Canute Ponders The Tides

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.

tidal force on three bodiesFigure 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:

tidal force is stretchingFigure 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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February 14, 2014 6:43 pm

millinewton (mN, or 10-6 newtons
1-^(-3) newton. ‘Milli’ means 1/1000 th

February 14, 2014 6:57 pm

Figure 2 and it’s accompanying explanation seem to imply that the major axis of the simplified ellipse of the ocean (in cross section), is always pointed towards the sun. But the major axis (of the simplified cross sectional view of the oceans) should surely be oriented towards the moon, which is why the tides change from day to day.

February 14, 2014 6:58 pm

Those Greek letters can be a real pain! Teachers may have a question with micro newtons, but micro is represented by the Greek letter mu, however if the symbol font is not available, then mu becomes m and the answer is off by a factor of 1000 and students get confused.

earwig42
February 14, 2014 6:59 pm

Willis, I learn something from each of your posts.
Thank you!

William Sears
February 14, 2014 7:06 pm

Willis, not sure about your numbers but also you seem to be calculating a force per unit mass (N/kg) and not a force in newtons. The tidal force (per unit mass) formula should also have D-cubed and not D-squared.

February 14, 2014 7:11 pm

Nicely done Willis, a very clear explanation. Thank you.

February 14, 2014 7:16 pm

Werner – alt+230 = mu = µ
I also like alt+0176 = °, alt+0153 = ™, alt+0177 = ± alt+0174 = ®

Baa Humbug
February 14, 2014 7:41 pm

For tidal forces on the Earth, 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 M2 at the points nearest and furthest from the sun

Possible typo, I believe the farthest unit mass is M3.
[Thanks, fixed. -w]

Richard D
February 14, 2014 7:55 pm

For me the math is not fine with paper, pen, calculator…..but entering on the compute really sucks. Thanks Willis.

noaaprogrammer
February 14, 2014 8:01 pm

With 3 bodies in play here, (earth, moon, and sun), are there not more than two bulges on opposite sides of the earth? In other words, if the earth was an all-water sphere with no land, what would be the length of the radius as measured from the earth’s center to the surface of the water as a function of its angles in a spherical coordinate system? With no continents to set any boundaries, it seems to me that it would be possible for more than one mode of a standing wave (bulge).

eyesonu
February 14, 2014 8:11 pm

Willis,
I must admit that I have never given much thought about the tides. I knew the sun was a factor and thought the moon was the other important factor but never really thought much deeper.
From your explanation, could the bulge opposite the sun be explained as a resultant force created by the centripetal force resulting from the earth’s orbit around the sun?
You’ve got me thinking and now I can’t rest until I fully understand the tides. Ohh, that need to know thingy.

William Sears
February 14, 2014 8:16 pm

Yes Willis, this would be true except for the fact that you do not have an m1 and an m2 in your formulae. You have only the sunmass with no earthmass. Also the numbers that you have calculated appear to have used only one mass, that of the sun.

William Sears
February 14, 2014 8:30 pm

Willis,
Also in your figure 1 caption you say “Solar tidal forces on a 1-kg mass at the earth’s surface average about half a micronewton (µN, or 10-6 newtons)” which means that you actually agree with me and were just a little sloppy elsewhere. Although I don’t think this is quite the right way of saying this as the tidal force that you have calculated is applied over the radius of the earth (per unit mass) and is not the force on a kilogram mass laying on its surface, tidal or otherwise. I’m in a pedantic mood this evening but will soon fall asleep. I think that I’m in a later time zone than you are.

eyesonu
February 14, 2014 8:33 pm

OK Willis, now my computer has refreshed after making the above comment and I see what you are expressing in a free falling earth. But even if outside the scope of your discussion, for my need to know, in reality would the bulge opposite the sun be from centripal force?

