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.
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Boiling it down, I’ve produced a two bulge tide model with no moon at all. Earth to Sun acceleration due to gravity is .006 M/sec^2 Plus 3% at midnight, minus 3% at noon makes it a change plus or minus .0002 m/sec^2 against our earth gravity of 10M/sec^2
Certainly tiny WRT to earth’s gravity at sea level but how much does that contribute to the formation of two lobes?
Willis, could you show the combined tidal forces on Earth from the Sun and Moon for both new and full moon positions.
Greg says:
February 15, 2014 at 1:33 am
Yeah, right … do the math again, Greg. Then do it again, again. At some point in there you might notice that they are roughly equal, but that they have the same sign.
I didn’t want to point out that you were the fool in the head post who made the same incorrect claim that the two forces acted in different directions. Instead, I said that perhaps you understood the tidal forces, but your words were just confusing … but now that you’ve restated your idiocy, I see you actually don’t understand.
DO THE FREAKING MATH, Greg. The two have the same sign.
This, of course, is as we’d expect. From the sun’s point of view, there is nothing to distinguish the first pair of masses (M1 & M2) from the second pair of masses (M2 and M3). So why would one be pulled in one direction by the tidal force, and the other be pulled in the other direction?
Math is your friend …
w.
In other words I’m stating that, simply by virtue of earth’s rotation, everything on the surface of earth weighs .002% less at midnight and noon than it does at 6am and 6pm.
wayne says:
February 15, 2014 at 1:53 am
I guess you missed the part where I said that GF1, GF2, and GF3 were the forces on a 1 kilogram mass. Obviously, if it were a two kilogram mass, the force would be twice that.
That is to say, the left half of your equations should be newtons times kilogrammes, not N/kg … I figured that most folks would understand that. However, it appears I underestimated you.
w.
Mike M says:
February 15, 2014 at 8:12 am
In other words I’m stating that, simply by virtue of earth’s rotation, everything on the surface of earth weighs .002% less at midnight and noon than it does at 6am and 6pm.
When is it noon at the North Pole?
Jan Kjetil Andersen says:
February 15, 2014 at 3:05 am
Jan (and others), the planet shown free-falling into the sun IS NOT THE EARTH. Here’s a clue to tell them apart … the planet in Figure 2 is free-falling into the sun. The earth is not.
As a result, although it is assuredly true, in this makes no sense to say “the distance between the sun and the earth is increasing” … we’re talking about planet free-falling into the sun, not the earth.
w.
Willis you say : “… because tidal force is always directed towards the sun.”
That sentence end could use a little correction/clarification I think.
Mike
A slight digression. Canute is mentioned in the title. May I put a word in for this king. When he said that he was commanding the tides he was not doing this out of hubris but to show his people that he had his limitations. Now there is degree of modesty that some warmists could benefit from.
lsvalgaard says: When is it noon at the North Pole?
Maximum kinetically induced differential is obviously at the equator so times the cosine of the latitude.
By all due respect, this is a rather misleading statement
The planet does not “fall into the Sun”. The gravitational pull is countered by the centripetal acceleration of the Earth in its orbit around the Sun.
The centripetal acceleration is the phenomenon one can feel in a carousel. One can feel an outward force away from the center of the spinning point.
The centripetal acceleration is given by the formula:
The planet stay in the orbit because the centripetal acceleration, a, is equal to the gravitational pull from the Sun
But as we see from the formula, a increases with the distance from the Sun. On the other hand, the gravitational pull from the Sun decreases with the distance from the Sun.
The centripetal acceleration is therefore greater than the gravitational pull in point M3 and weaker than the gravitational pull in point A1. In point A2 they are equal.
This means that the greater “carousel force” felt by the centripetal acceleration in M3 combined with the less gravitational pull, makes the oceans bulge out and away from the Sun.
In M1 we have the opposite situation. The smaller carousel effect and larger gravitational pull from the Sun makes the oceans bulge out from the Earth, but toward to the Sun.
/Jan
Mike M says:
February 15, 2014 at 8:35 am
Maximum kinetically induced differential is obviously at the equator so times the cosine of the latitude.
everything on the surface of earth …
And you statement is not correct anyway, as the rotation of the Earth causes things to weigh less at all times at the equator.
lsvalgaard says: “everything on the surface of earth …”
Okay, you got me, everything BUT an infinitesimal sized point at each pole – happy?
