The Moon has a four-week cycle; it is full every four weeks (actually every 29.5 days). But ocean tides exhibit a two-week cycle; they are large one week and then smaller the next.
Specifically, as in Fig. 1 below, ocean tides are stronger (“spring tides”) around New Moon and Full Moon than they are at First Quarter Moon and Last Quarter Moon (“neap tides”). The pattern is seen, roughly at least, all around the world (though the details are not simple, as they depend on the shape of the coastline and other factors.)
What’s behind this pattern? And what does it tell us about the Sun and Moon that we wouldn’t otherwise easily know? Perhaps surprisingly, it tells us that the average mass density of the Sun — its mass divided by its volume — is only a little smaller than the average mass density of the Moon. Here’s why.
Moon-Tide
The Moon creates a tide twice a day, for reasons I have explained here. This is not merely because it pulls on the Earth via gravity but because its gravitational pull on the Earth varies with distance. It pulls more strongly on the near side of the Earth, and more weakly on the far side, than it does on Earth’s center. The net effect (Fig. 2) is a tidal force which tends to distort the Earth into an oval shape.
The Earth’s rock is sturdy, and it barely responds to this tidal effect. But the Earth’s fluid oceans do respond by shifting slightly. The shift creates two small bulges in the ocean, one on the side facing the Moon and one on the other side, with dips between the bulges. When a bulge passes a coastline, a high tide results; when a dip passes, there’s a low tide.
But actually this is oversimplified, and not just because the coastlines and underwater topography affect the water’s flow and the shape of the bulges. We have forgotten the Sun.
Sun-Tide
The Sun, too, pulls more strongly on the noon-side of the Earth, and more weakly on the midnight side, than it does on Earth’s center. It, too, creates a tidal force that, on its own, would cause two bulges in the ocean. If there were no Moon, we’d still have twice daily tides, just smaller ones and always at the same time of day.
Together, the combination of the Moon’s tidal force and that of the Sun determines the locations and sizes of the two bulges in the ocean (Fig. 3). If the two forces point in the same direction, as at Full and New Moon, they squeeze the Earth in the same way, with an overall effect that is larger, creating larger bulges and dips, and thus more powerful tides. If they point at 90 degrees to one another, as at First and Last Quarter Moon, then one squeezes where the other stretches. As a result, they partially cancel, and the bulges and resulting tides are smaller.
This is where it matters that the tidal force of the Sun is neither much larger nor much smaller than that of the Moon. Let’s consider how different things could be otherwise.
If the Moon were, say, three times further away than it actually is, its tidal force would be less than 1/10th that of the Sun’s. Then the tides would have been almost the same every day, peaking around noon and midnight all year long. There would have been no significant monthly or daily changes.
If instead the Moon’s distance to Earth were a third of what it actually is, then its tidal force would be almost 50 times that of the Sun. Then the tides would be almost the same size all month, shifting by about one hour every day as the Moon orbits the Earth monthly.
(Actually, the tides would then show a subtle four-week pattern. That’s because the Moon, in its elliptical orbit of the Earth, comes slightly closer to the Earth once a month, making the tidal force larger, and reaches its furthest point from Earth two weeks later, making the tidal forces smaller. There’s some hint of this in Figure 1, where the two spring tides of the month aren’t the same size.)
But in fact the Moon’s tidal force is only about 2.5 times larger than that of the Sun, and so both objects are important in the tides. When their tidal forces align every two weeks, at Full Moon and New Moon, the overall tidal force is considerably stronger, while it is considerably weaker when they are 90 degrees apart, as seen in Fig. 2. The result is that the tides grow and shrink every two weeks, all while their timing shifts by about one hour every day.
What the Tides Teach Us
Thus the two-week pattern of the tides is remarkable in that it only occurs because of a special relationship between the Moon and the Sun, one which is not obvious. As I’ll show at the end of this post using simple math, if there are two spherical objects A and B, then
- the strength of the tidal force of object B on object A is proportional to
- the angular size of B in the sky of A, and
- the average mass density of B (its mass divided by its volume.)
