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.
RichardLH says:
February 17, 2014 at 10:01 am
Richard, you address me using words that people use to address a dog … and then you act all surprised and shocked, and you clutch your pearls and bitch and whine when I stand up for myself and I don’t take your insults lying down.
You would be funny, if you weren’t so damn pathetic. Please, before you start on another of your hilarious rants about how reasonable and calm and pleasant you are, go stand in front of the mirror and look up “passive-aggressive” in the dictionary.
w.
remmitt says:
February 17, 2014 at 10:31 am
Do we actually still speak of gravity as a force then? Or is it just another imaginary force like the centrifugal force?
In science we have the concept of an ‘effective’ theory. That is a theory that gives the correct answer within its domain of validity or application. Treating gravity as a force works very well for calculating the position of the planets and guiding spacecraft to land where we want them to land, so is ‘effective’.
Willis Eschenbach says:
February 17, 2014 at 10:33 am
“Richard, you address me using words that people use to address a dog … and then you act all surprised and shocked, and you clutch your pearls and bitch and whine when I stand up for myself and I don’t take your insults lying down.”
I am sorry that I stooped to your level. It is just that you have consistently and frequently called me names and other such things.
“You would be funny, if you weren’t so damn pathetic. Please, before you start on another of your hilarious rants about how reasonable and calm and pleasant you are, go stand in front of the mirror and look up “passive-aggressive” in the dictionary.”
This from a rock(et) scientist who will plot dT on a graph and then not understand the meaning of the word delta or that ID vectors are what you use to create the x/y graph in the first place. Get a mirror for yourself.
Remmitt,
The orbital path is not a straight line in curved space time although it is the path of least action given the planet’s angular momentum and the central force it is subject to. It is sometimes suggested that gravity is a fictitious force (imaginary is not the usual word). See the following.
http://en.wikipedia.org/wiki/Fictitious_force#Gravity_as_a_fictitious_force
Willis,
I guess that I’m never so certain of myself to risk calling someone else an idiot, at least not in print. Maybe I could use a pseudonym. How does Scotian sound? In general though, I think that one should be more polite in print than face to face, not less.
I’m impressed by your boundless energy and it staggers me just to see it.
I have spent three days on trying to fully understand the discussions pertaining to this thread and I may have it figured out now but need a little time to express it. I hope those with a full understanding will monitor this thread closely another day or two.
I gotta get this issue resolved in my mind.
Willis, where have you lead me? LOL
I got it. Well I may not have “it” but I got something. Was your simple diagram meant to be the passage between a small chamber and a really really big one?
I hope to try to put into words the way a simple mind as mine sorted this out. But then I could be wrong and my mind simply way out in space. Anyway, the dogs need to be fed now.
Short post, yes. Mental focus, not so short.
Thank you, Willis for your response.
Thanks, Kevin. As you will note in Figure 1 and Figure 2, the tidal effects do not depend in any sense on centrifugal acceleration. There is no centrifugal acceleration in the Figures, the masses and the planet are falling straight into the sun … and yet we still have tidal forces.
So any explanation that depends on centrifugal acceleration is incorrect.
I don’t want to belabor the point, but I would then ask how is it that the Earth can accelerate continuously toward the sun or Moon without ever reaching them? Answer: The acceleration is centripetal, with the bodies in orbit around their common center of mass. If you pretend the Earth is an inertial system, then you are obliged to do something with the centripetal acceleration from the orbit on the “ma” side of Newton’s 2nd Law. It goes on over to the other side (among the “Fs”) and is a “centrifugal force”. Not a real force, but perfectly useful for calculations. I’m not arguing with the correctness of your model; I’m arguing that there are alternative views of the situation that aren’t necessarily wrong but actually quite useful. For example, if you use include this centrifugal psueodoforce as part of the gravitational potential, then you can explain the relative extrema in potential that are the various Lagrange points (L1 to L5) in the Earth Moon System; and the stability of those. Doing it in other ways is more complicated.
When lsvalgaard says the orbit has nothing to do with the tidal effect, I might point out that lunar cycle and lunar semi-cycle tides illustrate the pertinence of the orbit to the longer period tides. The tides are dynamical (not static equilibrium) and all the dynamics of Earth-Moon-Sun have to be accounted for.
Willis:
From this rather lengthy and detailed look at the Earth Tide which is a non-trivial part of this whole picture.
http://www.gps.caltech.edu/classes/ge167/file/agnew_treat_tide.pdf
“The Diurnal tidal potential (once per day) is largest at mid-latitudes and vanishes at the Equator and the Poles.
