Guest Post by Willis Eschenbach
Short Post. You can skip this if you understand the tidal force. Some folks seem confused about the nature of tidal forces. Today I saw this gem: “The tide raising force acts in both directions (bulge on each side in the simplistic model)” … the author of that statement may understand the tides, but that isn’t how tidal forces work, they don’t act in different directions.
To clarify the situation, let me give an example. Imagine three one-kilogram masses M1, M2, and M3, connected by a blue rope, that are in free-fall directly into the sun. Figure 1 shows the situation.
Figure 1. Forces on three one-kilogram masses M1—M3, which are in free-fall straight towards the sun, and are connected by the blue rope. The equations at the bottom show the force of gravity on each of the three masses. The “tidal force” is the difference between two gravitational forces on two objects. Solar tidal forces on a 1-kg mass at the earth’s surface average about half a micronewton (µN, or 10-6 newtons), and the lunar tidal forces are about twice that, about a micronewton. By comparison, Earth’s gravity exerts a force of about 10 newtons on a 1-kg mass at the surface …
Clearly, the gravitational force on M1 is greater than the force on M2, which in turn is greater than the force on M3. The difference between those two gravitational forces is the tidal force. There are two different tidal forces of interest in the diagram, which are GF1 – GF2, and GF2 – GF3. As a result of these tidal forces, as the three masses fall towards the sun. the near mass M1 moves the furthest, the far mass M3 moves the least, and the blue rope is always under tension.
For tidal forces on a theoretical planet, just imagine that the surface of the planet is the gray dotted line, and that “r” is the radius of the planet. The tidal force is the difference between the force of gravity on the unit mass M2 at the center of the planet, and the forces of gravity on the unit masses M1 and M3 at the points nearest and furthest from the sun.
Calculating the Tidal Forces
As shown in Figure 1, to calculate the tidal force you can just subtract the gravitational force on the unit mass at the center of the planet from the force on the other unit mass, for example GF1 – GF2. So we can use the equations in Figure 1 to calculate the force.
However, when the distance D is much, much larger than the radius r, we can make some simplifying assumptions. First, we can assume that the force between M1 and M2 is the same as the force between M2 and M3. It’s not, as you can see from the equations … but it is not a large error. For tides from the moon, the individual tidal forces are about ±5% different from their average value.
The next simplifying assumption we can make if D is much larger than r is that the average tidal force can be calculated as
Tidal Force = 2 * G * sunmass * r / D^3
This leads to tiny errors with respect to the sun’s average tidal force on the earth, and slightly larger errors with respect to the moon. For the sun we have
Near side true tidal force on earth (GF1 – GF2) = 0.50590 µN
Far side true force (GF2 – GF3) = 0.50583
True average = 0.50587
Approximation = 0.50587
So the approximation of the average tidal force is good to five significant digits …
And for the moon we have:
Near side true tidal force on earth (GF1 – GF2) = 1.1350 µN
Far side true force (GF2 – GF3) = 1.0796
True average = 1.1073
Approximation = 1.1067
For the moon, the approximation of the average is good to four significant digits. For most results, of course, those differences are far too small to be meaningful, and we can use the approximation without concern.
Why Are There Two Tidal Bulges?
Moving on, let’s look at why there are tidal bulges on both the near and far sides. It’s not because the tidal force acts in different directions, because tidal force is always directed towards the sun. So lets look at Figure 1 again, and this time we’ll include a planet and an ocean. As in Figure 1, the planet is simply free-falling into the sun. That result is shown in Figure 2:
Figure 2. Tidal forces elongating the planet and the ocean. Note that the planet is elongated as well, but this is not shown in the diagram because obviously, tides in the solid planet are much smaller than tides in the ocean. NOTE THAT THIS PLANET IS NOT THE EARTH.
