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.
Bart says:
February 17, 2014 at 10:27 pm
Come on Leif. Put up or shut up.
Elementary, dear Watson. The correct derivation goes like this:
a = distance to Sun; r = radius of Earth; M = solar mass; difference between gravitational force on small mass m on either side of the Earth dF = GMm/a^2((1+r/d)^(-2) – (1-r/a)^(-2)) = -4GMm r/a^3 (for r much smaller than a).
lsvalgaard says:
February 17, 2014 at 10:48 pm
Not an answer to the question posed. And, wrong. You have neglected the centripetal acceleration with respect to the Sun.
We are talking about a circular orbit here, and to make things simple, we are talking about a tidally locked body.
What is the centripetal acceleration with respect to the Sun of the sub-solar point on the near side?
What is it on the far side?
Are they the same? Why or why not?
Don’t weasel your way out of it. Answer the questions.
Don’t give me any guff that I am making the tidal bulge go away when there is no orbit, and the body is falling straight toward the Sun. The accelerations are still there, just 33% less.
Don’t make the specious claim that the body, the Earth in this case, is in free-fall. As I explained above, only the center of the Earth is in free-fall.
Now, tell us exactly why the laws of physics do not hold for Earth tides. Or, man up and admit that I am, and have been, right.
I am waiting…
Bart says:
February 17, 2014 at 10:56 pm
Not an answer to the question posed. And, wrong. You have neglected the centripetal acceleration with respect to the Sun.
Your question was ill-posed to begin with [i.e. not even wrong]. For the Earth, the ‘orbital’ acceleration is the acceleration caused by the Sun at the center of the Earth. Instead of me making my own drawing, I will refer you to Figure 2 of http://www.gps.caltech.edu/classes/ge167/file/agnew_treat_tide.pdf
(substitute Moon with Sun for the Sun case).
lsvalgaard says:
February 17, 2014 at 11:15 pm
“For the Earth, the ‘orbital’ acceleration is the acceleration caused by the Sun at the center of the Earth.”
LOL. So, it is the same at the sub-solar point and its antipode? Please explain to us how the orbital tracks around the Sun at D-r and D+r radii are precisely the same circumference. This should be good…
Really, Leif. When you find yourself in a hole, stop digging. Man up, and admit you are wrong.
Bart says:
February 17, 2014 at 11:27 pm
LOL. So, it is the same at the sub-solar point and its antipode? Please explain to us how the orbital tracks around the Sun at D-r and D+r radii are precisely the same circumference.
The easiest would be for you to actually read from the paper . Start with section 3.06.2.1 and learn about the tidal forces.
Anticipating a lame response, in so many words:
“But, gravity is magical. It cancels the centripetal acceleration at both D-r and D+r, as well as at D.”
No. If it did, then the point at D-r would have to be moving at orbital rate omega = sqrt(mu/(D-r)^3), and the point at D+r would have to be moving at orbital rate omega = sqrt(mu/(D+r)^3). And, if that happened, the Earth would quickly tear itself apart.
The Earth center is pulling everything along at omega = sqrt(mu/D^3). The point at D-r is fighting being pulled along at that rate, both by the gravity at D-r which is stronger, and the centripetal acceleration at D-r, which is weaker. Mutatis mutandis at D+r.
Both effects influence the tides.
lsvalgaard says:
February 17, 2014 at 11:31 pm
“The easiest would be for you to actually read from the paper.”
No, the easiest thing for you is to defend your point of view, or admit that you do not understand the math, and must allow others to make your point for you.
But, then, that would also be an admission that you do not understand what they have written, and are taking it on faith that A) they did it right and B) you are sure they agree with you.
Bart says:
February 17, 2014 at 11:38 pm
No, the easiest thing for you is to defend your point of view
My point of view is based on what generations of scientists have pondered for centuries. Our modern treatment of tidal theory is based on the work by Munk and Cartwright (1966) and I happen to agree with them (understanding what they said): http://www.leif.org/EOS/Munk-Tides-1966.pdf You should study their work and learn something. Especially their derivation of the tidal potential starting on page 538, equation (2.4). Tell us where they go wrong in your opinion.
lsvalgaard says:
February 17, 2014 at 11:49 pm
If you cannot defend your POV, then you do not understand it.
