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.
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Before I get called on it I should have said centrifugal force instead of centripetal in my comments above.
========
Nope.
Its circular motion not apparent forces and inertia
Richard D says:
February 14, 2014 at 9:47 pm
Not so, Richard. In the example shown in Figure 2, there is no circular motion, and inertia is not an issue.
w.
Willis Eschenbach says: Not so, Richard. In the example shown in Figure 2, there is no circular motion, and inertia is not an issue.
++++++++++++++++++
You’re right re:figure 2, thanks……. I was looking at centripetal/centrifugal which got me off tract.
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.
+++++++++++++++++++++++++
Thanks for that Willis. Takes me back to the physics classroom.
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 .
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So, this will be hard to explain without drawing a diagram, but….
For simplicity, consider a two body system, with a fixed, Sol-centered reference frame and Earth as a moving body, however not yet in an orbit about Sol. As the Earth passes, the mutual gravitational attraction of Sol and Earth modifies the trajectory of Earth, which we call gravitational force. The gravitational force is equal to the change in momentum of earth with respect to time (F = dp/dt = d(mv)/dt = v*(dm/dt) + m*(dv/dt)). We assume both Sol and Earth are fixed mass, reducing the first term to zero and recognize dv/dt to be acceleration. Thus, the gravitational force is F = ma. Because of how the gravitational force equation is defined, the induced acceleration on Earth is towards Sol (the sign convention used in the derivations skipped in this and above). From this definition, you can see that Earth “falls” towards Sol, in the same way that an object dropped from a height “falls” to the ground.
You must remember the fact that momentum, velocity, and acceleration are all vectors. Even in a constant elliptical orbit, Earth will always be accelerating towards Sol. Were you to remove Sol (and, therefore, the force of gravity), Earth’s inertia will carry Earth along the same vector it was traveling at the moment of the disappearance of Sol. The continuing presence of the gravity force continuously accelerates the velocity of Earth about Sol. Even in a truly circular orbit (i.e. radial velocity is constant; rectilinear velocity magnitude is constant), the rectilinear velocity must be constantly accelerated to maintain the orbit. This is done by “pulling” Earth towards Sol. This is why we say it is constantly falling.
Now, for all of that, we traditionally think in terms of point masses, because it’s easier to think about (remember the physics student joke: a horse is a sphere, if it makes the equations easier). In reality, we should treat every body as a continuum of mass elements.
Properly speaking, Willis’s diagram would have a near infinity of mass elements, each working a different amount on each other, with a bunch of other interactions drawn on there. For ease, though, let’s leave the “solid” Earth as a single body. This is an easy idealization to do, since we assume that the frictional forces between the elements are such that they overcome the gravitational force differential between the elements and that the Earth is incompressible. Thus, we assume that the solids would move as one (i.e. the definition of a solid being that it holds its shape and volume in any container). In fact, there is a certain amount of deformation that takes place, as with all solids, in something called “creep”, but we deem this negligible and cancelled out by the rotation (think of it as a rotisserie chicken; we cook it evenly on all sides, so we neglect the temporal differences).
The fluid portions, however, do not have the same amount of friction between them, and therefore “slosh” more easily. The friction between the elements is insufficient to overcome the differential. As a bulk, it will still “fall” about the same as the solid Earth towards Sol (consider the magnitudes of each value input). The different individual elements, however, have are just different enough for us to perceive it.
In truth a fully solid Earth has the same “tidal force” as our blue and green Earth. It just manifests differently, due to the internal forces. The atmosphere “sloshes” far more than the oceans, but it doesn’t really matter to us (the pressure difference being small due to the low densities involved). Were the oceans made of molasses, the same tidal force would again be generated, but the oceans would deform far less (more viscosity = more internal friction to resist deformation).
I hope that helps a bit.
Still is the tide calculation based on Laplace’s tidal equations which states:
from http://en.wikipedia.org/wiki/Tide
“Ocean depths are much smaller than their horizontal extent. Thus, the response to tidal forcing can be modelled using the Laplace tidal equations which incorporate the following features:
1 The vertical (or radial) velocity is negligible, and there is no vertical shear—this is a sheet flow. 2 The forcing is only horizontal (tangential).