As the planet free-falls into the sun, the near side of the ocean is pulled the most. So it moves the most, pulling away from the planet. And the planet, in turn is pulled more than the far side, so it pulls away from the ocean. These relative motions create the “bulges”
In other words, the “bulges” on the two sides of the planet are simply a result of the tidal forces stretching the entire system. Indeed, the atmosphere is subject to the same phenomenon, so there are atmospheric tides as well. In prior centuries, these atmospheric tides were said to affect the weather, although in a cyclical rather than a secular manner. However, such connections, while certainly possible, have proven to be elusive and difficult to establish. One complicating factor is that the atmospheric tides that are caused by the sun occur in combinations with solar heating, which makes the atmosphere swell. Also, the variations in pressure caused by the atmospheric tides are small, on the order of a tenth of a millibar (or of a hectopascal), or a variation of only 0.01% …
Finally, I’ve said that the situation in Figure 2 shows that planet free-falling into the sun. As it gets closer and closer, the tidal forces stretching the system get greater and greater, stretching the system more and more. Eventually, at a distance called the “Roche limit”, the tidal forces can reach such strength that they rip the planet into pieces … dang, to have a ring-side seat right on the Roche limit line for that spectacle would be awesome, even if it were perhaps a bit warm …
My best to all,
w.
PS—As with gravity, every mass in the universe exerts a tidal force on every other mass, so you can personally take credit for raising a tide on the surface of the sun … yeah, it’s a small tide, but it’s real. Just sayin’, I had someone claim the opposite. Lots of tidal misinformation out there.
THE USUAL: If you disagree with me or anyone, please quote exactly what it is you disagree with, and tell us precisely why you disagree. That way, we can understand exactly what it is you object to.
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Both sides in this argument are correct. You don’t need to use the centrifugal force to derive the formula for tides. This is because there is a perfect balance between the centrifugal force and the gravitational force at the center of the earth when in orbit around the earth-moon barycenter. This balance also determines the strength of tides on earth. When in doubt see what Feynman says.
So a body in free fall into the sun experiences ever increasing tidal forces until it is torn apart. A body in orbit however experiences regular varying tidal forces depending on the eccentricity of the orbit.
So on a purely logical basis Willis is correct. But in order to calculate the variations of tides on earth you need to include orbital dynamics because they change the earth-moon distance.
The formula: “Tidal Force = 2 * G * sunmass * r / D^3” is wrong. This is the formula for the “Tidal acceleration”. To make it a tidal force, you need to have another mass. The usual way it’s written is with m for the little mass (you used 1kg masses before) and M for the big mass (the sun):
Tidal Force = 2G m M r/D^3
The sentence “It’s not because the tidal force acts in different directions, because tidal force is always directed towards the sun.” is incorrect. On the side nearest the sun, the tidal force is directed towards the sun and moves the water towards the sun. On the side away from the sun, the tidal force is directed away from the sun and moves the water away from the sun. It should be clear that these are opposite directions. On the other hand, it is true that the “gravitational force” is always directed towards the sun. Maybe that was the source of some confusion. “Tidal forces” are defined with respect to the center of mass of the body (earth in this case), not the gravitating object.
On “Eventually, at a distance called the “Roche limit”, the tidal forces can reach such strength that they rip the planet into pieces …” it might be useful to note that the Roche limit is achieved when the tidal forces exceed the gravitational force of the planet. This makes it clear that “tidal forces” have to act oppositely on opposite sides of the planet — as does the planet’s gravitational force. In fact, I got the impression that someone reading the article might conclude that tidal forces should cause people on the sunny side of the planet to fall off of the planet. The reason this doesn’t happen is that the tidal force is smaller than the planet’s gravitational force.
By the way, I teach this subject.
there is a correct explanation of tides, written by some scientists involved in that kind of thing, here:
National Oceanic and Atmospheric Administration
http://tidesandcurrents.noaa.gov/restles1.html
Carl Brannen says:
February 16, 2014 at 2:10 am
Carl, I calculated (as is common) the force on a 1 kg mass. So I could have written
Tidal Force = 2 G 1 M r/D^3
You happy now? It doesn’t make any difference to me.
That sounds good, Carl … but I’ve given the mathematical derivation above. Now if you have a mathematical derivation that yields a force that points away from the sun, then please, break it out.
Well … no. For example, the Roche limit is different for liquid and solid planets … how could that be if the only relevant limit is the gravitational force holding the planet together?
In fact, as a moment’s reflection will confirm, the tensile strength of the planet is also part of the calculations of where a planet will break up …
Since the claim about the Roche limit is wrong (e.g. a planet which is a giant diamond would have a different Roche limit than a liquid planet), no, it doesn’t “make it clear”.
Gosh … does that prove that you must be right? Now I’m really impressed.
