The Mystery of Earth's Opposing Tide

[tgt]

What is the explanation for the tide on the surface of the Earth opposite the moon?

 

[Danger]

It's due to the inverse-square law of gravity. The strongest pull of the moon is upon the ocean closest to it. The second strongest pull is upon the Earth as a whole. Less so is the pull upon the opposite ocean.
The gravity of the moon therefore pulls the closest ocean away from the Earth, and then pulls the Earth away from the farthest ocean.
That's a gross oversimplification, but the best that I can contribute.

 

[Georgepowell]

What is the explanation for the tide on the surface of the Earth opposite the moon?

The moon does not just simply spin around the earth. The moon and the Earth spin around their common center of mass. i.e. you can treat the moon and the Earth as one object spinning around its own center of mass.

What this leads to, is that the part of the Earth further away from the Earth is like the outer side of a ball being swung round your head, and so the ball elongates in the direction of the point it is spinning from.

Of course the moons gravity still pulls the opposite side of the moon closer to it, but it is not as strong as the pull on the closer side of the earth.

 

[Vanadium 50]

TGT, Danger's description is the correct one.

 

[Georgepowell]

TGT, Danger's description is the correct one.

So is mine wrong?

He said that the Earth is being pulled away from the ocean, but the Earth is not getting any closer to the moon. I am aware that the Earth is accelerating towards the moon and not getting any closer, but as the ocean on the other side of the Earth is also staying the same distance away from the moon I don't see how his description works.

His explanation would only work if the Earth was always accelerating towards the moon at a faster rate than the ocean was. Mine works because both the ocean and the Earth would naturaly carry on in a straight line at a tangent to its orbit, and the ocean at the other side of the Earth is trying to do this the same amount that the Earth is, but the gravitational pull of the moon has more effect on the Earth because it is closer. And therefore the ocean "bulges" on that side.

Correct me if I am wrong, but tell me why as well.

Thanks

 

[Ian]

George,
Imagine that you are swinging on a trapeze.
In this 'you' are the Earth and the 'moon' is the trapeze pivot. If you do a 'loop the loop' as you swing, your hands holding the bar will feel a force toward the pivot - this force represents the higher tide on the side of the Earth facing the moon.
Your shoes however, will try to fly off your feet - this the force represents the tide on the far side of the Earth from the moon.
Like the others said, the key to understanding the tides is to realize that the Earth and moon rotate as one body about a single point.

 

[Georgepowell]

George,
Imagine that you are swinging on a trapeze.
In this 'you' are the Earth and the 'moon' is the trapeze pivot. If you do a 'loop the loop' as you swing, your hands holding the bar will feel a force toward the pivot - this force represents the higher tide on the side of the Earth facing the moon.
Your shoes however, will try to fly off your feet - this the force represents the tide on the far side of the Earth from the moon.
Like the others said, the key to understanding the tides is to realize that the Earth and moon rotate as one body about a single point.

Yes, so we agree. You simply came up with another analogy (very similar to my analogy of swinging a ball around your head).

I don't know if you read my first post, or if you knew that it was me that posted it. I was the one who mentioned the Earth and moon spinning round their barycenter (centre of mass of the two planets). And I was asking why I was wrong... But by the looks of it you agree with me as well.

Danger said: "...and then pulls the Earth away from the farthest ocean" which I said I thought was wrong, or at least significantly incomplete.

I am willing to accept I am wrong, but I need some convincing arguments :tongue:.

 

[Danger]

Danger said: "...and then pulls the Earth away from the farthest ocean" which I said I thought was wrong, or at least significantly incomplete.

That caveat was acknowledged in my last sentence, so I certainly take no offense at your mention of it. I merely answered with my best understanding of the situation as it was explained to me previously. Since I'm not a scientist, I always defer to those who actually know what they're talking about.
Now I'm just going to lie back and learn from the ensuing exchange.

 

[tgt]

From the discussion it seems that on the far side of the Earth (compared to the moon), the centrifugal force resulting from the rotation of Earth and moon outweighs the gravitational pull from the moon, resulting in a net force away from the Earth hence the second tidal wave.

 

[tgt]

pulls the Earth away from the farthest ocean.
That's a gross oversimplification, but the best that I can contribute.

This pull on the Earth away from the farthest ocean must be the centrifugal force on the Earth since the Earth is not moving toward the moon as pointed out by george.

In this way both george and danger are both correct?

 

[D H]

I'll make a few observations first and then proceed with some math.

