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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Kevin Kilty,
There is only acceleration and deceleration involved. Get back in space in my thought experiment. There are only three forces involved. (1) Gravitational attraction between the two experimental bodies, (2) gravitational attractions with the distant body/force, and (3) inertia (decreasing and increasing) as a result of the gravitational forces involved. I used in my thought experiment a case where two weights/bodies were being effected by a distant gravitational force/attraction.The speed of the bodies would increase toward that distant force but would be in a sense a “stop and go” approach and kind of in a zig zag manner from an individual viewpoint.
My thought experiment didn’t include a thought experiment of the two weights/bodies already in orbit. The reason for that is that it’s not likely anyone would have stayed with this thread were they not pretty intelligent.. If one who has been following this thread this long and grasp something from my contribution you can “bet your ass” they immediately applied that to a planet/moon system in orbit around a much larger body/sun.The results would be similar, just that the distant force would now be a 90 degree angle from the general planet/moon direction at any given point in time. In this case the distant force is maintaining a relatively steady orbit (practically speaking but likely an ellipse of sorts) and not increasing the speed/velocity of this planet/moon system toward that distant force/body.
Forget the centrifugal and centripetal force. The only forces are only gravitational attraction and inertia. Look an earlier post where I commented that it would look like a big drink and a smaller one running around trying to keep a bungee cord taught while they walk down the road. As they go down the road they will need both lanes.
Willis and Lief are correct in the concept.
If you can grasp what I’m trying to show then the tides (as we want to relate to water) is a piece of cake
Lief and/or Willis, don’t go away. I need some clarification on the Roache limits. I’ll see if i can comment ASAP.
It’s one of those things I just gotta know.
Here are a couple of thoughts I have orbiting in the gray regions of my gray matter. I gotta ask rather than state, but my brain don’t wanna let go. Consider the case where a planet/moon system were to be on a trajectory towards the earth’s sun. It is not in orbit around the sun but headed directly (as much as can be stated in a warped time/space concept) towards the sun due to that gravitational attraction between the two (sun and planet/moon system). Now my brain is entertaining thoughts that as this planet/moon system approaches the sun that the orbital relationship will change to a more elliptical configuration and at some “sweet spot” near the point where the sun’s gravitational pull equals the planet’s gravitational pull will cause the planet and moon to collide. If this could be possible then would that occur before the Roache limit is reached? Oops, the Roache limit would be dependent on the axial/rotational velocity of the planet or moon and also on the composition of those so consider something like maybe Mars or Earth’s moon. In my desire to limit words on this I hope I have given you the idea/concept that is on my mind. So I guess the general question is: Would the planet and moon likely collide before the Roache limit of either is reached or would we we looking at a “sweet spot” where they could be close to the same point?
Then of course that would bring up the direction of an angle of approach of an hypothetical planet/moon system from the cosmic depths (with consideration of a relationship to the orbital plane that the sun’s (earth included) planets lie). These thoughts seem to be directed towards that orbiting (around the sun) belt of rocks between Mars and Jupiter. Does anyone really know where the rocks (in orbit around the sun) came from?
Time constraints (on my part) limits my expression of thoughts/explanation much. Please cut me some slack for an incomplete explanation, after all those still here are having issues with an explanation/consensus. But if you can interpret my thoughts I would really appreciate some discussion. Let me say this, I agree with Willis and Leif in what has been presented.
I just went somewhat off topic in my thoughts but those thoughts are directly related to the concept of this thread. Ohh, the unconstrained mind.
Gotta go, I saw a rabbit run into a brush pile over yonder.
eyesonU;
Ro[
a]che LimitYore spellin’ bugs me.
There’s a reason Willis’ scenario involved a stationary sun and moonless planet, falling directly (not orbiting). 3-body orbital mechanics are currently insoluble, beyond brief (cosmically speaking) periods.
For more phantastical speculations, come up with explanations for Venus’ retrograde and very slow (but not quite tidally locked, unlike Mercury’s) rotation (but conventional orbit). Probably involves flipping at some point. Maybe discarding a moon, which became Mercury?? Rabbits galore!
There’s also a Rock limit, which is the closest distance a even a mineral or metallic meteorite/comet can get to the sun without vapourizing.
Brian H says:
February 22, 2014 at 5:41 pm
==========
I have to agree with you about the spelling. Got a lot of thoughts in my head that weren’t there when this thread began. Trying to pull ’em out and write ’em down before this thread dies put me under pressure.
My brain has been having a great time with my new found knowledge as a result of Willis’ post. And it ain’t over yet!
Brian H says:
February 22, 2014 at 5:41 pm
==============
Now you got me thinking about Mercury/Venus idea to go along with my hypothetical possibility of a planet/moon collision as mentioned earlier. Venus does rotate about it’s axis opposite the other planets. A side swipe collision at an orbital location between Mars and Jupiter would have left a debris field and probably separated the two with momentum towards the sun and separate orbits. Now just have to figure out an angle of approach towards the sun and we got us a brand new theory. LOL
If anyone is still following this thread please stay tuned. My brain is acting like a good bird dog and may have sniffed out the track of a couple of celestial rabbits and seems to be on point now. Kept waking me up all night.