This is true both for the Moon’s tidal force on Earth and for the Sun’s.
As we see from monthly tidal cycles, such as that in Fig. 1, the Sun’s tidal force is roughly half that of the Moon. More precisely, it’s about 40% of the Moon’s tidal force. But we can all see that in Earth’s sky, the angular sizes of the Moon and Sun are about the same; that’s why the Moon can perfectly but briefly cover the Sun during total solar eclipses. The Sun is about 400 times further away than the Moon, but also its diameter (more precisely, the diameter of its opaque region) is about 400 times larger than the Moon’s, and as a result, their angular sizes appear the same.
The difference between the tidal forces can then only come from a difference in their mass densities. In particular, we can conclude that
- the Sun’s average mass density is about 40% that of the Moon.
(Meanwhile the Earth’s density is a bit larger than that of the Moon, and about four times larger than that of the Sun.)
It’s not surprising that the Moon and Earth have roughly similar densities; both are made of rock. But the Sun is made of a completely different material, a gas (really, an “electrically-charged plasma”) of electrons and atomic nuclei. It’s far from obvious that it should have even a vaguely similar density. Nevertheless, it does. We can return to the reasons for this on another day.
A Side Comment on Other Methods
I should point out that this is not the only simple way to learn about the relative mass densities of the Sun and Moon, although it is the simplest. Specifically,
- The Earth’s mass and density can be learned from the period of the Moon’s orbit, along with measurements of the radius of the Earth and the distance to the Moon, and Newton’s gravitational constant.
- The Moon’s density and mass can be inferred, albeit imprecisely, from the average size of Earth’s ocean tides and the Earth-Moon distance.
- The Sun’s density (but not its mass or size) can be learned (using Kepler’s third law) from the period of the Earth’s orbit around the Sun and the Sun’s apparent size in the sky.
From these results, the ratio of the Sun and Moon’s mass densities can be obtained in a second way, along with their relation to the Earth’s mass density. The fact that this approach and the tidal method give the same answer confirms that Newton’s laws of motion and gravitation are largely correct. Notice also that nothing in either of these two methods requires knowing the distance from the Earth to the Sun, which is one of the most difficult astronomical quantities to measure.
Why Tides Teach Us About Densities — Using A Little Math
Finally, for those of you who are interested in the details, here’s why the tidal effect of an object on the Earth is proportional to its angular size and to its mass density. The explanation just requires Newton’s law of gravity and some simple reasoning.
First, let’s define angular size and mass density. The angular diameter θB of an object B of diameter DB, as seen from object A at a distance LAB , is given by
If the angle θB is small then tan (θB/2) is approximately the same as θB/2, giving approximately
and so, approximately,
Meanwhile the mass density of B (usually called ρB) is its mass MB divided by its volume VB , which in terms of its diameter is
meaning that
or equivalently
The average gravitational pull of B on A, FA;B , is given by Newton’s law of gravitation, involving Newton’s gravitational constant, G:
which causes an acceleration aA;B of the object A due to object B that is equal to FA;B /MA :
But if the acceleration were constant across the object, there’d be no tidal force. The tidal force comes from the fact that B is closer to objects on the near side of A, which are separated from B only a distance LAB-½ DA , than to objects on the far side of A, which are a distance LAB+½DA from B.
Thus the acceleration due to B differs for objects on the near and far sides of A by an amount
But if LAB is much larger than DA , we can approximate the denominator by
which then simplifies the tidal effect to
Note that the tidal force of one object on another decreases as the cube (not the square) of the distance LAB between them, and so if the Moon were 3 times further away, its tidal force would be 27 times smaller than it is, while a Moon at 1/3 the distance would make a tide 27 times larger.