First point: The tides at some point in general are not aligned with the Earth-Moon orientation. The tides at some point have various frequency components. The magnitudes and times at which these components peak varies from place to place on the Earth. The dominant frequency at most places is the M2 tidal, with a period of 12.421 hours (half of a lunar day). A plot:

http://svs.gsfc.nasa.gov/stories/topex/images/TidalPatterns_hires.tif

The plot has a bunch of white curves emanating from various nodes where the M2 tidal effect is zero. The M2 tidal contribution will peak at the same time at any point along one of those white curves.

Second point: Some places on the Earth (e.g., most of the Gulf of Mexico and the Caribbean) essentially have but one tide per day. The M2 tidal is very small throughout the Gulf and the Caribbean.The ocean tides are a complex interaction between the driving tidal force and friction between the oceans and the solid Earth. Rather than dwelling on these complexities, I'll finish with the gravitational forces that drive the tides. The easiest way to look at things is from the perspective of an Earth-centered frame. As this is a non-inertial frame, pseudo-force terms are needed to give the appearance that physics follows Newton's laws. In particular, the Earth is accelerating toward the Moon and the Sun. Worrying about the Moon only, the Earth as a whole is accelerating toward the Moon:

$$\boldsymbol{a}_{e\to m,\,\textrm{inertial}} = -\,\frac {GM_m}{||\boldsymbol{r}_{e\to m}||^3}\boldsymbol{r}_{e\to m}$$

The acceleration at some point pon the surface of the Earth differs slightly from this:

$$\boldsymbol{a}_{p\to m,\,\textrm{inertial}} = -\,\frac {GM_m}{||\boldsymbol{r}_{p\to m}||^3}\boldsymbol{r}_{p\to m}$$

The acceleration toward the Moon at that point as perceived in an Earth-centered frame is the difference between these accelerations:

$$\boldsymbol{a}_{p\to m,\,\textrm{Earth-centered}} = -\,GM_m\left( \frac {1}{||\boldsymbol{r}_{p\to m}||^3}\boldsymbol{r}_{p\to m} - \frac {1}{||\boldsymbol{r}_{e\to m}||^3}\boldsymbol{r}_{e\to m}\right)$$

The acceleration at the point on the Earth's surface directly between with the Earth's and Moon's centers of mass is directed toward the Moon with magnitude equal to

$$\aligned & GM_m\left(\frac {1}{(r_{e\to m}-r_e)^2}-\frac {1}{r_{e\to m}^2}\right) \\ &\approx 2GM_m\,\frac {r_e}{r_{e\to m}^3} \endaligned$$

The tidal acceleration at the point on the opposite side of the Earth has nearly the same magnitude and is directed away from the Moon. The driving force has two nodes, and hence the two tides per day.

 

[D H]

The moon does not just simply spin around the earth. The moon and the Earth spin around their common center of mass. i.e. you can treat the moon and the Earth as one object spinning around its own center of mass.

What this leads to, is that the part of the Earth further away from the Earth is like the outer side of a ball being swung round your head, and so the ball elongates in the direction of the point it is spinning from.

That's not a good explanation. Think of it this way: The sun and Earth orbit (don't use spin here; that means something different) about their common center of mass, too, and the Sun-Earth center of mass is located *a lot* further from the Earth's center of mass than is the Earth-Moon center of mass.

The tidal forces result from the gradient of the gravitational acceleration, and this is to first order an inverse-cube effect. Even though the Sun is much more massive than is the Moon, the Moon has a much greater tidal influence than does the Sun because the Sun is so much further from the Earth than is the Sun.

 

[Georgepowell]

That's not a good explanation. Think of it this way: The sun and Earth orbit (don't use spin here; that means something different) about their common center of mass, too, and the Sun-Earth center of mass is located *a lot* further from the Earth's center of mass than is the Earth-Moon center of mass.

The tidal forces result from the gradient of the gravitational acceleration, and this is to first order an inverse-cube effect. Even though the Sun is much more massive than is the Moon, the Moon has a much greater tidal influence than does the Sun because the Sun is so much further from the Earth than is the Sun.

Both the sun and the moon exert a gravitational force on the earth, but the difference from one side of the Earth to the other is much greater for the moon's pull than it is for the sun's pull. If I have this right, then I already understood this.

What this explains is why the water is pulled towards the moon less, it is still pulled, just less than on the closer side of the earth. This alone therefore does not explain the tide on the far of the earth; Just because a force is smaller it doesn't make the ocean move further away from the moon.

The position of the earth-moon centre of mass means that this water is being swung outwards.

