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.
lsvalgaard says:
February 19, 2014 at 3:59 am
“No, this is irrelevant as I have been trying to explain to you.”
And, you are wrong. There is a factor of 3/2 amplification of the differential. You’ve admitted it.
“It is not the orbit that is tidally locked, but the rotation of the body that is locked. Here you show your confusion.”
It is not I who is confused. You are defining your fictitious forces with respect to an accelerated coordinate system, a big no-no in Newtonian mechanics.
“You are still confused.”
I have no more time to waste on you, Leif. Once again, you put on a big show, but it’s all hat and no cattle.
RichardLH says:
February 19, 2014 at 12:20 pm
You guys are clueless.
Jan Kjetil Andersen says:
February 19, 2014 at 12:51 pm
“Doesn’t this imply that one also have to take the centripetal force into account?”
Yes. In fact, the calculation of tension in such a structure depends very explicitly on the centripetal acceleration term. It magnifies the tension by a factor of… wait for it… 3/2.
Willis Eschenbach says:
February 19, 2014 at 1:04 pm
“This condition is equivalent, in every case, to the free fall condition.”
I explained previously why your source is trivially wrong. The entire Earth is not in free fall. Only the center of the Earth is in free fall. Everything else is being pulled along for the ride. This is what gives rise to tidal stresses in the first place.
This is pointless. I’m done with this thread.
Bart says:
February 19, 2014 at 9:26 pm
This is pointless. I’m done with this thread.
You could learn something from http://www.leif.org/EOS/tides_CHT_thesis.pdf
Bart says:
February 19, 2014 at 9:26 pm
You are defining your fictitious forces with respect to an accelerated coordinate system, a big no-no in Newtonian mechanics.
Not at all, it is much simpler than that, see e.g. section 5.2 of http://www.leif.org/EOS/tides_CHT_thesis.pdf
Study the thesis carefully and you’ll see.
lsvalgaard says:
February 19, 2014 at 10:13 pm
Bart says:
February 19, 2014 at 9:26 pm
You are defining your fictitious forces with respect to an accelerated coordinate system, a big no-no in Newtonian mechanics.
Not at all, it is much simpler than that, see e.g. section 5.2 of http://www.leif.org/EOS/tides_CHT_thesis.pdf
Study the thesis carefully and you’ll see.
lsvalgaard says:
February 19, 2014 at 10:16 pm
Poorly done.
Bart says:
February 19, 2014 at 9:26 pm
“You guys are clueless.”
What? To point out that all (well very, very nearly all) of any centripetal acceleration is exposed in the rock? That is the reality of the situation. If there were no ocean at all (and no Moon to boot) the globe would still be an oblate spheroid with approximately the same dimensions. Add in the oceans. Then there would be a small, easy to calculate Solar tide and that would be that. One application of the Solar gravitational field ala Wiki field vector plan. Easy stuff. Small variations due Earth orbital changes.
Now adding in the Moon is what adds the complications. But that is just a more complex form of what we have shown already. With complex, interleaved patterns that beat away with the Solar stuff.
And then we stand a chance of not getting things in the wrong proportions or from the wrong causes.
Bart says:
February 19, 2014 at 9:26 pm
Bart, the issue was not the conclusions. I provided a link, not for the conclusion which I quoted here, but for the elegant exposition of the mathematics of tidal forces. The math is derived using five different frames of reference. The paper is beautiful, with moving graphics illustrating each frame of reference. Heck, he even derives the result using the reference frame of an observer on the moon. I explained that in all five of them, the centrifugal/centripetal forces cancel out mathematically.
Your claim that your “explanation” and your unsupported statement that “the entire Earth is not in free fall” trumps five pages of math is hilarious. This is SCIENCE, Bart, not grade school. You can’t beat math with your mouth.
To beat math, you have to show where the man’s most clear and elegant math is wrong. Not claim that it’s wrong. You have to show where it’s wrong. All you’ve done so far is talk. Talk doesn’t trump math.