Then, using the boxed equations (1) and (2) above to replace MB and LAB, we find
and thus the tidal force of B on A is proportional to the angular size of B as seen from A and to the average density of B.
Therefore the tidal force of the Sun on the Earth is related to the tidal force of the Moon on the Earth by
and since, in Earth’s skies, θSun /θMoon is approximately 1,
which confirms the main claim of this post: the substantial variation in the size of the tides from week to week reflects the similar densities of the Moon and the Sun.
20 Responses
Sunday 14 January
Newcastle NSW, Australia
Dear Professor Strassler,
Sorry, I am a bit late on this,
comments are closed on your earlier tidal story.
This, in your earlier post,
is a bit misleading ;
“The air bulges too, but not only is there no easy
way for us to observe this, the Sun’s daily heating
creates a more important once-a-day bulge in
the atmosphere. ”
Atmospheric tides are a
lot easier to record (observe) than ocean tides.
A $3 barometric sensor (eg. BMP 280) and $15
Raspberry pi or Arduino works fine.
I have recorded atmospheric tides
for several months in Jakarta and near Manila.
Atmospheric Tides are semi-diurnal.
(not once-a-day).
This fact has a fascinating history,
Lord Kelvin fretted over it, and came up with
the wrong answer.
I think a future post on the
reason for semi-diurnal Atmospheric Tides
would be a good idea. It is not obvious.
Thank you for your clear
explanation of ocean tides.
Best regards
Jim Frazer
______________________________________________________
Atmospheric tides are indeed mostly solar thermal tides; the solar semi diurnal tide is about 1.2 millibars at the Earth surface. and is twice as strong as the solar diurnal thermal tides. (These tides increase with altitude.) It took decades for measurements and data analysis to improve enough to detect the very small lunar gravitational tides (the semi-diurnal tide is about 65 microbars) buried in the solar thermal noise.
Chapman and Lindzen ( (1970). Atmospheric Tides) wrote a book about atmospheric tides that is actually a joy to read. Chapter 9 concerns the history of the explanation of the thermal tides and, in particular, why the thermal semidiurnal tidal response is stronger at the surface when the diurnal thermal forcing is stronger (a question first raised by Kelvin). For some 70 years, the “resonance theory” was the leading explanation – the idea that the semi-diurnal response is stronger because it is exciting a semi-diurnal atmospheric resonance. C&L have the wonderful quote
“…Pekeris’s (1937) reexamination of the Krakatoa data demonstrates the very real dangers relating to the analysis of ambiguous and noisy data by theoreticians with vested interests in the outcome of the analysis.”
After World War II, the upper atmosphere started being explored directly (initially by captured V2 rockets) and the resonance theory died. After a bunch of theoretical and observational work, mode suppression and destructive interference turns out to be the answer to Kelvin’s question:
“it turns out that 80 percent of the diurnal forcing goes into physically trapped modes which cannot propagate disturbances forced aloft to the ground”
BTW, the thermal tides a relatively quite strong on Mars (also about a millibar on the surface, i.e., surface pressure changes of~ 15%) and Venus, where they may play an important role in changes in the Venusian length of day.
So if I understand correctly, Marshall, there are indeed twice-a-day atmospheric tides, but mostly driven by thermal rather than gravitational effects? And are you saying the diurnal atmospheric tide’s amplitude is about a millibar of pressure at Earth’s surface?
This all sounds too difficult to turn into a comprehensible post, but maybe I can figure this out someday. Right now I can’t even convincingly understand/calculate the solar tides on a spherical non-rotating planet with no Moon; do you know of any place where someone has carefully worked all out the simple cases, such as a solid planet, a liquid planet, a solid planet with a surface-covering ocean, etc.?
The semi-diurnal surface (sea level) tide has an amplitude of 1.16 millibars, while the diurnal tide has a sea-level amplitude of 0.593 millibars (in each case times sin^3 of the colatitude).