Are you saying that if their was no Centrifugal force (I say this with caution, I know it is not a force, just an illusion) then the tide on the other side of the Earth would still be there? If so, what pushes it?

 

[russ_watters]

Are you saying that if their was no Centrifugal force (I say this with caution, I know it is not a force, just an illusion) then the tide on the other side of the Earth would still be there? If so, what pushes it?

Yes! An object in freefall directly toward anotherh object (as opposed to being in orbit) feels tidal forces. If you model this as three objects connected via two springs and drop this system from a stationary point above a massive object (and the objects are oriented along that line), both springs will stretch as the three objects are pulled apart.

People often talk about orbits as situations where centrifugal force equals gravitational force, but this is a little bit of a misleading way of looking at it and tidal forces are an example of why. To be honest, I'm not sure if doing the math on your answer would produce the correct result, but even if it does, it doesn't give a full treatment to the concept of tidal force.

 

[D H]

Both the sun and the moon exert a gravitational force on the earth, but the difference from one side of the Earth to the other is much greater for the moon's pull than it is for the sun's pull. If I have this right, then I already understood this.

What this explains is why the water is pulled towards the moon less, it is still pulled, just less than on the closer side of the earth. This alone therefore does not explain the tide on the far of the earth; Just because a force is smaller it doesn't make the ocean move further away from the moon.

Sure it does. The relevant gravitational acceleration is that relative to the Earth. The relative acceleration toward some third body at the subbody point at its antipodes are

$$a_{\textrm{tidal}} = GM_{\textrm{body}}\left(\frac {1}{(R_{\textrm{body}}\mp R_\textrm{e})^2}-\frac {1}{R_{\textrm{body}^2}}\right)$$

where GM_body is the body's gravitational parameter, R_body is the distance from the center of the Earth to the center of the body, and R_e is the radius of the Earth. (Note: For non-aligned points the full vector form must be used. I am using the fact that the Earth and body centers of mass, and the subbody point and its antipodes are collinear.)

Expanding the above,
$$a_{\textrm{tidal}} &= \pm \frac{GM_{\textrm{body}}R_{\textrm{e}}}{R_{\textrm{body}}^3} \left(2\pm3\frac{R_{\textrm{e}}}{R_{\textrm{body}}}+\cdots\right) \approx \pm 2\frac{GM_{\textrm{body}}R_{\textrm{e}}}{R_{\textrm{body}}^3}$$

The tidal acceleration is directed toward the third body at the subbody point and away from the third body at the antipodes, and the accelerations are nearly equal to one another (the omitted term is $O((R_{\textrm{e}}/R_{\textrm{body}})^4)$).

 

[Vanadium 50]

DH has done a good job explaining how tides work. I'd like to take a different approach, and show why centrifugal force doesn't: the Earth is traveling 40x faster around the sun than the moon travels around the Earth - and the speed at which the Earth travels around the earth-moon system is smaller still. Centrifugal force goes as the square of velocity, so solar tides would be at least 1600x larger than lunar tides. This isn't what we see, so that theory can't be the explanation.

 

[Georgepowell]

Yes! An object in freefall directly toward anotherh object (as opposed to being in orbit) feels tidal forces. If you model this as three objects connected via two springs and drop this system from a stationary point above a massive object (and the objects are oriented along that line), both springs will stretch as the three objects are pulled apart.

People often talk about orbits as situations where centrifugal force equals gravitational force, but this is a little bit of a misleading way of looking at it and tidal forces are an example of why. To be honest, I'm not sure if doing the math on your answer would produce the correct result, but even if it does, it doesn't give a full treatment to the concept of tidal force.

I think that we both have the same theory behind the tide, just I might be explaining it differently...

As i have said, the centrifugal force is not a real one, just a tendadncy for objects to stay moving in a straight line. And in your analogy where the three objects on springs are in free-fall, the object on the further side has a tendency to 'stay in the same place' whilst the other parts are falling towards the masive object. So your explanation is really just another way of explaining the centrifugal force.

 

[D H]

George, your barycenter argument just doesn't work. See post #16.

You will get something closer to the gravity gradient model if instead of looking at the Earth orbiting the Earth-Moon barycenter, you look at it in terms of the Earth orbiting the Moon. It still won't be right, however.

This model also will not be able to explain the tidal force on point on the Earth's surface whose distance to the Moon's center of mass is equal to the distance between the Earth's and Moon's centers of mass. Here, the magnitude of the tidal force is about 1/2 that of the force at the sub-Moon point and at its antipode. While the force at the sub-Moon point and at its antipode are directed away from the Earth, the tidal force at these equidistant points (a small circle on the Earth) is directed inward.