In any case, I await your mathematical derivation of the force that pushes the water away from the sun, or as you put it:
That’s big news, Carl, that there exists a repulsive gravitational force, that instead of attracting the water towards the sun, actually moves the water away from the sun … that’s a neat trick. Break out the math, so we can see how that works. You teach this the subject, surely you can bring forth the math that substantiates your claim of gravitic repulsion …
w.
Ulric Lyons says:
February 15, 2014 at 6:54 pm
“I was remarking upon the described 54.7° angle rather than the diagrammatic representation of it.
It makes a sqrt 2, sqrt 3, sqrt 5 sided triangle.”
I hadn’t noticed that. Thanks.
Asmilwho says:
February 16, 2014 at 2:17 am
“there is a correct explanation of tides, written by some scientists involved in that kind of thing, here:
National Oceanic and Atmospheric Administration
http://tidesandcurrents.noaa.gov/restles1.html”
I know, I provided it to Willis on his previous thread about Tides. I don’t think it got through then.
And all this
“We must concentrate on the Tide Raising Force”
In an abstract way it IS interesting. But in a minutia sort of way really. How the planet reacts to that tiny force and the massive outcomes that occur here on the surface and below the surface is the real source of interest.
http://i29.photobucket.com/albums/c274/richardlinsleyhood/TheInternalTideatHawaii_zps7c7d5dbf.png
http://i29.photobucket.com/albums/c274/richardlinsleyhood/M2Tides_zps758f7faa.png
http://i29.photobucket.com/albums/c274/richardlinsleyhood/K1Tides_zps453d8381.png
clivebest says:
February 16, 2014 at 12:57 am
“The tractional force acting along the surface and therefore unaffected by the earth’s gravity is
Fy cos(theta) + Fx sin(theta) . The tractional force is zero at the central bulge and equal and opposite from either side.”
I have tried to point this out to Willis already.
The important thing here is that this tractional force is not a delta on a larger force, Gravity. That is what the vertical component Willis is so keen on concentrating on is. A delta, a small difference. One capable of affecting the planet in a way that is difficult to miss but a delta none the less.
The tractional or tangential to the surface force is NOT a delta. There is nothing opposing it. If operates in its full magnitude which makes it much more interesting IMHO.
Coldlynx says:
February 16, 2014 at 12:48 am
“That is valid only 2 days a year for the sun due to earth tilt. All other days of the year is the “Sun tidal zero poles” not at earth geographics poles. Moon inclination will also have large impact on where on earth the “moon tidal force zero poles” are located. And also where on earth the maximum horisontal tide forcing are located. The horisontal forcing that actually move air and water according to Laplace.
That latitude change between 45°N and 45°S +/- 28.36°
In a cyclic way. 18.6 years the most obvious.”
Again, something that has been pointed out to Willis before on another thread, by me and Greg.
On each occasion he has seen the force pattern, drawn on a vertically oriented globe and looked at it as though that was the true picture.
What he has never done is drawn the North Pole to South Pole line through that globe and considered how that changes the picture.
http://upload.wikimedia.org/wikipedia/commons/d/d8/Field_tidal.png
http://upload.wikimedia.org/wikipedia/commons/6/61/AxialTiltObliquity.png
Willis Eschenbach says:
February 15, 2014 at 11:41 pm
Steve Fitzpatrick says:
February 15, 2014 at 12:48 pm
“Steve, neither Leif or I are talking about actual tides on earth. We’re talking about how the tidal forces work on a planet. So whether Clive has it correct or not for the earth is material for a discussion of the actual Earthly tides … it means nothing to this discussion.”
Concentrate on what I’m talking about – not on how this is all relevant to the planet. We are off in la la land.
Asmilwho says:
February 16, 2014 at 2:17 am
Thanks for that link, Asmilwho. There is a famous line, “Science is the belief in the ignorance of the experts”, which I’d say applies to that explanation. Their graphic is below:
For the point “B” that is on the opposite side of the tidal creating body (the moon in their example), they say:
I disagree with their explanation entirely, because it claims that the tidal bulges are caused by centrifugal force exceeding the gravitational force.
To the contrary, as I show in Figure 2 above, the bulges are the result of differential gravitational attraction which simply “stretches” the planet-ocean system into more of a cigar shape (ellipsoid). Note that the same thing happens to the solid planet. It too is stretched by the tidal force into more of a cigar shape.
And since the bulges appear in the non-rotating system in free-fall shown in Figure 2, that means that centrifugal force cannot be the explanation for the bulges … so your citation is wrong. The explanation of the tidal bulges on both sides of the Earth is not centrifugal force. It is differential gravitational force.