“And you statement is not correct anyway, as the rotation of the Earth causes things to weigh less at all times at the equator.”
That’s a CONSTANT, (also per cosine of latitude)… I’m looking at variations in acceleration.
Jan Kjetil Andersen says:
February 15, 2014 at 8:37 am
By all due respect, this is a rather misleading statement
The planet does not “fall into the Sun”. The gravitational pull is countered by the centripetal acceleration of the Earth in its orbit around the Sun.
Amazing how many experts we have around here. Willis is completely correct [apart from a few typos in the beginning]. This tidal effect has nothing to do with rotation, centripetal forces, orbital movement or any of that.
Mike M says:
February 15, 2014 at 8:44 am
I’m looking at variations in acceleration.
Variations caused by what?
Willis, the idea that a person free falling into a black hole feet or head first will be at some point ripped apart lengthwise is a common one and consistent with your assertion.
“DO THE FREAKING MATH, Greg. The two have the same sign. ”
How about reading where I corrected that statement ???
“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 …”
OK, I didn’t realise you intended not to rotate anything. So you want to call be a fool for saying something about how planets moons and tides work by comparing to a situation where nothing rotates.
Should still be an interesting exercise.
It’s not clear what the mechanical properiteis of your piece of “string” are so I’ll stick with a steel cable.
As I think we are agreed GF1>GF2>GF3 so the outer mass would accelerate less and get left behind unless it was tethered. Inversely M1 would fall faster. So both cables are under (rougly equal) tension.
You show that we can ignore the asymmetry:
True average = 0.50587
Approximation = 0.50587
So the net force accelerating the three masses is 3GF2 acting 3kg, They experience a common acceleration (since they are tethered) equal to that of the centre mass alone.
But M3 is experiencing less gravitational attraction and has to be pulled by the tether to the tune of GF2-GF3. The tension in the tether is thus that value. Similary the other side which should accelerate more but is held back by the tether.
So at M2 the tensions in the tethers are equal and in opposite directions.
The cables will extent a little due to the tension and both masses will end up a little further away from M2 and continute accelerating as an ensemble. This is the equivalent of the two tideal bugles. The tidal force and the ‘tide’ (stretch in cable) will remain constant until they get significantly closer to the sun and the gravity gradient gets steeper.
All you have done is substituted free-fall acceleration to the sun , without rotation, for a centripetal acceleration towards the sun with rotation.
The difference terms remain the same and act in opposite directions. If they didn’t one of the cables would be lack.
Even a “fool” like me can see that.
slack.
The difference terms remain the same and act in opposite directions. If they didn’t one of the cables would be slack.
lsvalgaard says: “Variations caused by what?”
Read BACK further to my calculations of difference in heliocentric orbital velocity of a point (on the equator) due to earth’s rotation; fastest at midnight, slowest at noon. The kinematic difference compared to the acceleration due to Sun’s gravity, (at earth’s distance), compared to 6am or 6pm, is 3%.
Mike M says:
February 15, 2014 at 8:58 am
Read BACK further to my calculations of difference in heliocentric orbital velocity of a point (on the equator) due to earth’s rotation; fastest at midnight, slowest at noon. The kinematic difference compared to the acceleration due to Sun’s gravity, (at earth’s distance), compared to 6am or 6pm, is 3%.
Not so. The Earth is is free fall and knows nothing about orbital velocity.
So my comments are on moderation now. ?? Is “slack” a moderation tripwise or what?
REPLY: dunno, some triggers are wordpress controlled. May be some spam circulating that uses “slack”…maybe Viagra like – A
lsvalgaard says: Not so. The Earth is is free fall and knows nothing about orbital velocity.
I’m done with ya…
Ok, I see
I thought it was something totally wrong here.
I then see that my other comment on misleading description is also misplaced.
/ Jan
Mike M says:
February 15, 2014 at 9:05 am
I’m done with ya…
Good that you stop digging your hole deeper and deeper. According to your ‘theory’ a pendulum clock should vary its ticking rate during the day and it does not.