This article (I believe it is just Chapter 9 of Chapman and Lindzen) goes into the history of the thermal tides
https://ocw.mit.edu/courses/12-810-dynamics-of-the-atmosphere-spring-2008/9c695433ff2a222bdcf39ddc885f2e91_chapter_9.pdf
If you are interested in ocean tides, I would start with Lamb’s (1916) Hydrodynamics or
Adrian Gill’s Atmosphere-Ocean Dynamics. Lamb worked out ocean tides with a uniform ocean.
Solid body tides are a bit more scattered – I believe Solar System Dynamics by Murray and Dermott, would be a good start, along with Robin Thor et al. on lunar solid body tides ( DOI 10.1007/s00190-020-01455-8 ) and Bill Kaula’s 1964 review (DOI
https://doi.org/10.1029/RG002i004p00661 ). Also, for fun, you might look at “The tidal response of Ganymede and Callisto with and without liquid water oceans” (DOI
10.1016/j.icarus.2003.07.001 )’
Generally, with solid body tides you can do a good job fitting observations with Love / Shida numbers; the problems arise if you want to interpret those numbers.
Thanks, Marshall!
There is one gravitational thing that would definitely change if you replaced the current Sun by a Kerr (rotating) black hole, but only the celestial dynamics types would be likely to notice it.
The Sun has a set of gravitational spherical harmonics – not just J2 (C20), but higher zonal harmonics and also non-zonal harmonics (like C22). We know something of this from spacecraft tracking, and also from helio-seismology. These would all (except for the J2) vanish if the Sun was replaced by a black-hole, as a Kerr black hole can only support a J2 gravitational spherical harmonic.
To be more particular, if the Sun was replaced by a Kerr black hole with the current Sun’s full set of gravitational spherical harmonics somehow imposed, it would ring like a bell and radiate away the other, higher order. harmonics as gravitational waves in the next millisecond or so.
Matt is correct in stating that “far from the Sun (relative to its size), nothing changes,” but with current technology, “far” means at least 5-10 astronomical units.
Indeed, current technology is far beyond where we were a few centuries ago. But again, it’s not easy for anyone to explain what you just described to an average non-physicist. Meanwhile the methods I’ll be describing in this series of posts can be explained qualitatively to anyone, and can actually be carried out by anyone with pre-college math and one semester of introductory physics. Some of them don’t even require modern equipment, just a little bit of modern knowledge.
Thanks for this message of clarification, Jim. Some of this I did not know, and I’m glad to learn it.
When I wrote “The air bulges too, but not only is there no easy way for us to observe this,…” I meant that we can neither feel or see it with our senses (versus the visibility of ocean tides). I had not thought about how large the effect is in a barometer. You didn’t mention the amplitude of the effect — how large is it? [i.e., is it obvious over the effects of weather, or do you need to do a Fourier transform to pick it out of a complex signal? The latter would be too hard for the average person.]
Great post. Thanks.
Bill
Thanks for the wonderful explanation of another of Newton’s classics!
Do you have thoughts on the anthropic explanation for this “fine-tuning” of the Sun-Moon tidal ratio [Dynamical, biological and anthropic consequences of equal lunar and solar angular radii, Balbus (2014)]?
Stay tuned for some discussion of this — but I do not have an opinion on the matter. It’s an interesting speculation, but I don’t know how one would confirm it.
While we wait for a post discussion here, I note that it is an astrophysicist looking for a coincidence. The paper itself describes how there have been and still is ample opportunity for evolution of terrestrial locomotion, but amphibians evolved once. I’m not sure what is “anthropic” with the tidal hypothesis.
If we look for an evolutionary hypothesis it could be analogous to Darwin’s explanation why life evolved once, it filled the niche first. For modern phylogenetics the timing works out as well, both life and amphibians evolved at the first opportunity. There is just a few million years between the Devonian diversification of fish, the diversification of tetrapods, and the diversification of terrestrial locomotive tetrapods.
