# Why do we have two tides due to moon?

We have two tides because of the moon's attraction on the front and back side of the earth. I can understand the reason for the attraction of the moon on the front side which causes the water to move towards the moon but I can't understand the reason that the thing happens in the backside.

I can't understand this paragraph to be exact-

What do we mean by "balanced"? What balances? If the moon pulls the whole earth toward it, why doesn't the earth fall right "up" to the moon? Because the earth does the same trick as the moon, it goes in a circle around a point which is inside the earth but not at its center. The moon does not just go around the earth, the earth and the moon both go around a central position, each falling toward this common position. This motion around the common center is what balances the fall of each. So the earth is not going in a straight line either; it travels in a circle. 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". The result of this imbalance is that the water rises up, away from the center of the earth. On the near side, the attraction from the moon is stronger, and the imbalance is in the opposite direction in space, but again away from the center of the earth. The net result is that we get two tidal bulges.

I couldn't understand those bold sentences.

Thanks a lot for your help

Danger
Gold Member
Anyhow... the moon pulls on the nearside water, which sucks it away from the Earth. With less force, it pulls on the Earth, which sucks it away from the farside water. It's sort of as if you stretch an elastic band with 3 concurrent ink dot's on it. (Strangely similar to galactic recession.)

Anyhow... the moon pulls on the nearside water, which sucks it away from the Earth. With less force, it pulls on the Earth, which sucks it away from the farside water. It's sort of as if you stretch an elastic band with 3 concurrent ink dot's on it. (Strangely similar to galactic recession.)

Moon's field on the other side is weak. I agree with that. But what makes it to move away from the earth? This means that there is some other force acting in the opposite direction of the moon's force. What is that?

Danger
Gold Member
What is that?

That's just inertia. The water is reluctant to move unless forced to. It's sort of like how your head snaps back when you stomp on the gas pedal, but slower and a lot bigger.

rcgldr
Homework Helper
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A.T.
But what makes it to move away from the earth?
The Earth moves away from it, because it is pulled stronger towards the moon.
This means that there is some other force acting in the opposite direction of the moon's force. What is that?
If you go to the non-inertial rest frame of the Earths center, there is an inertial force pulling the other way. But this is not the "inertial centrifugal force", just a uniform inertial force.

We have two tides because of the moon's attraction on the front and back side of the earth. I can understand the reason for the attraction of the moon on the front side which causes the water to move towards the moon but I can't understand the reason that the thing happens in the backside.
As commented on above, the reason is because the moon is subject to what are known as tidal forces (this was described above using tidal accelerations/gradients rather than tidal forces).. The tidal force which the earth subjects the moon to causes the moon and anything on it to be stretched along the line connecting the centers of the earth and moon. The moon does the same thing on the earth thus forcing water away from the center from the earth on one side and towards the earth on the other.

Picture what would happen if three closely spaced objects were in orbit in the same plane around the sun each starting off along a radial line from the sun. Use the first one as a reference and let it be subjected only to gravitational forces. If you tried to restrain the object closer to you to remain along the line then you’d have to pull on it towards you. That is what tidal forces are all about.

It's also tidal forces that cause satellites to require adjustments to their alignment relative to the stars after a while.

That's just inertia. The water is reluctant to move unless forced to. It's sort of like how your head snaps back when you stomp on the gas pedal, but slower and a lot bigger.

I sort of thought like that in the beginning. But there is something in his book (Feynman). He says-

"Another school of thought claimed that the high tide should be on the other side of the earth because, so they argued, the moon pulls the earth away from the water! Both of these theories are wrong."

In some other place from what I understand he says that was because of the centrifugal force that was acting away from the center of the earth. This centrifugal force is because of the earth's rotation right?

Thanks a lot for your replies

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Danger
Gold Member
"Another school of thought claimed that the high tide should be on the other side of the earth because, so they argued, the moon pulls the earth away from the water! Both of these theories are wrong."

That was sort of a dumb thing for them to say, since we're talking about low tide.

A.T.
I sort of thought like that in the beginning. But there is something in his book (Feynman). He says-

"Another school of thought claimed that the high tide should be on the other side of the earth because, so they argued, the moon pulls the earth away from the water! Both of these theories are wrong."

In some other place from what I understand he says that was because of the centrifugal force that was acting away from the center of the earth. This centrifugal force is because of the earth's rotation right?

