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.
Discover more from Watts Up With That?
Subscribe to get the latest posts sent to your email.
Mike M says:February 15, 2014 at 3:38 pm
STOP! Right there! … that’s all I was ever talking about!
——————
lsvalgaard says: February 15, 2014 at 3:41 pm and what has that to do with the tides that are around all the time.
————–
So you forgot? I didn’t Let’s go over it shall we? – you never answered the question:
My question: If earth’s orbital velocity, (AT IT’S CURRENT ORBIT), around the sun slows down it will:
A) Go closer to the Sun
B) Go further away from the Sun
Answer the question, A or B?
Still waiting for an answer FROM YOU. (And when you can give a straight answer.. THEN we’ll move on to explaining more to you .)
Mike M says:
February 15, 2014 at 4:06 pm
(And when you can give a straight answer.. THEN we’ll move on to explaining more to you .)
That is very generous of you. But, as I said, your question is ill-posed and borders on nonsense, so you perhaps see my problem…
RichardLH says: …airless planet and a very flat …”
You’re missing the whole point. The orbit is earth around the sun, the thing stuck by earth’s gravity is NOT in orbit around earth, IT IS EARTH! Specifically a drop of water at the equator.
lsvalgaard says: “..perhaps see my problem…”
Yeas I do – you’re obstinate.
The answer is A and it happens to a mass M sitting on earth’s equator at noon – it’s going SLOWER than earth’s overall orbital velocity.
On the other side, at midnight, it is going FASTER than earth’s overall orbital velocity.
The slower particle gets dragged in more by the sun and the particle going faster is being flung away from the sun.
Now disprove it directly instead of trying to weasel around it. Did I state something that is not true?
Mike M says:
February 15, 2014 at 4:15 pm
Now disprove it directly instead of trying to weasel around it. Did I state something that is not true?
You have discovered the amazing fact that the Earth’s rotation [measured on a mass sitting firmly and fast on the equator] goes in the same direction as the orbital movement as midnight and in the opposite direction at noon. Fascinating! Now what has that to do with tides?
lsvalgaard says: Fascinating! Now what has that to do with tides?
What I already explained above but I guess you have a short attention span.
February 15, 2014 at 7:13 am: 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
February 15, 2014 at 7:51 amBoiling 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?
I’m stating that the difference EXISTS. Are you stating that it does not exist? Are my numbers way off?
Mike M says:
February 15, 2014 at 4:09 pm
“You’re missing the whole point. The orbit is earth around the sun, the thing stuck by earth’s gravity is NOT in orbit around earth, IT IS EARTH! Specifically a drop of water at the equator.”
Then the majority of the forces operating (to the first quite a lot of decimal places) will be the frictional, compressive, etc. components from the bits on water on either side of it.
Then will come the various forces that come from the Earth’s rotation alone, centripetal, coriolis, etc.
Then will come the 3d Tidal vectors.
That should cover the top 10 decimal places or so.
Mike M says:
February 15, 2014 at 4:09 pm
Oops. Sorry – left out tidal currents etc.
Mike M says:
February 15, 2014 at 4:28 pm
I’m stating that the difference EXISTS. Are you stating that it does not exist? Are my numbers way off?
As Bart has said many times, I am a dummy, embarrassing myself constantly, so please explain to this dummy how your differences produce tides. And how you would explain tides if the Earth were not rotating. I guess you would say that on a non-rotating Earth there would no tides ‘flung’ around, but explain that in some more detail, please.
RichardLH says: February 15, 2014 at 4:29 pm “Then the majority of the forces operating (to the first quite a lot of decimal places) will be the frictional, compressive….”
If you say so… I’m only pointing out the change in net acceleration force that goes up and down twice a day at the equator without any moon at all and simply because of earths’ rotation slowing/speeding earth’s crust slower/faster than earth’s solar orbital velocity.
lsvalgaard to his credit compartmentalized it best – you will weigh .002 less at midnight and noon than you will at 6am or 6pm. (Females using spring loaded bathroom scales take note.)
I would think seismographs register this effect but it isn’t noticed because it’s a much longer period than an earthquake and the amount of change is so small it’s probably ignored.
Mike M says:
February 15, 2014 at 4:51 pm
I’m only pointing out the change in net acceleration force that goes up and down twice a day at the equator without any moon at all and simply because of earths’ rotation slowing/speeding earth’s crust slower/faster than earth’s solar orbital velocity.