This small circle of equidistant points is very close to the great circle equidistant between the sub-Moon point and its antipode. One way to look at the tidal forces is to imagine the Earth as a rubber ball with strings attached to the sub-Moon point and its antipode. The tidal forces pull on the strings and squeeze on the middle.

 

[russ_watters]

I think that we both have the same theory behind the tide, just I might be explaining it differently...

As i have said, the centrifugal force is not a real one, just a tendadncy for objects to stay moving in a straight line. And in your analogy where the three objects on springs are in free-fall, the object on the further side has a tendency to 'stay in the same place' whilst the other parts are falling towards the masive object. So your explanation is really just another way of explaining the centrifugal force.

In my example, there is no rotational component, so no centrifugal force.

 

[Georgepowell]

OK, let's get this straight; I don't like arguing with someone that I agree with. Firstly, I think your explanation is completely valid, and is virtually the same explanation as mine:

In the following (correct) analogy:

If you model this as three objects connected via two springs and drop this system from a stationary point above a massive object (and the objects are oriented along that line), both springs will stretch as the three objects are pulled apart.

If you where standing on the middle object (the earth) you would see the object moving away from you, and it would look like something was pushing it away. The same thing happens when that object is in perpetual free-fall (orbit), and it is this exact same effect that we call the centripetal force, or at-least a form of it. Each part of our planet is subject to this centripetal force, and each part of it is 'trying' to escape of at a tangent. But because of the "gradiant" of the strength of the moon's gravity, the further side of the earth, and its oceans, are more effected by it, the Earth as a whole is less effected, and even less effected is the ocean closer to the moon.

I think you are all correct, and I know that the gradient of the moons gravitational strength on the Earth is part of the explanation of the tides. But it is the tendency for the parts of the Earth to keep on going in a straight line that causes the bulge:

If the barycentre of the earth-moon was directly in the centre of the earth, then there would be no tide on the far side of the earth, because the Earth would not be orbiting around anything, and would not be subject the the water at the far side of the Earth would not have the initial velocity to bulge out away from the barycentre.

I hope this ends our discussion.

Thanks - George

edit: This agrees with me as well: http://oceanlink.island.net/oinfo/tides/tides.html

 

[D H]

OK, let's get this straight; I don't like arguing with someone that I agree with. Firstly, I think your explanation is completely valid, and is virtually the same explanation as mine.

No, they are not. Please read post #16 by Vanadium 50, link https://www.physicsforums.com/showthread.php?p=1993649#post1993649".

To get a centrifugal force you must be working in a rotating frame, specifically a frame rotating with the Earth/Moon orbital angular velocity. The point with respect to which you wish to express the centrifugal force is the Earth-Moon barycenter, which is about 4670.3 km from the center of the Earth. Using centrifugal force as the explanation of the tides leads to several erroneous results, some of which are

1. The centrifugal force at the sub-Moon point, located about 6378.1 km - 4670.3 km = 1707.8 km from the barycenter, is less than 1/6 of the centrifugal force at the antipode of the sub-Moon point, located about 6378.1 km + 4670.3 km = 11048.4 km from the barycenter. The tidal forces at the sub-Moon point and its antipode are nearly equal in magnitude.
2. The centrifugal force is an order of magnitude larger than the tidal force even at the sub-Moon point: 1707.8 km * (2*pi / sidereal month) = 1.2×10^-5 m/s². (The acceleration at the antipode of the sub-Moon point is even further off.) Compare to the actual tidal acceleration at the sub-Moon point, 1.1×10^-6 m/s².
3. There is a non-zero centrifugal force at the center of the Earth. The tidal acceleration at the center of the Earth is identically zero.

edit: This agrees with me as well: http://oceanlink.island.net/oinfo/tides/tides.html

You're right: It agrees with you. And like yours, it's an incorrect explanation. The bulges on the sides of the Earth facing and opposite the Moon do not arise from two different forces, lunar gravity for the near side bulge and inertia (which isn't a force) for the far side bulge. Look at it this way: Lunar gravity extends to infinity. It doesn't stop at the surface of the Earth. Similarly "inertia" (Newton's first law) doesn't stop at the surface of the Earth. It applies to the near side as well as the far side.

If you develop the full mathematics in the rotating Earth-Moon frame you will eventually come up with the same answer as arrived at in an Earth-centered frame. Of course, you can also explain the behavior of distant galaxies from the perspective of an Earth-centered, Earth-fixed frame. In both cases, the math is atrociously nasty.