The Semi-diurnal part (twice per day) is largest at the Equator and vanishes at the Poles.
The Long-period tide is largest at the pole and (with reversed sign) at the equator.”
So your simple explanation only covers one small part of the overall picture, i.e. the Semi-diurnal part.
And this
http://www.whoi.edu/cms/files/lecture01_21351.pdf
for some calculations for working out the above. Including the rather important
Figure 7: Cotidal map.
Don’t know why the first link did not work – here is the full Google url for it
http://www.google.co.uk/url?sa=t&rct=j&q=&esrc=s&source=web&cd=3&cad=rja&ved=0CD4QFjAC&url=http%3A%2F%2Fwww.gps.caltech.edu%2Fclasses%2Fge167%2Ffile%2Fagnew_treat_tide.pdf&ei=L5YCU-j-C8Gv7AaJi4GgBg&usg=AFQjCNEqnt-ey0dfZpC_Zbc0Uw6Ekt86bA&sig2=H1xV2WbBW8Y_qV–gOtxAQ
Having a bad day with Google urls
Link 2.
http://www.google.co.uk/url?sa=t&rct=j&q=&esrc=s&source=web&cd=5&cad=rja&ved=0CFYQFjAE&url=http%3A%2F%2Fwww.gps.caltech.edu%2Fclasses%2Fge167%2Ffile%2Fagnew_treat_tide.pdf&ei=TpoCU_7bO8aQ7AaT1YHYCg&usg=AFQjCNEqnt-ey0dfZpC_Zbc0Uw6Ekt86bA&sig2=gTcccu9JbbBKJHEktqSASA
Kevin Kilty says:
February 17, 2014 at 2:51 pm
I don’t want to belabor the point, but I would then ask how is it that the Earth can accelerate continuously toward the sun or Moon without ever reaching them? Answer: The acceleration is centripetal, with the bodies in orbit around their common center of mass. If you pretend the Earth is an inertial system…
Perhaps the simplest explanation [without invoking centripetal acceleration] is how a satellite [or a cannonball] stays in orbit around the earth: http://www.physicsclassroom.com/mmedia/vectors/sat.cfm
RichardLH, can be frustrating can’t it. Stripped this out of your huge Google link. You have to find the actual url buried within the google’s link and use a paste to a text editor to dig for it. Here it is:
http://www.gps.caltech.edu/classes/ge167/file/agnew_treat_tide.pdf
wayne says:
February 17, 2014 at 4:46 pm
RichardLH, can be frustrating can’t it. Stripped this out of your huge Google link. You have to find the actual url buried within the google’s link and use a paste to a text editor to dig for it.
As I did for everybody in my post upthread:
lsvalgaard says:
February 16, 2014 at 10:00 pm
…A full [but lengthy] treatment is here http://www.gps.caltech.edu/classes/ge167/file/agnew_treat_tide.pdf
I see another “tide” quandry has broken out. A couple of points. (that may already be made in other comments):
There is no real centrifugal force. It is only gravity that matters to tides.
The “far side” does have a NET vector away from the planet, so it is reasonable to say the “Net Tide Raising Force” is opposite in direction on the two sides. The real gravity force is always one way though, and it is the ‘lessening with distance’ that leads to the net force having opposite directions on the vectors.
Would be a reasonable statement if they said “Net tide raising force”…
See the pictures here:
https://chiefio.wordpress.com/2014/02/16/tides-vectors-scalars-arctic-flushing-and-resonance/
Also has a nice link to a page that specifically addresses wrong ways to think about tides and bad examples from text books:
http://www.lhup.edu/~dsimanek/scenario/tides.htm
One other minor point: Since the moon orbits the sun, us not so much, it is not like other moons that neatly circle in the equatorial plane. It is tilted about 5 degrees to the ecliptic, and our equator is tilted about 23.5 degrees. So once a year the moon goes from a lunar monthly 23.5+5 = 28.5 to a 23.5 -5 = 18.5 degree excursion. That means it is bouncing “up and down”. As the lunar orbit precesses, just which season that more / less tides land in which hemisphere changes. So you really do have to allow for variations in the orbit changing where the gravity vector points, and how strong it is. It doesn’t just to around with a tidal bulge at the equator. That bulge wanders north / south too, and has a 9 and 18 year set of periods to how it aligns with the seasons / hemispheres and a 3 x that or about 56 year re-sync with exactly the same patches of dirt under it. As the reflections off of continents can be bigger than the actual tide directly raised, “this matters”. Rather a lot. It drives current formation and changes, and the degree and location of tidal mixing of deeper cool water to the surface. Periodically…
lsvalgaard says:
February 17, 2014 at 4:52 pm
As I did for everybody in my post upthread:
—
Missed that Leif, should have researched the thread to see if that embedded url already exited and that seems a paper well worth reading, thanks, pointed twice might get more to notice it if they missed your link on their pass through. Happened to me, time’s limited and can’t read every single thing.