Bart says:
February 17, 2014 at 11:53 pm
If you cannot defend your POV, then you do not understand it.
Nonsense. I think Maxwell’s equations, Newton’s law, etc are valid and that I understand them and can use them. I do not have to defend what they said beyond what they themselves did. But, in any event, a physical understanding can be got in the tide-problem by remarking that the equations are the same whether you assume the Earth going around the Sun or the Sun going around the Earth [in an otherwise empty Universe].
lsvalgaard says:
February 18, 2014 at 12:02 am
Nonsense. You haven’t the faintest idea of what to do with the centripetal acceleration terms.
Here’s a clue for you Leif: the effect is to amplify the accelerations by a factor of 3/2. They retain the same form otherwise, so can even be lumped in as scaling of the potential. And, since the papers you cite are looking for functions proportional to the equipotential surface, they have absolutely nothing to say on this score.
I showed you how the factor of 3/2 emerges in the calculation of the Roche Limit at the end of this section. Look at the equations:
See how the factor of 2 becomes a factor of 3? It’s right there.
You’re wrong, and you’re only arguing with me because we don’t like each other. Give it up. You’re hurting yourself more than me.
Bart says:
February 18, 2014 at 12:13 am
I showed you how the factor of 3/2 emerges in the calculation of the Roche Limit…
That was the rotation of the body itself, not orbital motion.Read Appendix A of http://www.leif.org/EOS/Munk-Tides-1966.pdf and learn.
You’re wrong, and you’re only arguing with me because we don’t like each other
That may be your motivation, mine is to educate you.
Some interesting facts about tides (the short version)
The Earth is an oblate spheriod.
The Earth’s equatorial radius is the distance from its center to the Equator and equals 6,378.14 kilometers
The Earth’s polar radius is the distance from its center to the North and South Poles and equals 6,356.75 kilometers
A 21.39 kilometers difference.
Angular velocity of Earth’s axial rotation in radians per second = 7.29 x 10^-5 rad/s
Angular velocity of Earth around the Sun in radians per second = 1.99 × 10^-7 rad/s
Centripetal (outwards) acceleration at Earth’s surface due to Earth’s axial rotation = 0.034 m/s^2
Centripetal (outwards) acceleration at Earth’s surface due to Earth’s orbit around the Sun ~ 0.0000952 m/s^2
Gravity at Earth’s surface = 9.8 m/s^2
The tiny differential between 9.8 m/s^2 (Polar) and 9.834 m/s^2 (Equatorial) creates the 21.38 kilometers difference in radii above.
The Oceans average depth is 3.79 kilometers so 3.79 / (6,378.14 – 3.79) =~ 1 / 1682 so the difference is mostly in the rock.
The Solar tidal force is 46% as large as the Lunar. More precisely, the Lunar tidal acceleration (along the Moon–Earth axis, at the Earth’s surface) is about 1.1 * 10^-7 * 9.8 m/s^2,
while the Solar tidal acceleration (along the Sun–Earth axis, at the Earth’s surface) is about 0.52 * 10^-7 * 9.8 m/s^2.
The theoretical amplitude of Oceanic tides caused by the moon is about 54 centimetres at the highest point, which corresponds to the amplitude that would be reached if the Ocean possessed a uniform depth, there were no landmasses, and the Earth were rotating in step with the Moon’s orbit. The Sun similarly causes tides, of which the theoretical amplitude is about 25 centimetres (46% of that of the Moon) with a cycle time of 12 hours. At Spring tide the two effects add to each other to a theoretical level of 79 centimetres, while at Neap tide the theoretical level is reduced to 29 centimetres. Since the orbits of the Earth about the Sun, and the Moon about the Earth, are elliptical, tidal amplitudes change somewhat as a result of the varying Earth–Sun and Earth–Moon distances. This causes a variation in the tidal force and theoretical amplitude of about ±18% for the moon and ±5% for the sun. If both the Sun and Moon were at their closest positions and aligned at New Moon, the theoretical amplitude would reach 93 centimetres.