So I can see why you want to say “I’m done with this thread” and go out the door. It’s because you can’t show where the math is wrong. Not because of your math-fu, however. That may be very strong, but it wouldn’t help. You can’t show where the math is wrong, simply because the math is correct, and has been well understood for a long time.
Now, that leaves you with a choice. What I would do in your position is first verify that I couldn’t find any mistakes in the math. Then I’d slap my forehead and say “Dang! I checked the math, I understand every step of it, and guess what? Looks like I was wrong after all!”.
That way everyone congratulates you, because that’s what scientists do—admit when they are wrong. That way, your credibility here at WUWT goes up.
Or you can just say “I’m done with this thread” and walk away. At that point, you might as well not post at WUWT again, because you’d be left without a scrap of credibility. Everyone here knows nobody can beat math with their mouth. And everyone here knows that you personally can’t demonstrate any errors in the math I linked to, because it’s basic stuff that’s been known for centuries.
So, the choice is yours. To encourage you to take the scientific direction, let me point to a past post of mine entitled Wrong Again … and I was, too. It’s one reason that even people who dislike me still read my scientific ideas and research—because they know that when I’m wrong, I stand up and say so loud and clear.
w.
I think a spacelift is a useful analogy for analyzing this topic.
Imagine a rod twice the length of the distance to the geosynchronous orbit. The rod is placed vertically in the geosynchronous orbit and stretching all the way from the Earth surface to twice the height of the geosynchronous orbit. A counterweight is placed in the upper end to stabilize the rod in an orbit where the lower end is hovering one meter above a point at the Earth’s surface at the equator.
This will be a satellite to the Earth and will be subject to the same tidal forces as any satellites. However, the extreme dimensions of this structure highlight the tidal effects.
The lower end will have gravitation equal to one G. The gravitation at the upper end will be about 1% of that. A 100 kg body will weigh 100 kg at the lower end, and the same body will have a gravitational pull of 1 kg force in the upper end. But the centripetal force in the upper end will drag the body outwards with approximately 4 kg force, giving a net outward force of 3 kg.
That means that the tidal force, calculated in the way Willis does above, give a force of 99 kg on the 100 kg body, but when taking centripetal acceleration into account we get a force of 103 kg.
As I see it, this must mean that we have to take centripetal acceleration into account when we calculate tidal force.
Or can anyone find an error in this reasoning?
For spacelifts see:
http://www.spacelift.org/
/Jan
Jan Kjetil Andersen says:
February 20, 2014 at 12:23 pm
But the centripetal force in the upper end will drag the body outwards with approximately 4 kg force, giving a net outward force of 3 kg.
some confusion: the centripetal force is directed towards the Sun being just the gravitational force]. The centrifugal force is directed outwards, but is the same for all points of the Earth.
lsvalgaard says:
February 20, 2014 at 12:33 pm
Thank you Leif, I should have used the word “centrifugal”, not “centripetal”
I would appreciate if you also would comment on my reasoning of the combined tidal effect of 1.03 G on the spacelift.
Do you agree that an astronomer on the spacelift would experience a tidal variation of 1.03 G between top and bottom?
If so, how can this be explained by gravity alone?
/Jan
Jan Kjetil Andersen says:
February 20, 2014 at 12:42 pm
Thank you Leif, I should have used the word “centrifugal”, not “centripetal”
I would appreciate if you also would comment on my reasoning of the combined tidal effect of 1.03 G on the spacelift.
It is a bit more complicated than that as gravity itself decreases with altitude. The lift would not be standing on the ground, but hang from the anchor point at geostationary orbit where the total force on it would be zero, and built simultaneously up and down to balance the weight of the cable.
Now as to the tides: The rotation of the Earth already raises an equatorial bulge of 21 km [or a 100,000 times greater than the solar tide]. This is not a tide and the tension in the cable is also not due to a tide.
lsvalgaard says:
February 20, 2014 at 12:56 pm
Now as to the tides: The rotation of the Earth already raises an equatorial bulge of 21 km [or a 100,000 times greater than the solar tide]. This is not a tide and the tension in the cable is also not due to a tide.