Barry L.
February 14, 2014 8:33 pm

Willis,
Love reading your stuff.
Ever about the tides and if they could induce some kind of spherical harmonics within the ocean?
http://en.wikipedia.org/wiki/File:Spherical_harmonics.png

Steve Reddish
February 14, 2014 8:33 pm

Thanks for presenting a clear explanation of tidal forces and thus why there is a tidal bulge on the backside of the Earth for both the Sun and the Moon, Willis.
I would like to add 2 points:
1. Atmospheric contraction and expansion contribute to its tidal bulges, but not so for the oceans. Sea tides rise and fall solely due to the lateral flow of sea water.
2. While the tidal bulge on the “backside” of the Earth is due to the Earth being pulled toward the Sun and the Moon, and NOT due to any force directly acting on the seas, to an observer it appears as though the waters are flowing away from the Sun and the Moon. It is the same type of effect that makes the Sun appear to rise or set when it is actually the horizon that is moving. Just as it is normal to speak of the Sun rising, it is normal to speak of the seas flowing towards the backside bulge.
SR

February 14, 2014 8:36 pm

Thanks Willis, very good explanation, but I think this one is also very clear:
Ocean Tides:
The tides that we see in the oceans are due to the pull of the Moon and the Sun. The simplest explanation is that the water on the side of the Earth closest to the Moon is pulled, by the Moon’s gravitational force, more strongly than is the bulk of the Earth, whereas the water on the side furthest from the Moon is pulled less strongly than the Earth. The effect is to make bulges in the water on opposite sides of the Earth. The effect of the Sun’s pull is similar, and the tides that we see are the net effect of both pulls.
When the pull from the Sun adds to that of the Moon, the tides are large and we call them Spring tides, whereas when the pulls are at 90 degrees, the tides are small and we call them Neap tides. The heights of spring tides are governed by the distance of the Moon from the Earth, being largest at Perigee (when the Moon is closest to the Earth) and smallest at Apogee (when the Moon is at its furthest).
Because the Sun’s pull is aligned with that of the Moon at New Moon and Full Moon, these are the times when Spring Tides occur. The pull of the Sun is less than half that of the Moon, and so the frequency of the tides is determined by the apparent passage of the Moon around the Earth, which takes just over a day. We, therefore, in most places on the Earth have two tides a day, with the time of each becoming later from one day to the next by just under an hour a day. (The actual period is, of course, determined by the rotation of the Earth and the orbit of the Moon).
Produced by the Information Services Department of the Royal Greenwich Observatory (1996).
From http://www.oarval.org/tides.htm

Angech
February 14, 2014 8:42 pm

Problems
One the earth is not falling into the sun it is semi steady in orbit around it.
Therefore the gravity would be attracting each piece of the earth and sea at the right amount to keep it rotating around the sun. Hence nothing should be moving in respect to anything else.
Imagine 4 balls going around in orbit next to each other. There is no reason for them to move out of their orbits relative to each other.
Hence how do tides form?
Know I am wrong but look forward to explanation .
Second the moon is actually the greater attractor than the sun by a small percentage (WIki) and both together give king tides and both apart 45 degrees give neap (low) tides.
Not that this detracts from your explanation of tides, just rounding out the edges.
Third is the rotation of the earth important in tide formation, not mentioned here.

William Sears
February 14, 2014 8:57 pm

Angech,
Consider that the four balls are next to each other but at different distances from the sun. They will have different orbital periods and thus will separate from their initial alignment. Connect them with a string as in Willis’ analogy and a tensile force is required to keep the alignment. This is the tidal force. The rotating earth will have an effect both because of the centrifugal and Coriolis forces of a non-inertial frame and the rotation of the tidal bulge. Willis has discussed the latter elsewhere.

Jerry Gustafson
February 14, 2014 9:10 pm

For a great explanation of the tides as caused by the moon on Earth watch the series The Mechanical Universe And Beyond. It is a physics course created at Cal Tech for high school or introductory physics. Look at Program 25 Kepler to Einstein. http://www.learner.org/resources/series42.html#
Great explanation and computer graphics illustrating what they are talking about. This whole series is really worth watching. It is available in the US and Canada through Annenberg Learner, I think it’s on Youtube as well.

Richard D
February 14, 2014 9:11 pm

in reality would the bulge opposite the sun be from centripal force?
++++++++++++++++++++==
Nope.

eyesonu
February 14, 2014 9:15 pm

Before I get called on it I should have said centrifugal force instead of centripetal in my comments above.
Long day, time for bed. 🙂

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