Real amplitudes differ considerably, not only because of depth variations and continental obstacles, but also because wave propagation across the ocean has a natural period of the same order of magnitude as the rotation period: if there were no land masses, it would take about 30 hours for a long wavelength surface wave to propagate along the equator halfway around the Earth (by comparison, the Earth’s lithosphere has a natural period of about 57 minutes). Earth tides, which raise and lower the bottom of the ocean by less than 1 metre,and the tide’s own gravitational self attraction are both significant and further complicate the ocean’s response to tidal forces.
In most locations, the four largest amplitude tidal components turn out to be:
M2 Principal lunar 12.42 hr
K1 Luni-solar diurnal 23.93 hr
S2 Principal solar 12.00 hr
O1 Principal lunar diurnal 25.82 hr
S2 is largest at mid-latitudes and vanishes at the Equator and the Poles.
M2 is largest at the Equator and vanishes at the Poles.
The Long-period tide (not listed above) is largest at the pole and (with reversed sign) at the equator.
A list of other components can be found in Knauss (1978) table 10.1. Particularly important is the fortnightly (2 week) tide, often written Mf.
See Figure 10.15 in Knauss (1978) for plots of partial tides.
Tides in different locations are classified based on the predominant frequency of the tide using a function called the form ratio which measures the relative strength of the diurnal and semi-diurnal tides.
F = (K1 + O1) / (M2 + S2)
F > 3 Diurnal 1 High, 1 Low per day
0.25 < F < 3 Mixed 2 Highs, 2 Lows per day, but of different strength
F < 0.25 Semidiurnal 2 Highs, 2 Lows per day, similar strength.
Bart says:
February 17, 2014 at 9:48 pm
“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.”
Because it you take the Earth’s rotation value you can add and subtract the Earth orbital value to get the difference Sun side to Dark side?
0.034 m/s^2 ± ~0.0000952 m/s^2
which is what I thought you were discussing.
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. Here it is:
http://www.gps.caltech.edu/classes/ge167/file/agnew_treat_tide.pdf”
Even more frustrating is that the shortcut you have and the shortcut I have are apparently the same (copied to Notepad) but yours works and mine doesn’t!
lsvalgaard says:
February 18, 2014 at 12:18 am
“That was the rotation of the body itself, not orbital motion.”
The body is orbiting at synchronous rate, i.e., with the orbital motion. The effect can thus be attributed either way – to rotation or to orbital motion. If you really want to separate the two, you need to tilt the axis of rotation away from the orbit normal.
In any case, the fact remains that the differential acceleration from the sub-solar point to the antipode is 6*(mu/D^3)*r, and not 4 times. This clearly has an effect on the Roche limit, and my equations are correct.
Bart says:
February 17, 2014 at 9:48 pm
Or were you talking about
Angular velocity of Earth’s axial rotation in radians per second = 7.29 x 10^-5 rad/s
Angular velocity of Earth around the Sun in radians per second = 1.99 × 10^-7 rad/s
and
Earth’s average orbit of 149.59787 * 10^6 kilometres
and Earth’s equatorial radius of 6,378.14 kilometres
Having read the links provided it seems to me that we are dealing with tidal predictions now which is not the same thing as asking why there are tides at any given point in time on a specific place on the earth.The word potential I think would best describe the effect of all the forces acting on the earth,moon and the sun etc. which determines there exact position and tidal forces in the future rather than imaginary.My simple understanding is.
http://en.wikipedia.org/wiki/Tide
Bart says:
February 18, 2014 at 3:07 am
The body is orbiting at synchronous rate, i.e., with the orbital motion.
Is an irrelevant straw man, only used to give an example of a possible rotation rate.
don penman says:
February 18, 2014 at 3:42 am
“My simple understanding is. http://en.wikipedia.org/wiki/Tide”
Totally correct. The rather limited view that a single vector treatment of the rather more complex true nature of tides can be seen from the M2 tidal plot in that link alone. That shows just how complex this all is in reality. Even if you understand correctly the top 4 ways in which the field changes, the way the Earth reacts to them is hardly recognisable from even that slightly more complex viewpoint.
Mike M,
“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.”