On the other hand the space lift is influenced by both the lunar and solar tides as everything else is. Here is an analysis of what the normal tides will do to the lift:
http://repositories.tdl.org/ttu-ir/bitstream/handle/2346/21793/Kaithi_Vijaya_Thesis.pdf?sequence=1
lsvalgaard says:
February 20, 2014 at 12:56 pm
I agree on all that, but it does not answer my question so let me rephrase.
I am talking about the spacelift as a satellite to the Earth, and I am discussing the tidal forcing experienced on that satellite from the Earth.
For simplicity, let us forget about the Sun and the Moon because that is not the tides I want to analyze with this example.
The spacelift is a rod or a cable and let us say the anchor point is in 2x geosynchronous height, that is 71 600 km above the Earth surface. In that anchor point the Earth’s gravity will be approximately 0.01 G.
That means that an astronaut with a weight of 100 kg will have a gravitational pull of 1 Kg towards the Earth in that point.
But the centrifugal force caused by the Earth’s rotation will push the astronaut outwards with a force of 4 kg. The net force experienced by the astronaut will therefore be 3 kg outward, or -0.03 G.
In the lower point the gravity will of cause be 1 G.
That means that the total tidal force on board the spacelift satellite is 1.03 G.
/Jan
Leif:
“Now as to the tides: The rotation of the Earth already raises an equatorial bulge of 21 km [or a 100,000 times greater than the solar tide].”
You would think that at that point people would stop and go ‘duh – the rock centripetal is 20 km, the water tidal is less a than a meter plus/minus variation on that’ but no……
Jan Kjetil Andersen says:
February 20, 2014 at 1:27 pm
“I am talking about the spacelift”
The problems with the ‘spacelift’ concept is one that is rarely discussed. In order for the far end anchor point to stay immediately above a point on the equator requires the lift rod to push it faster than orbital sped at that height. It cannot be a cable as otherwise it will drift back against the spin in some form a curve. That extra, outwards, centripetal force has to come from somewhere you know, it doesn’t just create itself. It must come from tensions/torsions in the ‘rod’ that maintains its position vertically above the surface anchor point. The rod also has to supply the tangential acceleration required to any object travelling up it to get it to the required ‘top end’, tangential velocity.
RichardLH says:
February 18, 2014 at 4:45 pm
http://i29.photobucket.com/albums/c274/richardlinsleyhood/EarthSun_zps14b35168.png
So where are your two particles?
————————-
I appreciate you drawing that. Thanks and I intend to produce a free body diagram for particles at, per your sketch, that are at positions “A” or “B” versus either “C” or “D”.
Doing this via FBD is more involved but there are only two general factors – sum up the forces acting on the particle that produce the radius of curvature of it’s path at that point in time and at it’s velocity WRT to the center of the sun at that point in time. We already have the velocity per all the above discussion so providing a summation of forces and a value for radius of path curvature comprise the task ahead. It will reveal why we weight less at noon and midnight than at 6am/pm by an acceleration factor of .00018 meters/sec^2 at the equator.
I haven’t had time to do it yet – I promise I will this weekend.
I’m just a country bumpkin and continue to drop in on this discussion from time to time ’cause I needed more shotgun shells to shoot the squirrels and rabbits I need for dinner and have an inquiring mind.
I want to bring up a thought experiment that is only a very simple and elementary view that may help to explain what Willis and Lief are showing. Maybe it’s enough to get the very intelligent minds here focused in the same direction.
You must understand the concept of Willis’ three weight (with blue rope) diagram and an external gravitational force applied with the direction of attraction being applied running in the same direction as the blue rope.
I may wish I never post this, but here it goes. You will not need any concepts of centrifugal/centripetal force. Freeze frame the entire universe (i.e. nothing is in motion/moving). Consider that there is only one very distant body in that universe and it (due to it’s mass) has a gravitational force. The other bodies are your experimental weights/bodies.