3 The Coriolis effect appears as an inertial force (fictitious) acting laterally to the direction of flow and proportional to velocity.
4 The surface height’s rate of change is proportional to the negative divergence of velocity multiplied by the depth. As the horizontal velocity stretches or compresses the ocean as a sheet, the volume thins or thickens, respectively.”
I repeat: The forcing is only horizontal (tangential).
More here:
http://en.wikipedia.org/wiki/Theory_of_tides
“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% …”
I fear there may be a misconception here, or perhaps 2. Take a column of air, or water, and measure the pressure at the bottom. Now heat the air, or the water, so it expands upwards. Unless any can leak out sideways the same mass of air or water is till there in the column so the pressure at the bottom must be constant. So whether the atmosphere swells or not, the pressure remains constant.
Now for atmospheric tides. As the earth rotates so the bulge of water tries to remain directly on the line connecting the centre of the earth to the attractor. But friction, inertia, and the continents prevent this happening. So water piles up on the edges, and we see the phenomenon of tides. The atmospheric bulge also tries to stay in the same place, but still there is friction, and hills and mountains get in the way. So the effect of the atmospheric tide is amplified in some places and is minimized in others. I believe it true to say that were the earth a perfect sphere, with a uniform layer of water, the actual rise and fall of the sea level due to tides would be as little as a foot (the surface of the solid earth is also reputed to distort about a foot as well). As all well know, in certain places the tidal effect is greatly magnified, eg, Bay of Fundy with tides up to 40 ft. Just so with the atmospheric tide, and a barograph will record the daily rise and fall of air pressure of up to 3 mb in particular locations. So while the theoretical variation in pressure doe to the atmospheric tide may be 0.1 mb, the observed variation can be 20 or 30 times as great.
The tidal force is affecting every part of the planet.
The force that actually move fluid as water and air is according to Laplace the horizontal (tangential) force.
The vertical force can not move the fluid.
The result of the tangential force is the bulges. It is not the vertical force that create the bulges since water does not expand.
The reason the vertical force are at maximum at bulge maximum is causing the common misinterpretation that the vertical force is the reason for the bulges,
The horisontal forcing is the reason why the earth rotation and tilt and moon inclination have an impact on the direction of tidal induced movement of earths fluids.
Moon inclination is 5.145° to the ecliptic which is between 18.29° and 28.58° to Earth’s equator.
That is then also the direction of the tidal induced fluid movement direction. Fluid is our ocean water and atmosphere are depending on how high the moon are above (or below) the horizon in mornings and evenings when vertical tidal force are at maximum.
Same for solar tide.
Quite a nice explanation Willis. Since there is a huge amount of confusion about this subject, it’s good to have this over-view.
However, one thing is inconsistent in your description.
“There are two different tidal forces of interest in the diagram, which are GF1 – GF2, and GF2 – GF3. ”
That is correct and you will see that are roughly equal and of opposite sign.
” It’s not because the tidal force acts in different directions, because tidal force is always directed towards the sun.”
There you are confusing the tidal raising force (a difference) which you had correctly explained up to that point and the gravitational attraction which is indeed in the same direction, just slightly different in magnitude.
Your piece of blue string is of course the earth’s own gravitational field that is many orders larger than the tiny tidal forces. That’s why tides are only a few metres on a planet thousand of kilometres across.
“string” is not a particularly good idea, a strong steel cable may be better. The cable would hold M1 and M3 anchored in place, but would stretch a little, allowing the ‘tides’ to rise.
So as the proud author of the gem you opened the article with, I note your explanation is in agreement with what I wrote, if you don’t confuse tidal force, the difference in gravitational attraction with gravitational attraction itself.
“The tide raising force acts in both directions (bulge on each side in the simplistic model)”
( GF1 – GF2 ) is negative ( GF2 – GF3. ) is positive. They are roughly equal in size.
Otherwise this nice and clear and easily digestible to a more general readership. Nice work.
Here is my attempt at presenting it. It’s somewhat more technical in language
http://climategrog.wordpress.com/?attachment_id=776
Clive Best has also written a good description with some fairly detailed equations rather then just considering the straight line forces along the earth-moon axis.
http://clivebest.com/blog/
( GF1 – GF2 ) is negative ( GF2 – GF3. ) is positive.