You have to keep in mind that the centripetal and centrifugal forces that you refer to do not exist together but are used to describe the situation in different reference frames. In the inertial frame the centripetal force disappears when you lose contact with the door as the reaction force of the door pushing on you is gone and you then travel in a straight line as per Newton’s first law. In the non-inertial (rotating) frame the centrifugal force does not disappear as it is not caused by the reaction force of the door but was balanced by it, thus producing the static situation as seen inside the car. Remove the door and the centrifugal force then causes the person to leave the car. The statements that the centrifugal force pushed you out of the car versus the car slid away from you are not contradictory. They are just the different views of the two reference frames.
I am not sure what you mean by “inertial reaction”. You clearly wish to view things from the inertial frame and there is nothing wrong with this but at the same time you wish to retain the term centrifugal. This is a problem. Non-inertial frames are used because in many situations they greatly simplify the mathematics and make it easier to interpret results. Try to explain what is going on when a bug crawls around on a horizontal non-uniformly rotating disk without them. For that matter try explaining the motion of the earth’s atmosphere or oceans on a global scale in an inertial frame of reference. You have to get the details right.
Bart says: February 17, 2014 at 9:48 pm ” 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.”
This is exactly why I stole earth’s rotation away from ‘them’ with my free body model above (February 17, 2014 at 5:25 am). It allowed me to demostrate that an additional force is needed to keep a particle on earth’s orbital path IF it is going slower than earth’s free orbital speed. (One of ‘them’ actually stated that staying on earth’s orbital radius at some other speed than earth’s free orbital speed was “impossible” – which is pure BUNK!) In this slower case, that ~other~ force must act away from the sun to counter sun’s gravity to thereby decrease the centripetal force down to the reduced amount needed to keep the particle traveling on earth’s free orbital path at the slower speed. If the velocity dropped to zero, that amount of extra thrust needed would become EXACTLY – the force of sun’s gravity on the particle and the particle would simply hover above the sun at a altitude from the sun equal to earth’s orbital radius.
And, for any instant in time, it could be a long rope to Jupiter, it could be a rocket thruster or it could be … a weenie bit taken away from the inertial reaction to earth’s gravity – which would indeed register on a very sensitive bathroom scale or accelerometer.
Conversely, when a particle is going faster than free orbital speed, again some ~other~ force, but this time acting toward the sun, is required to keep it on earth’s free orbital path. It could go 10, 100, 1000 times earth’s velocity and yet stay on earth’s orbital path, it would just require a lot more of thrust to produce the required centripetal force over that provided by sun’s gravity.
Bart, you and I are of course speaking of exactly the same effect, one with net force differentials acting in the same direction as Willis’ effect but, (as I believe to be), a ~couple orders of magnitude stronger than his.
I suspect that we are getting nowhere here because we are bucking … the consensus.
(It’s why my eyes lit up when I saw denniswingo’s comment, February 17, 2014 at 10:03 pm, because there may empirical evidence to back us up. Hope he returns with more details… )
Ohh, how much simpler it would be to explain the gravitational effects with a series of simple diagrams. It could be relatively easy to explain how an orbit of one body around another could come to be. Use of only gravitational force and inertia.
A picture is worth a thousands words.
For those having a hard time with this discussion on this thread, get a few big rocks and go off into deep space and play around with them (in your mind only of course). From Willlis’ simple diagram you can find the source/direction of gravitational forces far far away.
I hope to find time to put it all in words but the mortgage company is not interested my educational endeavors.
William Sears says: “I am not sure what you mean by “inertial reaction”.
We’ve already been over this. Gravity is pulling you down and the inertial reactive force is the ground pushing you up – equal and opposite. Take the latter one away and down you go!
How about I construct a trap door over a vertical mine shaft, have you stand on it and then you scream to me about non-inertial reference frames after I pull the lever?
If you are on rusty roller skates and I’m pushing 50 pounds against you to keep you moving at 2 feet/sec I feel that same 50 pounds, as a reaction, pushing BACK on me thus I am doing REAL work = F*V REAL heat comes from your rusty roller skate bearings in ANY frame of reference.
Mike M:
Earth’s average orbit = 149,597,870.00 kilometres
Earth’s equatorial radius = 6,378.14 kilometres
Remind me again about how much difference this makes as to the forces involved.