There is a correct explanation of tides in the head post. If you think not, please tell me exactly where you think I’m wrong.
w.
Willis Eschenbach says:
February 15, 2014 at 11:30 pm
“And at the poles, that difference is zero, so the tidal force is zero.”
If, and only if, the Earth’s rotational axis and orbit were not tilted to the line of both of the items in question, the Moon and the Sun.
If everything was vertical, then things would be trivially simple and not nearly as complicate as they really are.
I have given this to Willis before
“And this for explaining why it is not due to centrifugal forces :-)”
http://www.lhup.edu/~dsimanek/scenario/tides.htm
And this provided by Greg, on the other thread about tides, about how the heat (missing or otherwise) dances to a tidal pattern down deep in the oceans when no-one is looking 🙂
https://sites.google.com/site/climateadj/argo-animations”
EDIT – damn ” got in there by mistake.
And this provided by Greg, on the other thread about tides, about how the heat (missing or otherwise) dances to a tidal pattern down deep in the oceans when no-one is looking 🙂
https://sites.google.com/site/climateadj/argo-animations
Add to that the earth moon barycenter is only average 1707 km below earth surface. Compare with earth radius of 6,371 km.
Coldlynx says:
February 16, 2014 at 4:03 am
“Add to that the earth moon barycenter is only average 1707 km below earth surface. Compare with earth radius of 6,371 km.”
You do have to keep a sense of proportion here though when talking about the barycentre. Sure the whole thing does revolve around that point. But at a rate that is 28 times slower than it is running round the Earth central spin axis.
So the effect is spread out over that much faster rate and the higher rate centripetal force is correspondingly that much higher as well. 28 times as much in fact.
clivebest says:
February 16, 2014 at 12:57 am
“Yes you’re right
The correct equation is more complicated. Taking x as the direction from the centre of the earth to the moon and taking y the vertical axis then the net tidal force has 2 components.
Fx = Gm( \frac{ \cos \phi}{R^2 + r^2 – 2rR \cos \theta} – \frac{1}{r^2})
Fy = \frac{ -Gm \sin \phi}{R^2 + r^2 – 2rR \cos \theta}
where
\cos \phi = \frac{r-R \cos \theta}{\sqrt{R^2 + r^2 – 2rR \cos \theta}}
and
\sin \phi = \frac{R \sin \theta}{\sqrt{R^2 + r^2 – 2rR \cos \theta}}
The tractional force acting along the surface and therefore unaffected by the earth’s gravity is
Fy cos(theta) + Fx sin(theta) . The tractional force is zero at the central bulge and equal and opposite from either side.”
There are three components to the tidal vector space as oriented to the Earth’s surface at any single point on that surface.
There is x, a vertical up and down vector that varies from 1 day to 1 year depending on Latitude.
There is y, a North/South vector that varies likewise
and z, an East-West vector that also varies likewise.
Now that will turn into a 3 channel, 3 colour normalised Mollweide projection movie of how this all varies over the 4 * 18.6 year solar/lunar cycle.
Any one with movie making skills out there?
RichardLH says:
February 16, 2014 at 3:16 am
I fear I don’t understand what you mean by “a delta on a larger force”. Usually, a “delta” means a change in something, that is to say the value at time 2 minus the value at time 1. Obviously, you are using “delta” in some non-conventional way.
And once again, you haven’t quoted what you disagree with, which makes it hard.
It seems that what you are calling a “tractional force” is the horizontal component of the tidal force. It is not a separate force. A force can be divided into a vertical and a horizontal components, but this is a vectorial representation, not separate forces. The horizontal and vertical components of the tidal force are real, but they are not separate forces.
At any instant, at two locations on the earth (points A and B in the graphic above), the tidal force is pointed directly vertically. Everywhere else, the tidal force is pulling at some angle to the surface. Using vectors we can divide this angled tidal force into a vertical and a horizontal component.
Very near the spots A and B in the graphic above, the tidal force is almost directly vertical, and the horizontal forces are smallest. Now given that, you’d think that near the north and south poles (top and bottom of the earth in the above diagram), the horizontal component of the angled tidal forces would be the largest. And you’d be right.
The problem is that as you move towards the poles, the horizontal forces indeed are a larger and larger component of the tidal force … but at the same time the tidal forces themselves become smaller and smaller as you move towards the poles.