No, that badly put. The point is M3 is accelerated faster towards the sun , faster than the solid earth , M1 less fast than the solid earth. So _when viewed from the point of view of the solid earth_ the forces act in opposite directions.
“I think the units are correct. Consider gravitational force. The units of G are N m2 / kg2.
As a result, G * m1 * m2 /D^2 (gravitational force) has units of
G ( N m2 / kg2) * m1 (kg) * m2 (kg) / D^2 (m2)
The kg and the meters cancel out leaving newtons for the unit of gravitational force. Since tidal forces are the difference of graviational forces, they also are measured in newtons.”
Yes Willis but drop your logic and just look at your tidal force equation that need to be explained:
Tidal Force (N/kg) = 2 * G ( N m2 / kg2) * sunmass (kg) * r (m) / D^3 (m3)
Cancel away. m2 is cancelled by the m/m3 and only one of the kg’s cancel.
Are the units not N/kg?
Two other commenters above seem to have come up with this answer.
You are familiar with these tidal force equations and not myself, yet, so maybe you can explain the discrepancy in either your logic or the tidal force equation itself.
This is all about what scientists refer to as “frame of reference” , it depends up on where you’re sitting as to how you view things.
Willis adopts a frame of reference outside the earth which is rotation about the sun but he does not discuss the centrifugal and Coriolis “fictitious forces” that are necessary if you do that. The other way is regard it as an instantaneous snapshot of a system in rotation where gravity provides the centripetal force that causes the roughly circular orbital motion.
That is the approach I adopted because you can avoid the complication of fictitious forces.
http://climategrog.wordpress.com/?attachment_id=776
Sorry folks, but bits of string and billiard ball mechanics don’t work correctly if the billiard table is on a rotating platform. All this “frame of reference” stuff is not a simple idea but you can’t just stop the earth going round the and forget about it.
I like the simplicity of Wilis’ diagram but he needs to add the centrifugal “fictitious force” acting outwards (to the left). When you do that you find the net force on M1 is to the left and the net force on M3 is to the sun . We then see the equivalence with my statement about opposite forces.
Wayne.Your analysis is correct.
Tidal Force (N/kg) = 2 * G ( N m2 / kg2) * sunmass (kg) * r (m) / D^3 (m3)
“Are the units not N/kg?”
Well, no. It’s because Willis had written his equations for a unit mass of 1kg. and thus left out one of the m’s
The force should be of the form GmM/d^2 , he leaves out the m which is 1kg. It would be clearer to leave it in.
Music for physicists, outside of my benchtop chemistry and lately jewelry comfort zone. Thanks to Willis, WUWT is becoming Khan Acadamy, an online college, so here I can at least at times study my old man eyes out, as an empiricist, delving into new ways to prove those rat bastard theorists of Nature wrong again. Sometimes correctly.
I still think there’s a better groking of Maxwell’s equations out there somewhere in rebel controlled space.
Willis? Torqued space? Wither the aether?
Lubos won’t go there. He’s not a classicist, even in spirit, ever.
The three body problem alone, but add to it the severe inability of overheated supercomputers to even fold simple Tinkertoy proteins right, the stuff of life, and theory becomes just an adjunct to real lab work. Benchtop chemists always bust limits, lately the diffraction limit of traditional optical microscopes.
Oh, they’ll catch up, dragged kicking and screaming, into Reality.
Willis,
I usually enjoy reading your articles because I think you have a gift for explaining complex matters in a correct and simple way, but unfortunately I think this is an exception.
You say:
From this one could think that the distance between the Sun and the Earth is decreasing, but that is not the case, it is increasing.
The tidal force interacts with the rotation of the Sun and causes the Earth to be pushed slightly outwards every year. The energy for the push is taken from the Sun’s rotational energy.
We have the same effect between the Earth and the Moon. The earth is rotating slower and is pushing the moon away.
See:
Sun – Earth distance increasing one micrometer per year:
http://curious.astro.cornell.edu/question.php?number=317
Earth – Moon distance increasing 3 cm per year
http://www.newton.dep.anl.gov/askasci/ast99/ast99639.htm
/ Jan
Angech ponders the tides.
