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.
I’ve been grappling with this tide issue for two days now. I fully understand what Willis has shown in free falling bodies under gravitational forces. Then when I apply my new found knowledge to a physical application of some liquid covered planet in orbit around a much larger body (sun) I am having difficulty scrubbing centrifugal force from the mix in adding to an appreciable bulge to the far side of the planet (away from the sun).
It will soon click (I hope) but my brain is being quite stubborn on the issue at this point.
====
Truthseeker says:
February 16, 2014 at 5:55 pm
Thanks for the link. That and others have helped.
BTW, the earth and the moon do not rotate around each other, or around the barycentre. The earth and moon are a two planet system, not a planet and satellite system.
This may sound like heresy, but I believe if you check you will find that the moon’s orbit is always concave wrt the sun, just as is the earth’s. I call upon that great scientist, Isaac Asimov, for support (see: http://en.wikipedia.org/wiki/Double_planet )
“Tug-of-war definition
Isaac Asimov suggested a distinction between planet–moon and double-planet structures based in part on what he called a “tug-of-war” value, which does not consider their relative sizes.[4] This quantity is simply the relationships between the masses of the primary planet and the Sun combined with the squared distances between the smaller object and its planet and the Sun:
tug-of-war value = m1⁄m2 × (d1⁄d2 )2
where m1 is the mass of the larger body, m2 is the mass of the Sun, d1 is the distance between the smaller body and the Sun, and d2 is the distance between the smaller body and the larger body.[4] Note that the tug-of-war value does not rely on the mass of the satellite or smaller body.
This formula actually reflects the relation of the gravitational effects on the smaller body from the larger body and from the Sun. The tug-of-war figure for Saturn’s moon Titan is 380, which means that Saturn’s hold on Titan is 380 times as strong as the Sun’s hold on Titan. Titan’s tug-of-war value may be compared with that of Saturn’s moon Phoebe, which has a tug-of-war value of just 3.5. So Saturn’s hold on Phoebe is only 3.5 times as strong as the Sun’s hold on Phoebe.
Asimov calculated tug-of-war values for several satellites of the planets. He showed that even the largest gas giant, Jupiter, had only a slightly better hold than the Sun on its outer captured satellites, some with tug-of-war values not much higher than one. Yet in nearly every case the tug-of-war value was found to be greater than one, so in every case the Sun loses the tug-of-war with the planets. The one exception was Earth’s Moon, where the Sun wins the tug-of-war with a value of 0.46, which means that Earth’s hold on the Moon is less than half the Sun’s hold. Since the Sun’s gravitational effect on the Moon is more than twice that of Earth’s, Asimov reasoned that the Earth and Moon form a binary-planet structure. This was one of several arguments in Asimov’s writings for considering the Moon a planet rather than a satellite.[4]
We might look upon the Moon, then, as neither a true satellite of the Earth nor a captured one, but as a planet in its own right, moving about the Sun in careful step with the Earth. From within the Earth–Moon system, the simplest way of picturing the situation is to have the Moon revolve about the Earth; but if you were to draw a picture of the orbits of the Earth and Moon about the Sun exactly to scale, you would see that the Moon’s orbit is everywhere concave toward the Sun. It is always “falling toward” the Sun. All the other satellites, without exception, “fall away” from the Sun through part of their orbits, caught as they are by the superior pull of their primary planets – but not the Moon.[4][5][Footnote 1]”
— Isaac Asimov”
So bear that in mind when quibbling as to whether the earth is falling into the sun, or the moon is falling into the earth, and whether or not tides wait for no man.
if you would understand tides, realize that any reference to ‘Centrifugal force’ is wrong. Also any reference to ‘inertia’. See: http://www.lhup.edu/~dsimanek/scenario/tides.htm
The general view from Earth is here:
http://chiefio.wordpress.com/2014/02/16/tides-vectors-scalars-arctic-flushing-and-resonance/
But even then, reality is more complicated and “things change”…
Thanks for that link willis
This was also helpful
http://www.lhup.edu/~dsimanek/scenario/tides.htm
To add to my comment above (February 16, 2014 at 8:06 pm)
I am only concerned with the force(s) causing the far side bulge on a non axial rotating planet in orbit around another much larger body. The only centrifugal force (if any) I need to clarify in my mind would be that due to the orbit around the sun.
Maybe to put it another way. Why is there no centrifugal force involved with regards to a planet’s orbit around a much larger body (sun).