So you are claiming that if the Earth were not rotation there would be no tidal effects…
Good news for the proverbial astronaut falling into a black hole: just be sure that you don’t rotate and you’ll not be shredded by tidal forces…
Mike M says:
February 15, 2014 at 4:51 pm
“I would think seismographs register this effect but it isn’t noticed because it’s a much longer period than an earthquake and the amount of change is so small it’s probably ignored.”
I think you will find that the Earth Tide, the bulge in the rock itself from the Moon and Sun is an order to two higher but….
http://en.wikipedia.org/wiki/Earth_tide
Mike M says:
February 15, 2014 at 4:51 pm
you will weigh .002 less at midnight and noon than you will at 6am or 6pm.
Here is get-rich scheme using your insight: buy gold at noon, e.g. 100 kg, file off 0.2 kg [=6 oz @ur momisugly US$ 1300/oz = $7600], then sell the gold at 6 pm as it at that time will weigh 100 kg and recover what you paid for the 100 kg at noon, pocketing $7600/day. In a year that is almost $3 million [tax free].
lsvalgaard says: “So you are claiming …”
I did NOT discount or preclude other factors! Stop putting words into other people’s mouths!
lsvalgaard says: Here is get-rich scheme
Ummm, only an idiot would use anything but a MASS BALANCE scale for stuff like gold – works on the moon.
Wow. Leif never stops amazing me with his infinite patience here. I’d be into the whiskey by now.
Mike M says:
February 15, 2014 at 5:05 pm
I did NOT discount or preclude other factors! Stop putting words into other people’s mouths!
So, answer the question: if the Earth did not rotate would there be tidal effects? If so, how large would they be compared to the ones caused by rotation? Huge or tiny? Or don’t you know?
Mike M says:
February 15, 2014 at 5:07 pm
Ummm, only an idiot would use anything but a MASS BALANCE scale for stuff like gold
And Mother Nature does not do that when it comes to create tides. How do you know?
RichardLH says: “I think you will find that the Earth Tide, the bulge in the rock itself …”
Oddly, that’s what I generally guessed at above. ( February 15, 2014 at 12:15 pm)
lsvalgaard says: February 15, 2014 at 5:14 pm “So, answer the question: if the Earth did not rotate would there be tidal effects? ”
From the moon, of course.
I think there is a gross misunderstanding of “free fall”. Willis is correct in his analogy and may even underestimate his own clarity.. the planet is actually “free falling” and is plunging toward the sun as described in his picture. There is also a sideways motion (orbital velocity) so we (earth) manage to keep missing the sun as we fall into it. Since the sideways motion is perpendicular it has no effect on tidal force so we can ignore it and are left with only the free fall condition. So the bodies M1, M2 and M3 are each different distances from the main attractor so each has a slightly decreasing pull. Lets call the forces magnitudes of 3,2, and 1 .All are positive pulls toward the sun ( or moon) in this frame of reference. The closest is hardest, the middle less and the extreme even less again. so you end up with tension in the theoretical string (local gravity replaces the string) and this results in a front-side and back-side bulge relative to the middle . Rather simply presented I think.
Good one Willis!
Trying to bring in orbital and rotational centripetal force is not required and has insignificant affects.
lsvalgaard says: “And Mother Nature does not do that when it comes to create tides. How do you know?”
Ho do I know what exactly? How to spot non-sequitur blather?
Mike M says:
February 15, 2014 at 5:25 pm
“Oddly, that’s what I generally guessed at above. ( February 15, 2014 at 12:15 pm)”
I Googled it with much less effort (to get the value – I already knew it existed). As I said., an order of magnitude or two greater than the discovery you are looking at.
RichardLH says: “I Googled it with much less effort ”
Keep up the good work!
Kirk c says: “Trying to bring in orbital and rotational centripetal force is not required and has insignificant affects.”
Willis’ is pointing out a gravitational differential producing a difference of ~1.1 MICRO Newtons, (.0000011 Newtons) between two 1KG masses.
I’m pointing to a .0002 m/sec^2 difference in acceleration from a rotational induced difference in sun’s gravitational field which, for a 1 KG mass, is .0002 Newtons. Insignificant?
Mike M says:
February 15, 2014 at 5:42 pm
“Keep up the good work!”
Just to point out that it existed and was significantly greater than you were looking for……
I can’t see why you didn’t try that route first anyway.
And looked at the internal tide which is much more significant than all the stuff going on at the surface anyway.