There is a minute change in the value of g which leads to a minute bulge .
The g value is the gravity of the moon not the sun.
It is actually twice as strong as the gravity of the sun on the earth tides as it is much closer to the earth.
The moons gravity is a lot weaker than the earth’s so any effect on bulges (tides) is a lot less than I originally contemplated and the effect of the sun is even less!
Why are the tides longer than 12 hours
The moon orbits the earth every 28 days so the tides are 24 hours divided by 28 means about 50 minutes later every day.
Are their minor tides due to the effect of the sun alone separate to the main 2 we realise.
Once actual movement occurs rotational effects and shallowness of sea beds and Coriolanus forces may make the waves much bigger
Willis
Hi
not strictly on topic but I saw an interesting post on Bishop Hill by a commenter called Paul K which I set out below. The connection to your piece is the gravitational effect on the ‘solid earth’.
Regards
“Personally, I think that it wouldn’t do to underestimate the importance of the England paper – even if it is founded on poor data.
As Nic correctly points out, from the observed data, the total global ocean heat flux shows a peak around 2001-2005 depending on which dataset one takes. TOA radiative measurements show a peak in net radiative incoming flux somewhere around 1997-2000, driven largely by SW changes in net albedo. Modern MSL data from satellite altimetry (or indeed from tide gauge data) shows a peak in its derivative function around 2001-2003, which should also be a proxy for net heat flux going into the ocean. (Using gravimetric data from GRACE, we can rule out the possibility that the peak in MSL derivative was caused by mass addition – it is a peak clearly driven by thermosteric expansion. There is a useful presentation here by Nerem: http://conference2011.wcrp-climate.org/orals/B3/Nerem_B3.pdf) So there is a consistent story from three data sources which says that the net incoming flux hit a peak and has since been decreasing overall for about a decade. This is not compatible with increasing forcing from GHGs and flat or declining tropospheric temperature – a mini paradox, if you will.
The mini-paradox becomes a major paradox when we consider the historical behavior of MSL from tide-guage data. The derivative function of the MSL data shows a dominant and remarkably consistent quasi-60 year cycle. It shows dominant peaks around 1750, 1810, 1870’s and 1940s. (See Jevrejeva 2008.) In other words, the modern peak in the MSL data came in right on time relative to previous recorded oscillatory cycles which date back to 1700. Using the modern peak for calibration, which we know relates to a peak in incoming net flux, we can very reasonably infer that the previous peaks were also due to peaks in net heat flux. The paradox is that these dates for peak incoming flux correspond closely to peaks in the multidecadal oscillations of surface temperature. This is a major bust. This is exactly pi radians out of phase with what we would expect if these cycles were caused by an unforced redistribution of internal heat. (High surface temperatures should induce an increase in outgoing radiation which translates into a decrease in net incoming radiation.) I think that we are therefore led to the inevitable conclusion that these are forced climate oscillations, which means that we have to look for a new flux forcing to explain them, since the current selection box does not have any forcings of the correct frequencies. I now return to the work of Matthew England. His work adds an important piece to the jigsaw puzzle, even if he himself is failing to appreciate the implications. We saw from Kosaka and Xie 2013 that a large chunk of the late 20th century heating as well as the modern temperature hiatus could be captured by the simple expedient of prescribing sea surface temperatures in a small area of the eastern Pacific. Those temperatures are in reality controlled by ENSO events which are in turn controlled by equatorial trade wind strength and direction. England’s work confirms at least in skeletal form that controlling the wind stress tensor in the same area gives a similar result, even if he is wrong on some of the details.
The question it leaves is: what then controls the equatorial trade winds? The answer was actually known more than 40 years ago when science was still relatively unsullied, but it will not be accepted easily by mainstream climate science today, since the answer makes not one but two major breaches in fundamental assumptions of climate science.
