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.
Kevin Kilty says:
February 18, 2014 at 2:00 pm
We may be talking past one another here, but I don’t see that anyone on this thread denies that gravity is the “cause” (forcing) of the tides, but you, yourself, used the term “tidal effect” which suggests to me what we ultimately observe as the tides–i.e. things beyond the cause.
By ‘tidal effect’ I meant the effect derived from the static difference between gravity on the two sides of the planet. Using fictive forces for ‘ease’ of calculation is OK as long as you get the same result. The standard treatment in tide theory has no need of fictive forces, see e.g. page 537 [equation (2.4) ff) of http://www.leif.org/EOS/Munk-Tides-1966.pdf for the derivarion of the potential. It is sad that this thread has turned into a pissing contest.
lsvalgaard says:
February 18, 2014 at 2:15 pm
Actually I don’t think it is sad even if if occasionally it looks like a pissing contest, because there are things to be learned from reading the thread. For instance your reference to the paper by Munk, which I was not familiar with, is my take-away here. It has become a sort of peeve of mine that people have no trouble with dynamical terms appearing on the right-hand side of F=ma as inertial terms, but then suggest that it just wrong to put them on the left-hand side and give them funny names. There are many ways to work problems, and, as you say, what matters is that one get the correct answer.
Mike M. says:
February 18, 2014 at 1:13 pm
“Yeah… okay. It’s a circular orbit for the purpose of illustration so they can be on opposite sides of the sun for all I care.”
It that case they will exhibit the same behaviour as regards speed/radial velocity, etc. Stable orbit anyway.
RichardLH says:February 18, 2014 at 3:36 pm “It that case they will exhibit the same behaviour as regards speed/radial velocity, etc. Stable orbit anyway.”
Nope, one of them is going slower by 463 M/sec so it has to have another force acting on it, (pointing away from the sun), to keep it in a stable orbit because the sun’s gravity is providing too much centripetal acceleration for that orbital radius at that slower speed.
It’s all about the gravity, strictly in the interests of science.
SI swimsuit edition cover girl Kate Upton in Zero G (string):
http://www.today.com/entertainment/kate-upton-flips-spins-bikini-zero-gravity-sports-illustrated-2D12124620
Mike M says:
February 18, 2014 at 4:00 pm
“Nope, one of them is going slower by 463 M/sec so it has to have another force acting on it, (pointing away from the sun), to keep it in a stable orbit because the sun’s gravity is providing too much centripetal acceleration for that orbital radius at that slower speed.”
I obviously do not understand what it is you are trying to plot. Where does the 463 M/sec come from?
Can you provide a diagram marked up with what it is you are tying to show? Just a simple vector plot should do.
Mike M says:
February 18, 2014 at 4:00 pm
Try this for starters.
http://i29.photobucket.com/albums/c274/richardlinsleyhood/EarthSun_zps14b35168.png
So where are your two particles?
denniswingo – I forgot to ask, at what latitude are the accelerometer measuring that?
It was at 34.85 degrees N 86.99 W (Huntsville Alabama)
I have my experimental hardware back and will be hooking it up in a few weeks. I have been able to “watch” the Moon go overhead by watching the output of our accelerometers change (sensitive to ~+/-1 micro gee).
I used to have all of the numbers in my head on these things but I could be off. Will play with my hardware soon.
Personally I don’t think of these things as “tides” in the way that Willis does. We look at the equipotential surfaces and that the water bulges because it is slightly farther “uphill” in the gravity well. That is just us physicists talking though…..
denniswingo says:
February 18, 2014 at 5:53 pm
Personally I don’t think of these things as “tides” in the way that Willis does. We look at the equipotential surfaces and that the water bulges because it is slightly farther “uphill” in the gravity well. That is just us physicists talking though…..
—
Now that’s a great way to look at it and is more the way I view it also, not only speaking of just the bulge but the tangential acceleration component that starts the water to flow which then builds momentum and inertia that when obstructions, like shores or opposite flows, are encountered can cause the actual time of high tide locally to either precede or lag behind the moon’s exact passage directly overhead even also adjusted for the sun’s influence, due to that momentum that may have already reversed to outflow, depending on the actual topography that disperses that built up momentum, or that the view I have held for a long, long time. It all depends on the slosh and how steep and deep horizontally the obstructions are that counter the flow. Sorry for the running sentence.
lsvalgaard says:
February 18, 2014 at 9:59 am
“Under the assumption that the rotation rate is that of a tidally locked moon [not a bad assumption], the rotational role of the breakup is to raise the factor from 2 to 3.”