Centrifugal forces due to Earth's rotation can be used to explain the equatorial bulge, but not the tides. Going into a non-inertial frame, where there are inertial forces, can be usefull to explain something. But in an inertial frame the gravity gradient alone must explain all of the effect, so the inertial forces can't add anything to the deformation.

What Feynman probably means by "centrifugal force" is not the inertial force due to rotation of the reference frame, but a linear inertial force due to non-linear translation of the reference frame (circular motion). The difference is explained here:
http://vialattea.net/maree/eng/index.htm

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It's easier to imagine with the sun instead of the moon. You have the sun's gravity that pulls on the earth and you have the centrifugal force that pulls the earth away from the sun. The two are cancelling each other. This is also the reason for why astronauts on the ISS are weightless even though the gravity of earth is nearly as strong there as it is on the ground.
So on average the forces cancel out but one side of the planet is a little closer to the sun than the other.
On the near side of the earth you have a slightly stronger gravity that pulls the earth towards the sun. The centrifugal force is slightly weaker there so in total you have a net force towards the sun. On the far side the gravity is slightly weaker and the centrifugal force slightly stronger so you have a net force away from the sun. Therefore you get two bulges in the ocean.
Of course the force on the near side and the force on the far side have to be identical in magnitude for the earth to stay in orbit. So the two tides are equally strong.
With the moon it works exactly the same as with the sun. It may seem different since it appears that the moon is circling around the earth and not the other way around but actually earth and moon are both circling around each other. There is a centrifugal force here acting on the earth and it is identical in magnitude to the gravity that the moon excerts on the earth so it really is exactly the same situation.

That's just inertia. The water is reluctant to move unless forced to. It's sort of like how your head snaps back when you stomp on the gas pedal, but slower and a lot bigger.

You are actually right there. It is inertia. And the effect of inertia on a rotating system is called "centrifugal force".

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A.T.
So the two tides are equally strong.
No, they are not. The opposite side tide is weaker.

sophiecentaur
Gold Member
2020 Award
No, they are not. The opposite side tide is weaker.

It's weaker but only 'a bit' weaker. If you look at the graphs of tidal heights, at the peak of springs, the difference is only a few percent (at least on the South Coast of the UK). The tidal height, in practice, is complicated by many factors like resonances in the oceans and the effects of depth near the land. The Sun's effect is significant, too and clouds the issue a bit - is it springs of neaps that give the best indication of what's going on due to the Moon / Earth system.
I have just looked at the curves for Ascension Island (mid Atlantic) and they show a variation of tidal height of about 0.5m max, interestingly (1/10 of the range in UK). Also, the difference in height between the two tides seems to be a bigger proportion than in UK.
Correction: for some daft reason my tide predicting software changed from metres to fathoms so the difference is not so great!
It would be good to know what would happen with a planet covered in water and with a smooth ocean floor. I guess the depth of water would be a factor. I'm sure someone must have done a model.

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A.T.
It's weaker but only 'a bit' weaker. If you look at the graphs of tidal heights,
I'm not talking about the actual measured heights, which are affected by many other factors. Just about the idealized tidal deformation.

D H
Staff Emeritus
It's easier to imagine with the sun instead of the moon. You have the sun's gravity that pulls on the earth and you have the centrifugal force that pulls the earth away from the sun. The two are cancelling each other. This is also the reason for why astronauts on the ISS are weightless even though the gravity of earth is nearly as strong there as it is on the ground.
Aside: Using centrifugal force to explain why astronauts on the space station are feel "weightless" is fundamentally wrong. The reason is that you can't feel gravity.

Using centrifugal force to explain the tides is also fundamentally wrong. Every explanation I've seen of the tidal forces that invokes centrifugal force makes at least two errors, but with the happy result that two or more wrongs in this case do make a right. It's a terrible fudge.

Here are a couple of correct ways of looking at things. One is from the perspective of an inertial frame. Because gravity is a 1/r2 force, the Moon pulls the water closest to the Moon away from the Earth and it pulls the Earth away from the water furthest from the Moon. This is exactly what Feynman wrote in Six Easy Pieces:
Feynman said:
It actually works like this: The pull of the moon for the earth and for the water is "balanced" at the center. But the water which is closer to the moon is pulled more than the average and the water which is further away from it is pulled less that the average.

Another valid way of looking at things is from the perspective of an Earth centered frame. From the perspective of this frame, the Earth as a whole accelerates towards the Moon due to gravity (and towards the Sun as well, and also Jupiter, Venus, etc.; these also induce (very small) tides). This Earth centered frame is an accelerating frame; this results in a constant inertial force at every point. Compare with the centrifugal force, which is not constant. It is instead varies linearly with distance from the axis of rotation. From this perspective, the tidal force is roughly a 1/r3 force, where r is the distance between the Earth and the Moon.