But there are many competing hypothesis and they are not easily amenable for testing. For example, the diversification of fish is hypothesized to be due to evolution of jaws, but the evolutionary advantage is unclear. And so on.
Evolution often erases its history, and history often erases evolution’s causes. It’s very difficult to imagine how we will ever know the answers to these types of questions, even though they are fun to think about.
Another interesting thing that depends only on the angular size in the sky and density is the observed (angular) acceleration between any two objects under gravity. For example if an astronaut held two marbles out at arms length (so they subtend a half degree of arc) they will appear to fall towards each other under gravity exactly as quickly as if two moons (assuming same density) in the sky (also half a degree of arc), starting from the same angular distance, fell toward each other under gravity.
Agreed; this is something I’m touching on in the next series of posts.
Dr.Strassler:
Another great teaching article! The “gravity gradient” (tidal force) goes as 1/r^3
Isn’t this the reason that if our sun compressed down to the size of a black hole, nothing really would happen to our orbit, as at the distance the earth is from the sun, the gradient really wouldn’t change much?
The answer is “not really”, I think. While it’s true the gradient wouldn’t change, it’s stronger than that: far from the Sun (relative to its size), nothing changes. In quasi-Newtonian language, one would say that the graviational forces outside a spherical body don’t depend on its size — which is a statement about the 1/r^2 nature of the force. In Einsteinian language, one would say that the long-distance shape of spacetime around a massive object is independent of its size. So there is a stronger and more general reason than merely the 1/r^3 tidal law that is at at work for this question.
Where the 1/r^3 law is really important is that it explains why the Moon can have a larger effect than the Sun on Earth even though one might imagine that the Sun’s larger Newtonian 1/r^2 force would somehow be more important.
Ahh, ok that makes sense. The distance, RELATIVE to the size, is a key factor. However, even from a “Newtonian argument “, if I was standing on the earth, and the earth….with its same mass, shrunk down to the diameter of say the moon, gravity would be stronger, if I was still standing on the surface of this shrunken earth. But this would be because of how CLOSE I’m to the mass…..as a consequence of the 1/r^2 law. Would that be a valid statement?
If you stand on a shrunken Earth of the same mass, then yes, the forces your body needs to exert (a force on the ground, and internal forces within your muscles) to remain standing on the surface — countering the Earth’s gravity — will increase as a result of the 1/r^2 law.
Tidal forces would also increase, by even more. But despite this, the tidal force on your body is currently so small that you’d have t0 shrink the Earth enormously before the tidal forces would be noticeable.
An active area of research here (which I am actually working on) is the tidal recession of the Moon. The ocean tides have a drag on the solid Earth from moving all that water around, and that causes the Length of Day (LOD) to slowly lengthen. We cannot see that in our modern LOD research, as it is overwhelmed by higher frequency LOD variations, but by using eclipse data from the Arabs (in Baghdad), the Chinese and Japanese together with the ancient Greeks and the Babylonians, we have a record of LOD measurements stretching back to 720 BC, and that is sufficient to see the LOD rate of 2.152 +- 0.050 msec / cy (milliseconds / century).
That means that the Earth is losing rotational angular momentum, and that angular momentum is going (mostly) into the lunar orbit, increasing its size (and slowing the Moon’s orbital speed and slowly increasing the lunar Month). Thanks to Lunar Laser Ranging measurements of the Earth-Moon distance, we can also see _that_ effect; the orbital recession of the Moon is 38.204 +- 0.090 mm / yr. We can thus compare the angular momentum gain of the Lunar orbit and the angular momentum loss of the Earth’s rotation, and (after taking into account post-glacial rebound and global warming, both of which change the Earth’s J2) we get an agreement (expressed in LOD units) of 0.208 +- 0.113 msec / cy (or a 1.82 sigma disagreement), which I think is pretty phenomenal.