I will take that as concession that the differential acceleration factor is, indeed, 6 and not 4, as I have been trying to explain to you.
” If it does not rotate at all, then only the tidal effect plays a role.”
As we are dealing with “fictitious” forces, the frames of reference play a key role, and we must define them carefully. In the particular case of a tidal locked orbit, the force can be looked upon as due to centripetal acceleration in one frame, and pure spin in another. They are entirely equivalent in terms of their ultimate effect.
What you call not rotating at all, I can as easily call the effect of angular momentum being cancelled by spin in the rotating frame.
“This has nothing to do with the orbital centripetal acceleration in causing tides, and is thus irrelevant.”
There you make the elementary mistake of assuming that first order necessary conditions for an equilibrium are also sufficient. Although uniform expansion is an equilibrium solution of the dynamic equations, it is an unstable one. Failure to take into account second order sufficient conditions for stability has played a role in many engineering disasters. A prominent one, often used as a cautionary tale for aspiring engineers, is the failure of the Explorer 1 satellite, the first artificial satellite launched by the United States.
In space, bodies always seek their minimum energy configuration. And, that minimum energy configuration is with the steady state minor axis of inertia pointing towards the gravitating body, and spinning about the major axis, i.e., a fluid body will always end up bulging along the +/- radial direction in steady state. And, part of that bulge will be due to the centripetal acceleration which, as you have conceded, produces 3/2 larger differential stress due to the centripetal acceleration.
denniswingo says:
February 18, 2014 at 5:53 pm
“Personally I don’t think of these things as “tides” in the way that Willis does. We look at the equipotential surfaces and that the water bulges because it is slightly farther “uphill” in the gravity well. That is just us physicists talking though…..”
They are just trying to break that varying equipotential surface into its sine wave components – now where have I heard of that approach before 🙂
Bart says:
February 19, 2014 at 1:46 am
I will take that as concession that the differential acceleration factor is, indeed, 6 and not 4, as I have been trying to explain to you.
No, this is irrelevant as I have been trying to explain to you. If a body is rotating fast enough, the rotation in itself will cause a bulge which has nothing to do with tides.
As we are dealing with “fictitious” forces, the frames of reference play a key role, and we must define them carefully. In the particular case of a tidal locked orbit
It is not the orbit that is tidally locked, but the rotation of the body that is locked. Here you show your confusion.
And, part of that bulge will be due to the centripetal acceleration which, as you have conceded, produces 3/2 larger differential stress due to the centripetal acceleration.
You are still confused.
lsvalgaard says:
February 19, 2014 at 3:59 am
Bart
“And, part of that bulge will be due to the centripetal acceleration which, as you have conceded, produces 3/2 larger differential stress due to the centripetal acceleration.”
Leif
“You are still confused.”
So confused so as not to realise that most of the centripetal acceleration shows up in the rock, not the water anyway.
RichardLH says:
February 19, 2014 at 4:18 am
“And, part of that bulge will be due to the centripetal acceleration which, as you have conceded, produces 3/2 larger differential stress due to the centripetal acceleration.”
The Roche Limit is how close a body can get to another [larger one] without breaking up. If the body rotates fast enough it will break up regardless of any tidal effects. In the general case, the breakup is determined by both fast rotation and tidal forces working together, having nothing to do with any orbital accelerations. I can only encourage bart to study the Munk reference I gave him to learn about how to calculate the shape of the ‘gravity well’ Dennis referred to.
lsvalgaard says:
February 19, 2014 at 4:31 am
“The Roche Limit is how close a body can get to another [larger one] without breaking up”
I do know what the Roche limit is.
I was just pointing out that most people seem to be asserting that centripetal acceleration affects the oceans, whereas it mostly affects the rock on which the oceans are but a very thin skin.
RichardLH says:
February 19, 2014 at 4:36 am
I was just pointing out that most people seem to be asserting that centripetal acceleration affects the oceans, whereas it mostly affects the rock on which the oceans are but a very thin skin.