You don’t even need to know what direction that distant gravitational force is coming from because with only two of those weights/bodies (Willis’) using a red rope between them you can find out. Consider that you have no mass so as to not affect this thought experiment.
Please note that in the first case I am only considering the effect of releasing the weights/bodies inline with a distant gravitation pull. In the second case I’m releasing them with the rope perpendicular to that distant force. For simplicity in this thought experiment let’s skip what would happen with the weights being released in an orientation other than the red rope being aligned in an orientation that is not perpendicular or inline (parallel with) the distant gravitational force. That makes things a little more difficult and would detract from this most basic example (but by all means do it later as that’s where things get interesting).
Now release the two equal size/mass weights/bobies (with the red rope taught) and watch them. If the the rope begins to slacken then you have placed the weights perpendicular to the direction of the distant source of gravity. They are now accelerating in the direction of the distant force equally and at the same time attracting each other equally. They will come together. As with Willlis’ simple diagrams there will be a bulge with no centrifugal force necessary. Some inertia as a result of the difference in acceleration due to the diameter of the two weights/bodies as the two weights/bodies (with red rope) accelerate towards one another but remember they are NOT spinning/orbiting. Now think this all through again using bigger weights/bodies. The bulge will become more obvious.
Willis has already explained the case where the weights/bodies are in alignment with the forces of gravity and used a blue rope. You have already thought about that so I’ll skip it.
Now for the fun part. Consider releasing the two weights/bodies with the red rope at a 45 degree angle in orientation to direction of the distant gravitational force. Then for fun do all this again and use two different size/mass bodies. Close your eyes and let your brain play. Then un-freeze the universe and watch what happens.
There a lot of warts associated with this vague description . Cut me some slack until you read the next comment from me.
This not so organized bunch of words that I have placed on digital paper may not be absolutely clear but hopefully will help some get a start to the concept of what Willis has correctly explained. Now back to the woods with another box of 20 ga. shot shells.
I’m stunned that this thread is still running, albeit slowly, so many days after its birth, but I am glad that you brought this thought-experiment up, because you are one commenter here who now believes that you do not need centripetal/centrifugal forces to make proper explanations of things.
Let me make the following points.
1) If centrifugal force wasn’t a useful concept, then Coriolis force wouldn’t be either, because they both originate in the same way. Moreover, why do mechanics textbooks used in upper division physics courses, and also engineering mechanics courses, have sections of the text devoted to this topic of fictitious forces, complete with example problems, if the concepts weren’t needed or useful?
2) I have consulted the Barger and Olsson (B&O) textbook, which is a very good upper division text from which to learn classical mechanics by the way, and found that to explain the tides on Earth as raised by the Sun they actually state the impact of the “centrifugal force” of Earth’s orbital motion around the Sun. Then they derive the tide-raising potential by adding a potential function based on this centrifugal force to the gravitational potential.
3) I decided to consult Horace Lamb’s book “Hydrodynamics” (1932) on the subject. Leif has been pointing everyone, including me, to a paper by Munk as proof that one needs no more than gravity to make the tide-raising potential, but Munk in turn references Lamb, so let’s go straight to the horse’s mouth on this subject. Lamb derives the tide-raising force in a different manner than B&O, but it is explicitly clear that he uses also uses the centrifugal force to make a contribution to the gravitational potential.
4) Why do these eminent authors use the centrifugal force? Because it is absolutely essential to getting the potential function correct, that’s why. The reason is this: If you write the gravitational potential of the Moon in a Taylor series referenced to the center of the Earth, you will obtain a term proportional to “radius of the Earth times cosine of latitude” that is not part of the tide-raising potential. The centrifugal force just happens, not coincidentally, to provide what’s needed to cancel this.There is no way to eliminate this term otherwise and obtain the correct form of the tide-raising force.
5) Willis uses free fall rather than centripetal acceleration to provide the offsetting term, which I have no objection to, but keep in mind that free-fall and centripetal accelerations are not the same thing. Free-fall acceleration is an increasing speed of an object tangential to its path; centripetal acceleration results from changing direction and is perpendicular to the path. In both instances they provide something needed to get the tide-raising force right; but I feared that people would learn from all of this that centrifugal forces are fictional and never needed because they complicate the picture, and there is something fishy about centripetal acceleration as well. You appear to have done just that.