Both the inertial and Earth-fixed points of view yield valid explanations of the tidal forcing functions. They do not explain the tidal bulges for the simple reason that Newton's tidal bulges do not exist. They can't exist for a number reasons:
• Newton's equilibrium theory of the tides dictate that high tide occurs when the Moon is closest to zenith / closest to nadir and that low tide occurs when the Moon is on the horizon. This happens only rarely, and it's sheer dumb luck when it does. One problem is that there are two huge north-south barriers to this supposed tidal bulge, the Americas and Africa/Eurasia. If Newton's equilibrium theory of the tides was correct, the timing and magnitudes of the tides at the Pacific and Atlantic sides of the Panama canal would be more or less the same. They're anything but. The tides on the Pacific side of the Panama canal are huge, over an order of magnitude larger than predicted by Newton's equilibrium theory of the tides. The tides on the Atlantic side are rather small, smaller even than the smallish tides predicted by the equilibrium theory. The tides on the Pacific and Atlantic sides of the canal also differ significantly in timing. The tides don't magically pick up where they would have been had Panama not existed.

Another place of interest is the North Sea. The North Sea is rather small; per Newton's equilibrium theory of the tides, high tide should occur more or less simultaneously across all of the North Sea. That's not what happens. Instead, regardless of time of day, you can always find some point in the North Sea where it's high tide. At the exact same time, you can also find some other point in the North Sea where it's low tide. The equilibrium theory of the tides cannot come close to explaining the tides in the North Sea.

• Suppose those continental barriers didn't exist, with the Earth covered everywhere by four kilometer deep oceans (the median depth of the oceans). You still wouldn't get Newton's tidal bulges. The problem is the Earth's rotation. The tidal bulges would be an ocean wave with a wavelength equal to half the circumference of the Earth. This wavelength is much, much greater than the four kilometer depth. The tidal bulge (if it existed) would be a shallow wave. Shallow waves travel at a speed of ##\sqrt gd##, or about 200 km/second in the case of a 4 km deep ocean. The tidal bulge on the other hand would need to travel at 465 km/second. The tidal bulge cannot exist even in a water world with oceans only 4 km deep.

• The oceans would have to be over 20 to 30 km deep to support Newton's tidal bulge, and even with that Newton's tidal bulge still wouldn't exist except near the equator. The problem this time is the coriolis effect. The tidal bulge can't exist as a world-wide phenomenon thanks to Kelvin waves. Any bumps in the bottom of this deep, deep ocean will eventually result in amphidromic systems forming, even in this very deep ocean.

The correct theory of the ocean tides started with Laplace, about 100 years after Newton's time, and culminated in the early 20th century with the works of George Darwin and A T Doodson. This dynamic theory of the tides accounts for the issues described above. There is no tidal bulge, at least not in the ocean tides. Newton's equilibrium theory does work (to some extent) for the Earth tides.

Hello. In his books he says "Earth and moon go around a central position, each falling toward this common position". This common position is inside the earth but not at its center. I can't understand this fact. Is that because they both go around the center of mass? The centrifugal force because of this seems to be the reason for balancing the moon's gravity at the center of the earth right?

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turbo
Gold Member
Can you Google on tides? It would be less intrusive than asking PF members to type out a 2-3 page dissertation on tides - something that is already well-understood. I'm not trying to be mean, just trying to keep stuff in balance and keep extraneous materials off the forum. Good luck.

sophiecentaur
Gold Member
2020 Award
It may be worth making the point that 'things" (which includes sticks, wheels, funnels and groups of objects in space) will always rotate about their centre of mass when in iisolation. ThEarth is so much more massive than the Moon that the CM is actually below its surface and not a point 'in between'. There are plenty of objects with one big bit and one small bit , which will balance about a point inside the bigger bit - think of some 'mobiles' in kids' bedrooms.

Danger
Gold Member
Sophie, I often suspect that there are a lot of people out there who think that gravity originates at the surface of an object.

sophiecentaur
Gold Member
2020 Award
PF exists to put them right, then. :-)

Danger
Gold Member
PF exists to put them right, then. :-)

Yeah. Try it at 2 am in a bar full of cowboys...

(I never gave up, but the urge to kill often became incredibly strong.)

sophiecentaur