If I tied a brick to a rope and spun it around my head it would produce an outward force (centrifugal force) that would need to be countered acted by the rope. If we relate that concept of the orbiting brick and rope to a planet (brick) and gravitational attraction (rope) and call it a tidal force then would it be fair to say that the tidal force is the net result of gravitational attraction combined with the centrifugal force all wrapped into one package?
Bart says:
February 16, 2014 at 1:09 pm
The place where centripetal effects are not insignificant is solar induced tides on the Earth. I keep explaining this, and it seems you simply disregard it, and continue on, as though I am claiming something I am not. If “stop being stupid” is snide, then let me clarify: stop telling me I am wrong about something I am not claiming.
Prof. Dale E. Gary, NJIT, has a good simple explanation that you could learn from:
http://web.njit.edu/~gary/320/Lecture12.html
A full [but lengthy] treatment is here http://www.gps.caltech.edu/classes/ge167/file/agnew_treat_tide.pdf
Yet another [elementary] derivation is http://oceanworld.tamu.edu/resources/ocng_textbook/chapter17/chapter17_04.htm
and finally here is a good explanation dispelling the influence of centripetal forces: http://www.vialattea.net/maree/eng/index.htm
AlecM says:
February 16, 2014 at 11:33 am
No, it is not. Read the links, especially the one posted here, or you could just read their conclusion.
w.
RichardLH says:
February 16, 2014 at 12:17 pm
Perhaps your world works that way.
On my planet, people commonly study simplified models of complex situations, precisely to draw conclusions from them Happens all the time, I’m in mystery why you think we can’t learn valuable lessons from simple models.
w.
wayne says:
February 16, 2014 at 12:32 pm
Ok … since you haven’t disagreed with me in any points, I’ll join you, and I’ll stop there as well.
w.
Dudley Horscroft says:
February 16, 2014 at 8:13 pm
Thanks, Dudley. Your misunderstanding seems to be that the “this” in your quoted words of mine doesn’t refer to the tidal force. In the paragraph above the one you quoted, the “this” is specified as “the horizontal component of the tidal force”, not the tidal force itself.
And indeed, the horizontal component of the tidal force is zero at both the poles and the equator.
w.
Bart says:
February 16, 2014 at 12:31 pm
Got it … so, can we assume that your method is from the ultra-new field of investigation called “Science By Assertion”?
I remember that one from the playground, where when someone says “you’re wrong”, you say:
There have been a number of citations posted saying that centripetal force is immaterial to tides … not just one, but a bunch. So when you continue to repeatedly claim that centripetal force is a player in the tidal game, you’re obviously either:
1. Not reading the links provided, or
2. Not understanding the links provided, or
3. Just not paying attention.
In other words, although I hate to say it … you’re wrong.
Leif provided links that back up the claim that centripetal force isn’t a player. Mosher did the same. Others did the same.
Or you could just read, and re-read, and read again, the conclusion to one of those many links that I posted above, viz (emphasis mine):
See the part where he says that centripetal forces “do not produce any tidal deformation”? He backs that up with pages of analysis, diagrams, and equations.
You back your claims up by repeating you’re wrong, you’re wrong, you’re wrong …
I know which one I agree with … but then, I’ve done the math.
eyesonu says:
February 16, 2014 at 9:39 pm
Nope. Centrifugal (actually centripetal) force is entirely separate from the tides. See my previous comment.
w.
Willis Eschenbach says:
February 16, 2014 at 11:56 pm
“On my planet, people commonly study simplified models of complex situations, precisely to draw conclusions from them Happens all the time, I’m in mystery why you think we can’t learn valuable lessons from simple models.”
OK. Then at least do it correctly. So as to allow understanding instead of confusion. I obviously need to put my teaching hat on and explain how I think that this would best have been done so that there would have been a lot less squabbling all round.
Let us start from a very simple concept. A nearly featureless Globe, oriented vertically in the orbit, with a constant depth ocean, and a big ‘pin’ stuck down into the surface at 0 Lat, 0 Long.
Derive the gravitation forces that apply from the field diagram as below
http://upload.wikimedia.org/wikipedia/commons/d/d8/Field_tidal.png
Fig 1.
Let us then take some pictures of that from 3 directions starting from the longest possible intervals and working down towards human time frames.
The pictures are
1. looking down on the Earth from above at right angles to the orbit around the Sun,
2. looking along the orbit with the Poles North/South
3. looking out along the line drawn fro the Sun though the Earth.
This gives us x, y, z
Let us just deal with the Sun first, no Moon at all.
And we start taking pictures. At 4 year intervals to start with and at the Periapsis in the below.
http://en.wikipedia.org/wiki/File:Seasons1.svg
Fig 2.