Of course you know, but bart does not, so a bit of explanation for his benefit. In addition, the bodies for which the Roche Limit is important have no oceans, being just piles of rock [which may be made of ice]. Comets are such bodies and are not ‘tidally locked’.
lsvalgaard says:
February 19, 2014 at 4:47 am
“In addition, the bodies for which the Roche Limit is important have no oceans,”
The chance of a body still having an ocean by the time it reaches the Roche limit is likely to very small 🙂
I suppose if one were to posit a very elliptical orbit you might be able to get one with the other though I suspect it would have boiled away long before closest approach!
I grabbed my old college textbook on classical mechanics on the way out the door (Barger and Olsson) and had a few moments to read the section on the tides before my Thermo class this morning. They derive the tidal potential by 1)switching to an inertial reference frame, 2)write a potential gotten from the centrifugal force and using the binomial theorem to write an approximation for distance from the moon to an ocean particle on Earth. This is exactly the approach I suggested doing days ago, which got me a rebuke from both you and Willis. This was 42 years back in my memory, but old habits die hard. I rest my case about the dynamics of the system being pertinent to the problem. I now intend to read Munk, who I expect follows Lamb in this regard, and I bet dimes to donuts that there is an implied use of the dynamic “ma” term in there someplace.
Kevin Kilty says:
February 19, 2014 at 12:09 pm
The problem with using centrifugal force and the oceans is that almost all of the centrifugal force goes into the rock not the water.
6,378.14 kilometres of rock (average).
3.8 kilometres of water (average)
is a very hard ratio to contest!
This has been a most interesting thread. It was very difficult to toss out centrifugal and centripetal forces from my mind. Now I can’t constrain my thoughts when thinking outside the earth, Every time I close my eyes my brain jumps in gear analyzing a planet/moon system traveling to or away from a distant gravitational force. If not that then analyzing the same planet/moon system orbiting that same distant gravitational force. And then there’s the issue with elliptical orbits. And of course changing the size relationship between the planet/moon. Yeah, of course that brings up the angle of the center line through the ellipse off from the general direction of travel of the various planet/moon systems my brain wants to play with.
It sort of looks like a big drunk and a little one trying to walk around an incline while keeping a bungee cord tight. LOL
Willis, thanks for the insight to that which I was blind.
Just wonder how the tidal forces on a space-lift would be. You know the thing described by Arthur C. Clarke; a lift which goes from equator all the way up to the geosynchronous orbit and beyond.
Let’s for simplicity say that the chosen length is twice the distance to the geosynchronous orbit. This structure can in theory be built as free floating satellites just hovering 1 meter over the Earth’s surface and stretching 2x geosynchronous orbit into space.
The tidal effect caused by the gravity would be close to one G. It is 1 G in the lowest point, and nearly nil in the upper point.
But an astronomer in the lift would experience twice that tidal effect. In the upper point it would be one G outwards, giving a total tidal effect of 2G.
Doesn’t this imply that one also have to take the centripetal force into account?
/ Jan
Kevin Kilty says:
February 19, 2014 at 12:09 pm
l
Kevin, I re-posted this link above, and I re-post it here, for a simple reason.
It distinguishes clearly between fictive forces and real forces, and it does so using five different frames of reference. One of them is the “inertial reference frame” you speak of above.
The key insight in the question is encapsulated in his conclusion, viz (emphasis mine):
Note that his statement, that rotating around the CM doesn’t produce any tidal deformation, is not an unsupported claim. It is supported and explained by the detailed mathematical calculations that he provides for each of the five different reference frames. In every case, the result is the same … “centrifugal force” doesn’t produce tidal deformation. It just makes the calculations more complex.
w.
Willis, you, Leif, and I will have to agree to disagree on this; or maybe we are arguing against different things here. The derivation in Barger and Olsson settled the issue in my mind this morning. It was abundantly clear. By the way, I never argued that you had done this wrong–rarely do I find much to argue about in your efforts. I like your essays. I was arguing initially against what I perceived as a blanket statement that use of a repulsive force (“centrifugal”) is just plain wrong. I may have misunderstood that.
Best wishes always.
How can something so simple be so hard to explain?
I tried several times to make the simple explanation to help those here and every time it don’t come out right or at least not the way I intend.
WOW, Willis said it was a short post!
I’ve got some text docs (mime) and hope to put /combine/edit/compile/etc with regards to this discussion.
I had a difficult time with this and when it finally clicks you may say WTF was I thinking.
Think “corkscrews”.