually dragging the tidal bulge against the direction of the rotation of the Earth, and is the prime cause of the slowing down of said rotation. There is no "catching up"; the tidal bulge is always leading ahead of what would be a perfect line from the center of the earth to the moon.

[@AtomicSchool]

All true Chuck. You can find about the sun's effect at kzbin.info/www/bejne/fpWskIOvgquBqNE​.. The tidal dragging is beyond the scope of this elementary school video.

[@janj9650]

@@AtomicSchool lol elementary school video....and I’m sitting here 24 yrs old fully learning this 💩 finally for the first time today 🤣😭