Now we can see the ‘pin’ in the centre for 3, and pointing at the Sun for 2 and 1.
We can now deal with that oblate spheroid that the Earth’s rotation gives. We can also point out that most of the flattening is in the rock because the water skin is so thin. So all the centripetal forces, spin and orbit can be dealt with before we get into other stuff.
Now we speed up to once a year. And the first complication shows up. The ‘pin’ does not stay steady. It moves around those field diagrams in Fig 1.
So we draw out the first small cycle. One at 4 years. Then we move to 4 times a Year. Describe what happens there.
Now we have done the Solar components. Time to add in the Moon.
I am sure you can get the rest. Or do I need to do the full slide and description set?
Now is that a better way to do it or not? You tell me.
Oops sorry, Skipped the 4 times a year step to get all the Solar components. Must remember not to hurry. Overlook things if we do.
Willis:
That http://www.lhup.edu/~dsimanek/scenario/tides.htm link. I gave it too you on the previous thread about tides. I gave it to you on this thread about tides. Three other people have given it to on this thread alone as well,
Steve Mosher, E.M. Smith, Truthseeker.
Is it not about time you referenced it as well?
Giving as it does quite a reasonable explanation of how this all works and dispels a lot of misconceptions.
Mike M says: February 16, 2014 at 4:39 pm
Going slower at noon and faster at midnight WRT to solar orbital velocity
============
ferdberple says: February 16, 2014 at 7:15 pm the orbital speed of the earth around the sun is simply a change of reference field. Since the centrifugal force is exactly opposite the centripetal force (gravitational attraction of the sun) you can get the correct answer using the centrifugal force in place of the gravitational force.
==============
“the centrifugal force is exactly opposite the centripetal force” – I’m asserting that that is NOT true for the particle at noon and at midnight.
Forget earth’s rotation and all the canceling ‘on-earth’ forces for a moment and just assume there are three independent particles P1, P2 & P3 all present at earth’s orbital radius R from the sun. At “T ZERO”, P1’s velocity, V, exactly match’s earth’s tangential/orbital velocity so it’s V^2/R inertial reaction matches the centripetal force of sun’s gravity to make a perfect orbit.
Now consider P2 which going exactly in the same direction as P1 but it’s velocity is, instantaneously, (we are at T ZERO), slower than V by an amount S so, P2’s velocity is (V-S). For P2 the sun’s gravity is GREATER THAN the centripetal force needed to cause P2 to track the trajectory of earth’s orbit.
Left on it’s own, P2’s trajectory would decay from earth’s orbital trajectory toward the sun but we’re not going to allow it to do that… we’re going to add an outward acceleration, M, (away from the sun) to balance against sun’s gravity to make the centripetal force on P2 that required to keep P2 on earth’s orbital trajectory despite it’s slower speed. (Ask Buzz Aldrin what you do with a LEM to maintain an orbit around the moon at less than a free orbital speed…)
So I am contending that acceleration M, the additional amount of outward radial acceleration needed to keep P2 on earth’s orbital trajectory at the slower speed is: V^2/R – (V-S)^2/R
So where does M come from…? EARTH! Earth’s gravity G is way more than enough to keep P2 stuck to earth and keep traveling on earth’s orbital trajectory.
Particle P2 therefore weighes less by the amount:
(V^2/R -(V-S)^2/R) which I calculated to be approximately .0002M/sec^2
P3 is going faster so it’s velocity is (V+S) so you add M in the other direction … ETC.
No moon at all, two opposite acting amounts of differential acceleration at noon and midnight by the amount .0002M/sec^2 and, of course, S is the tangential velocity of a rotating earth’s equator WRT earth’s center.
There ain’t no such thing as a centrifugal or a centripetal force. There is only one relevant force, gravity, which accelerates the earth or moon, or any other planet, towards its primary. But as Newton’s Law (Which?) says that if a thing is moving it keeps on going in a straight line at the same speed UNLESS an external force acts on it, the earth (other planet, satellite, etc) moves subject to the result of momentum and acceleration roughly at right angles to the current direction of motion. Hence, the path is bent, result an ellipse.
Because of the size of the earth, the force of gravity is different on the side of the earth nearest to the sun to that on the side away. Hence, unless otherwise constrained, the two sides would then follow different orbits. But the two sides are constrained by the earth’s own attraction, and hence the earth does not fly apart. Still, it does its best, and the shell distorts, more so in the case of water which is more mobile, less in the case of the rocky portion. Hence tides and two bulges. Or, more properly, two bulges and, because of the earth’s rotation, two tides. QED.