6) Leif, Willis, and some others here believe the forcing function for the tides is just gravity and this is a problem in mechanical static equilibrium–or at least that’s how I interpret their reaction to all I say. But the essential function of either free-fall acceleration or centripetal acceleration in deriving the tide-raising potential shows that even equilibrium tides are a dynamic phenomenon.
I hope this clears up some of your confusion.
Kevin Kilty says:
February 22, 2014 at 10:49 am
4) Why do these eminent authors use the centrifugal force? Because it is absolutely essential to getting the potential function correct, that’s why.
Since the centrifugal force is the same for every point on or in the Earth, it cancels out when you compare the forces on any two points, e.g. the subsolar and the anti-solar points, and hence does enter into the expression for the difference between the forces and thus also not for the height of the tidal bulge.
Here we are all talking about the special case of an object(earth,moon) in orbit around another object (sun-earth-moon barycentre). In this special case the centrifugal force is perfectly balanced by the centripetal force(gravity). This is the same case for the space station in orbit around the earth where astronauts are essentially weightless. As a result of the earths orbit(s) we being attached to the earth don’t feel the gravitational force of either the sun or the moon. We only feel the gravitational attraction of the earth.
Now consider an asteroid passing close by the earth. It gets deflected by the earths gravity and feels a centrifugal force which is not equal to the centripetal force(gravity). Now work out the tidal force exerted on the object of radius R. Now you must include the centrifugal force !
Kevin Kilty says:
February 22, 2014 at 10:49 am
“Lamb derives the tide-raising force in a different manner than B&O, but it is explicitly clear that he uses also uses the centrifugal force to make a contribution to the gravitational potential.”
Oh there is a component to the water tides that is centripetal. But it is very, very tiny.
The majority of the centripetal force is expressed in the rock. All 21 km or so of it. Then we get approximately 1 m or so of Earth tide. Then we get a few mm or so of water centripetal. Then we get a few 10’s cm of water tide (open ocean – no land).
Do the maths. You’ll se I’m right.
Remember, if there were no oceans, the Earth would still be an oblate spheroid with a 21 km bulge.
Richard LH,
Thanks for your reply. You are speaking of the centrifugal force pertaining to the Earth’s rotation on its axis; I am speaking of the centripetal acceleration pertaining to the orbital path around the tide-raising body. They are utterly different things.
Kevin Kilty says:
February 22, 2014 at 11:26 am
I am speaking of the centripetal acceleration pertaining to the orbital path around the tide-raising body.
Since the centripetal acceleration [directed towards the sun] is just gravity [what else holds the Earth in orbit?], you must mean the centrifugal force.
Lsvalgaard,
Good grief, you got in here even before RichardLH. You and I are saying almost exactly the same thing in two different ways, I think, but I was really refering to the potential, not to its gradient, and I think one ought to say really that the centripetal force equals the gravitational attraction of the tide raising body at its centroid. By the way Lord Kelvin spoke not just of lunar and antilunar points, but of a moon and antimoon, at least according to Lamb. I have no idea if such a concept helps someone understand the tides or not–I’d say not.
Kevin Kilty says:
February 22, 2014 at 11:39 am
and I think one ought to say really that the centripetal force equals the gravitational attraction of the tide raising body at its centroid.
The centrifugal force due to orbital revolution is the same at all points of the Earth and is equal [but in opposite direction] to the centripetal acceleration = the gravitational acceleration at the center of the Earth.
Kevin Kilty says:
February 22, 2014 at 11:26 am
“Thanks for your reply. You are speaking of the centrifugal force pertaining to the Earth’s rotation on its axis; I am speaking of the centripetal acceleration pertaining to the orbital path around the tide-raising body. They are utterly different things.”
You want to do the maths?
Earth rotation 0.034 m/s^2
Earth orbital ~0.0000952 m/s^2