wayne says:
February 17, 2014 at 5:17 pm
Happened to me, time’s limited and can’t read every single thing.Saying it again and again may make more people read it, so is not a bad thing. Unfortunately, geniuses like bart will not even try.
RichardLH says:
February 16, 2014 at 2:03 pm
“…or did I get something wrong?”
Who cares? Centripetal acceleration from Earth spin is the same everywhere at the surface and radially symmetric with respect to the center of the Earth. It has no net effect. Centripetal acceleration due to movement about the Sun is greater on the far side, and weaker on the near side. That is the kind of stress inducing delta-acceleration which causes tidal bulges.
lsvalgaard says:
February 16, 2014 at 10:00 pm
I’m sorry that you cannot understand the math, and must try to find references to back you up without apparently being able to argue on your own. I’ve given the very simple and obvious equations above.
Willis Eschenbach says:
February 17, 2014 at 12:25 am
“Got it … so, can we assume that your method is from the ultra-new field of investigation called “Science By Assertion”?”
What a load of crap. How many times have I posted here explaining things? I gave you the equations right up above. You havn’t done the math at all.
“The Earth is in free fall towards the Moon, attracted by the gravitational force.”
No! The Earth center is in free-fall towards the Moon. The rest of the Earth is being pulled along for the ride. Only the very center is in free fall. The fact that the rest of the Earth is being pulled differntly than the center is the very thing that causes the tidal bulges.
lsvalgaard says:
February 17, 2014 at 5:47 pm
Do the math, Leif. Can you?
………………Thanks, Leif, fixed. It’s micronewtons ………………….
Willis, are you absolutely sure that it is micronewtons? Could you express it in milligee? I think that it is around 3.1 x 10-3 g. We actually measure this with extremely sensitive accelerometers.
In Low Earth orbit the number is about 1 x 10-7 g per meter. Yes, I have measured this term and this is why the most stable orientation of the space shuttle 18.5 meters long, was nose down or nose up vertical…
Mike M says: February 17, 2014 at 6:42 am
“You ARE being pushed against the door by “centrifugal” force in the opposite direction of the centripetal force … as an inertial reaction to the centripetal force. The door is pushing against you in order to impart the same centripetal acceleration that the roadway is imparting to the tires.”
William Sears says: February 17, 2014 at 7:29 am “It is only a centrifugal force in the non-inertial (accelerated) frame that you view as reality inside the car. ”
I disagree. The reality inside the car is that when you open the door, BOTH the centripetal force that was causing you travel in a circle – AND – your inertial reaction to it, centrifugal force of you pushing against the door, … disappear.
My consternation over the use of “centrifugal force” is ONLY when it is used to describe what happens when there is no door (and no friction between you and the seat), and you go flying out of the car when it turns. Many people would say “centrifugal force” “pushed” you out of the car and that is indeed a thoroughly fictitious concept – it was the car that slid out from under you.
Bart says:
February 17, 2014 at 9:48 pm
Do the math, Leif. Can you?
Obviously, but why bother? the links I gave you to study [have you?] should be enough.
denniswingo says: February 17, 2014 at 10:03 pm “Willis, are you absolutely sure that it is micronewtons? Could you express it in milligee? I think that it is around 3.1 x 10-3 g. We actually measure this with extremely sensitive accelerometers.”
Are you saying that you measure that much of a reduction of gravity occurring (maximally) at noon and midnight? If so, could you please verify the magnitude because I calculated .0002 meter/sec^2 (.00002 g) earlier on by another effect.
lsvalgaard says:
February 17, 2014 at 10:11 pm
Come on Leif. Put up or shut up. Which of these equations is wrong, and why?
denniswingo – I forgot to ask, at what latitude are the accelerometer measuring that?