Dudley Horscroft says: February 17, 2014 at 5:48 am “There ain’t no such thing as a centrifugal or a centripetal force. ”
Rubbish. HOW does anything turn without centripetal force? Do explain….
Dudley Horscroft says: February 17, 2014 at 5:48 am “There ain’t no such thing as a centrifugal or a centripetal force. ”
Centripetal force is that acting PRECISELY at a right angle to the velocity of a particle at all times hence – gravity on an object in orbit… always at a right angle … no change to the particle’s speed … only to it’s DIRECTION.
eyesonu,
“If I tied a brick to a rope and spun it around my head it would produce an outward force (centrifugal force) that would need to be countered acted by the rope. If we relate that concept of the orbiting brick and rope to a planet (brick) and gravitational attraction (rope) and call it a tidal force then would it be fair to say that the tidal force is the net result of gravitational attraction combined with the centrifugal force all wrapped into one package?”
You are confusing reference frames. From your point of view, spinning the brick, there is only one force which is the tension that you impart to the rope. This is called a centripetal force (centre directed) and it produces a centripetal acceleration (Newton’s second law). There is no balance of forces since this is a dynamic and not a static system. The concept of a centrifugal force (centre fleeing) arises when you place your point of view on the brick. Here you experience a “weight” directed outwards and to explain this, ie to make Newton’s second law work in this accelerating system, you have to invent a fictitious centrifugal force. You can’t have both the centripetal and centrifugal force at the same time. Compare this to the sensation of weight which is not the sensation of gravity which you can not feel, tidal effects aside, but the sensation of the normal force of the ground pushing you upwards. This is why centrifugal forces are not needed to explain the tidal effect since most explanations are done in a reference frame where they do not exist.
Willis,
I can’t seem to get a rise out of you. I guess that I’m just not trying hard enough. 😉
William Sears says: February 17, 2014 at 6:24 am “There is no balance of forces since this is a dynamic and not a static system. The concept of a centrifugal force (centre fleeing) arises when you place your point of view on the brick.”
A car is going around in a right hand circle. The centripetal force is supplied by the road surface sideways against the tires. The force is transmitted by structure to the left door which pushes on you centripetally to change your direction along with the rest of the car.
You ARE being pushed against the door by “centrifugal” force in the opposite direction of the centripetal force … as an inertial reaction to the centripetal force. This is no different than the force of the floor holding you up to counter the force gravity is imparting on you downwards. Two equal and opposite forces CANCELING OUT at the point of contact resulting in ZERO relative velocity between you and the floor. If you don’t think so – take way the floor. You begin accelerating down and … earth begins accelerating up.
Mike M,
“You ARE being pushed against the door by “centrifugal” force in the opposite direction of the centripetal force … as an inertial reaction to the centripetal force.” The door is pushing against you in order to impart the same centripetal acceleration that the roadway is imparting to the tires. It is only a centrifugal force in the non-inertial (accelerated) frame that you view as reality inside the car. The only force that you feel is the reaction (normal) force of the door or seatbelt. Centripetal versus centrifugal can be viewed as third law reaction forces but they apply to different objects in different reference frames.
You have misinterpreted my gravity – normal force statements which was a reference to the fact that in free fall you can not feel your own weight, an insight of Einstein. The same thing would apply to electrostatic attraction. Contact (normal, reaction or whatever term you prefer) forces are what you directly experience. Again tidal or tidal-like forces on an extended object are another question.
RichardLH says:
February 17, 2014 at 2:15 am
Why? It’s been referenced four times, and it says exactly what I said in the head post … was your citation somehow inadequate, such that it needed my assistance?
In any case, while that link is quite accurate, I found it much less complete than the reference yirgach gave and that I cited above … so sue me.
w.
Willis Eschenbach says:
February 17, 2014 at 10:00 am
“Why? It’s been referenced four times, and it says exactly what I said in the head post … was your citation somehow inadequate, such that it needed my assistance?”
Obviously not.
“In any case, while that link is quite accurate, I found it much less complete than the reference yirgach gave and that I cited above … so sue me.”
Down boy.
William Sears says:
February 17, 2014 at 6:24 am
Your problem is that you are far too reasonable, and your science-fu is strong, and you haven’t attacked me personally, or called me an idiot for not agreeing with you …
w.
I’ve known the Simanek page for some time and learning of it was a real eye opener.
Sorry if the following question is a needless distraction at this point in the discussion. Isn’t it so that a planet in orbit is following a straight path in (curved) spacetime? Do we actually still speak of gravity as a force then? Or is it just another imaginary force like the centrifugal force?
Thanks, Remmitt.