This means that the horizontal component of the tidal force is strongest at 45°N and S. At that point, the horizontal component of the tidal force is half the total tidal force.
Now, of course on the real planet it is more complex, because the earth’s axis is tilted, and because the moon wanders above and below the ecliptic, and in and out as well, plus there’s the sun as well …
But none of that changes the fundamental relationships.
w.
Willis Eschenbach says:
February 16, 2014 at 4:42 am
“RichardLH says:
February 16, 2014 at 3:16 am
The important thing here is that this tractional force is not a delta on a larger force, Gravity. That is what the vertical component Willis is so keen on concentrating on is. A delta, a small difference. One capable of affecting the planet in a way that is difficult to miss but a delta none the less.
The tractional or tangential to the surface force is NOT a delta. There is nothing opposing it. If operates in its full magnitude which makes it much more interesting IMHO.
I fear I don’t understand what you mean by “a delta on a larger force”. Usually, a “delta” means a change in something, that is to say the value at time 2 minus the value at time 1. Obviously, you are using “delta” in some non-conventional way.”
No in its normal usage. That is a small relative change in a larger force. Sort of like ‘Normals’ rather than ‘Absolutes’!
EDIT: Make that No in its normal usage. That is a small relative change in a larger force. Sort of like ‘Anomalies’ rather than ‘Absolutes’!
Willis Eschenbach says:
February 16, 2014 at 4:42 am
“It seems that what you are calling a “tractional force” is the horizontal component of the tidal force. It is not a separate force. A force can be divided into a vertical and a horizontal components, but this is a vectorial representation, not separate forces. The horizontal and vertical components of the tidal force are real, but they are not separate forces.”
I was re-using the terms used by others so as to not add further complications….but you’re getting there.
“At any instant, at two locations on the earth (points A and B in the graphic above), the tidal force is pointed directly vertically. Everywhere else, the tidal force is pulling at some angle to the surface. Using vectors we can divide this angled tidal force into a vertical and a horizontal component.”
As I have been trying to point out to you for the last week or so.
“Very near the spots A and B in the graphic above, the tidal force is almost directly vertical, and the horizontal forces are smallest. Now given that, you’d think that near the north and south poles (top and bottom of the earth in the above diagram), the horizontal component of the angled tidal forces would be the largest. And you’d be right.
The problem is that as you move towards the poles, the horizontal forces indeed are a larger and larger component of the tidal force … but at the same time the tidal forces themselves become smaller and smaller as you move towards the poles.”
But now we are not adding or subtracting a small delta from a larger constant force. We are balanced on the centre of a see-saw. The slightest change around the centre will have a large effect on the position of the plank.
That is why you need to consider the 45-60 degree band as relative to the two orbital planes and its changes.
The force is not opposed by any other force.
Willis Eschenbach says:
February 16, 2014 at 4:42 am
“Very near the spots A and B in the graphic above, the tidal force is almost directly vertical, and the horizontal forces are smallest. Now given that, you’d think that near the north and south poles (top and bottom of the earth in the above diagram),”
WRONG. This is only true if the North/South Earth axis line is at the same angle as the orbital planes. That is only true for 2 days a cycle, one each for the orbital planes on Sun and Moon. That damn Saros cycle again!
http://upload.wikimedia.org/wikipedia/commons/d/d8/Field_tidal.png
http://upload.wikimedia.org/wikipedia/commons/6/61/AxialTiltObliquity.png
Willis: See
http://www.nfo.edu/tilt.jpg
But you knew that right?
Willis:
Let me try to come up with a presentation that you can see what I mean.
Take some heavy planks (like your buckets)
Align them vertically on the Earth’s surface suspended/pivoted in a gimballed/bungee cord sense so that they float just above the surface.
Assume that the Earth, Sun and Moon are orbit/spin axis wise aligned to start with.
Now watch the planks as things rotate.
At the Equator they bounce up and down on a yearly, monthly and daily pattern.
At the Poles they wave left to right (say) and bounce up/down on a yearly and monthly pattern.
At 54.7° angle to the orbital plane they only wave left/right on a yearly, monthly and daily pattern, no vertical bounce at all.
Now make the orbital planes and the spin axis how they are in reality and…..
EDIT: Damn it.
At the Poles they bounce up/down on a yearly and monthly pattern.