The first part of the answer is that the climate oscillations are triggered by gravitationally forced changes in the angular velocity of the solid Earth. These changes transmit a (non-radiative) momentum flux into the hydrosphere and atmosphere via frictional torque and conservation of angular momentum. These changes explain the fluctuations in trade winds and, just as importantly, the latitudinal meanderings of the jet streams. Before anyone starts calling for the men in white coats, I would suggest that you have a look at this 1976 paper: http://gji.oxfordjournals.org/content/46/3/555.full.pdf and this: http://gji.oxfordjournals.org/content/64/1/67.full.pdf . For the excellent correlation apparent in the higher frequency data between Earth’s rotation velocity, atmospheric angular momentum and ENSO events, you might also try this paper: http://trs-new.jpl.nasa.gov/dspace/bitstream/2014/17186/1/99-0613.pdf .
So it seems that England has probably confirmed that the multidecadal oscillations are driven by atmospheric tides which are driven by a non-radiative orbital forcing. He just hasn’t realised yet that what he has done is to demonstrate that the GCMs are all missing a massively important piece of physics which was considered small enough to be neglected on energetic grounds. The story doesn’t end there. The orbital forcing is a triggering and control mechanism, but it is “energetically deficient” to explain the full amplitude of the climate oscillations. On my sums the trough-to-peak transfer of energy via momentum flux and friction amounts to something less than 2*10^22 joules during the 60-year cycles. The amplification factor comes from the cloud response to the change in phase of the orbital forcing, which is why we note the dominant effect of SW changes in the radiative signature. This is a feedback mechanism of sorts, but it is not a “temperature dependent” feedback mechanism; it does not correlate simply with global surface temperature, but rather with the phase of orbital forcing. This post is already too long for me to try to explain how that works. I am hoping if I live long enough to try to get some of this stuff down in more detail in an article for Lucia, but I do keep getting distracted, not to mention beaten up by my wife for wasting time on that climate change rubbish instead of doing something useful.
Feb 14, 2014 at 2:35 PM | Paul_K”
Let’s test this theory. How does one go about measuring the “tidal force”?
The logical consequences of the tidal theory (such as elongation and Roche limit) have been falsified by observations. Metis and Pan orbit inside the Roche limit. Pan seems to be accreting. Far from being elongated, the Moon has a deficit of material on the near side.
http://burro.cwru.edu/Academics/Astr221/SolarSys/lunaint.gif
Other than that, can any 2- or 3-body theory (plus rope) explain the real ocean tide? As understand, tide prediction is not physical. It uses harmonic curve-fitting.
http://web.vims.edu/physical/research/TCTutorial/tidepredict.htm
The idea that there are two bulges on the planet is also too ideal. It is true that there are places where one sees the passage of two maxima every day, but that does not mean there are two bulges on the planet.
I would love to see a similar animation with the position of the Moon plotted on it.
For some reason my instinct says that this explanation cannot be completely correct because I ask that if we take away the moon altogether, (leaving only the sun and earth), would there be an ocean tide at all?
If the earth wasn’t rotating then the answer is easily no. If there was any bulge at all it would be very slight, at the same spot and never vary. But earth is rotating and rotating so that the outermost material at midnight is going the fastest above earth’s center’s orbital velocity resulting in it bulging outward away from the sun and the material at noon time going the slowest below earth’s center’s orbital velocity resulting in it bulging toward the sun thus resulting in a double bulge – without any moon at all.
Is this rotational factor significant? I’ll have run the numbers or wait for someone else to do it.
I really meant, instead of “If the earth wasn’t rotating”, – – “If earth rotated once per year”
Fascinating, Willis.
Isn’t it amazing how such things are often (always?) more complex than they seem at first glance.
No wonder the “Post-Normal” science lot suffer from physics envy.
A quick look gave me a result for a “no moon” variation of “midnight fling” acceleration over average center V^2/R of 3%. For now I’d say it looks significant.
Earth’s average orbit velocity V1 as 29885 m/sec. Crust WRT center V2 as 463 m/sec. So combined velocity in orbit around the sun at midnight V1+V2=V3 as 30349 m/sec V3^2 / V1^2 = 1.03
More on the Roche limit.
Images for the Roche limit
NASA on the Roche limit: Tidal Forces: Let ‘er Rip
Greg says:
February 15, 2014 at 1:55 am
Not true at all. I have a heliocentric (sun centered) frame of reference, which does NOT rotate about the sun. Neither does the planet in my example. The clue is where I say it is free-falling into the sun …
w.