 

[Georgepowell]

I don't think there are two different forces involved! Inertia, or the centrifugal force, are not forces.

I'm going to explain my point via various quotes from Wikipedia:

Tidal Force:

The tidal force is a secondary effect of the force of gravity and is responsible for the tides. It arises because the gravitational acceleration experienced by a large body is not constant across its diameter. One side of the body has greater acceleration than its center of mass, and the other side of the body has lesser acceleration.

Inertia:

Inertia is the resistance of an object to a change in its state of motion.

The tidal force, and the overall inertia of the Earth balance out equally, so the moon and the Earth are in a nearly perpetual orbit. At the far side of the moon, the inertia is larger than the gravitational pull, and at the closer side of the moon the gravitational pull is greater. The overall effect is perfectly balanced, and hence the perfect orbit.

Post #16 does not take into account that (in the sun-earth orbit) the Inertia is balanced with the sun's gravitational pull almost perfectly throughout the planet because it is so far away, and the pull of the suns gravity does not change from one side of the Earth to the other (significantly). So the centrifugal force is not apparent in the earth-sun system.

This image shows that on the right (the moon side of the earth) the gravitational force is greater. And on the left, the inertia is larger, and so that side of the planet does not accelerate as fast towards the moon. And in the centre, they balance out perfectly.

*This also includes the Earth's gravity.*

 

[chis]

Centrifugal or gravity or both?

 

[D H]

I'm going to explain my point via various quotes from Wikipedia

Rather than argue with words, argue with math. Show us, in math, exactly what you are talking about. Doing physics without math is just sophistry.

This image shows that on the right (the moon side of the earth) the gravitational force is greater. And on the left, the inertia is larger, and so that side of the planet does not accelerate as fast towards the moon. And in the centre, they balance out perfectly.

*This also includes the Earth's gravity.*

That image most certainly does not include the Earth's gravity. Think about it: If the net force was outward the oceans would long ago have flown off the face of the Earth. You took this image from the Wikipedia article on http://en.wikipedia.org/wiki/Tide#Forces". The caption below the image reads

Figure 7: The Moon's (or Sun's) gravity differential field at the surface of the Earth is known as the tide generating force. This is the primary mechanism that drives tidal action and explains two tidal equipotential bulges, accounting for two high waters per day.​

Since you are so won't to quote Wikipedia, why did you not use this, from the same section of the Wikipedia article referenced above?

The tidal force produced by a massive object (Moon, hereafter) on a small particle located on or in an extensive body (Earth, hereafter) is the vector difference between the gravitational force exerted by the Moon on the particle, and the gravitational force that would be exerted on the particle if it were located at the center of mass of the Earth. Thus, the tidal force depends not on the strength of the gravitational field of the Moon, but on its gradient (which falls off approximately as the inverse cube of the distance to the originating gravitational body; see NASA).​

Note well: That differential force is exactly what I described with mathematics in post #11. That differential force is also exactly what is depicted in the referenced image.Edit:
The cited image is a tad too symmetric. The author of the image (source attribute on the Wiki commons page is "(Differential gravity field that causes tides Source:me {{GFDL}} en:Tide)") plotted the first-order approximation to the tidal force. Assuming a non-rotating frame with the x-axis instantaneously co-aligned with the Earth-Moon vector, the tidal force at some point r on the Earth's surface to first order is

$$\mathbf{a}_{\textrm{tidal}} \approx \frac {GM_{\textrm{moon}}}{R_{\textrm{earth}\to\textrm{moon}}^3} \bmatrix 2 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \endbmatrix \mathbf r$$

 

[Georgepowell]

Your right, it doesn't include the Earth's gravity. (I added that at the end because I saw that the arrows at the side of the Earth where pointing inwards, but I now see why this is)

I'm fed up of this discussion anyhow, it is going nowhere and I'm getting tired and bored.

My point is this, and only this:

If it wasn't for the inertia of matter, then a difference in gravitational acceleration from one side of the Earth to another would not make a difference.​

Thanks for the discussion anyway, you will be glad to know my point of view has slightly changed slightly. I still think I understand this though, It is just the wording of my explanations that causes the debate.

 

[Georgepowell]

Centrifugal or gravity or both?

In terms of forces - Just gravity.

It is the inertia of the water that resists the acceleration due to gravity.
The inertia is constant for each part of the earth. But the gravity is not constant, so the acceleration of the oceans towards the moon is different from one side of the Earth compared to the other.