I’m going to try to pull my thoughts from the deep regions of my mind and put ’em on digital paper today.
I might be nuts so it could be interesting to watch me go off the deep end right here on WUWT. But then maybe not. Hard to say as I’m talking to a computer screen at the moment! ROFLMAO
Leif, where are you?
eyesonu says:
February 23, 2014 at 9:37 am
Leif, where are you?
Still here, but ‘im Westen nichts neues’.
clivebest says:
February 23, 2014 at 3:01 am
Thanks for that, Clive, but I’m sorry—you still don’t need centrifugal force. Here’s a thought experiment to show why.
Suppose you are in an elevator in free-fall in space. Now, suppose you end up “passing close by the earth”. Do you experience any centrifugal force? For example, as the elevator’s path is curved by passing the earth, are you slammed up against the elevator door like you get slammed up against the car door in a car going around a tight corner?
Nope. Not in the slightest. There is no discernible force exerted on you that would sling you to the outside. You could slingshot around the earth and get slung in the direction of Jupiter … and from inside the elevator, you’d never know it. As far as you are concerned you are in undisturbed free fall. Getting slingshot past the earth has absolutely no impact on you … EXCEPT for the tidal effect. That tidal effect you can measure from inside the elevator.
But it is totally separate from the centrifugal force, which you cannot feel at all inside the elevator. This means that from the perspective of the elevator, we do NOT need to include the (undetectable from within the elevator) centrifugal force exerted by passing close to the earth.
Regards,
w.
Willis,
Sorry to drag this on yet again ! However lets consider that the asteroid consists of your 3 masses joined together with springs. As they approach the earth an observer on the first mass begins to see that the spring tension gradually increase as the other two masses apparently start to repel each other and move apart. At closest approach the separation reaches a maximum and so does the tension in the spring. However gradually the tension begins to reduces until eventually the separation of masses are exactly the same as they were before.
Then the observer looks outside and finds that the star constellations have dramatically changed. Everything has now been rotated by 70 degrees. He therefore concludes that a torque has been applied to his system.
The real mystery here is the equivalence of inertial mass and gravitational mass. General relativity kind of explains this by explaining that mass/energy distorts space-time. Your lift shaft or free falling object then becomes a geodesic in curved space.
Would a spin dryer still dry clothes in an empty universe ? Centrifugal forces are not completely trivial.
This is the only point I want to make.
Willis Eschenbach says:
February 23, 2014 at 10:38 am
“Nope. Not in the slightest.”
Wrong. This would only be the case if your CG were precisely coincident with the CG of the elevator. If it is off even a smidgeon, you would be pushed to the wall in the direction of your CG offset.
Take two satellites in orbit. Draw an imaginary box around them, and call that your elevator. The satellites are at two distinct altitudes. Are they going to stay in your “elevator”, locked in the same relative positions? Hardly.
There is no difference here in your thought experiment. You and the elevator are two different satellites. Unless you are in the same orbit, your paths are going to diverge.
I dislike it when people who are untutored and unaware of the subject upon which they are pontificating make unsupported assertions based on fallacious mental pictures. I really hate it when they are smug about it. Smug ignorance is what led us to the climate change fiasco. You would be well advised to choose a different path.
clivebest says:
February 23, 2014 at 1:25 pm
“General relativity kind of explains this by explaining that mass/energy distorts space-time. Your lift shaft or free falling object then becomes a geodesic in curved space.”
Yes, and the curvature tensor tells you how neighboring geodesic lines will diverge. According to Willis (and, Leif), spacetime magically flattens in the vicinity of a distributed mass. I feel like I am arguing with cavemen.
Bart says:
February 23, 2014 at 1:43 pm
You make my point perfectly. The difference is the tidal effect, and it exists in Figure 2, nothing to do with centrifugal force.
w.
Bart says:
February 23, 2014 at 1:52 pm
I’ve given you a link to five different mathematical calculations of the tidal force, using five different frames of reference. In every one of them, the centrifugal terms cancel out. I’ve pointed out that you can’t beat math with your mouth, and invited you to show where the math was wrong.
Rather than even attempt that, you just give us more mouth … prate all you want, Bart. Until you can show the math is wrong, you’ve done nothing. Your mouth doesn’t beat the math and never will. So either get onto the math, or shut the mouth—it’s not doing you a bit of good.
I give you math. You give us mouth. Just who is the caveman here?
w.
Willis Eschenbach says:
February 23, 2014 at 6:34 pm
“You make my point perfectly. The difference is the tidal effect, and it exists in Figure 2, nothing to do with centrifugal force.”
You are making my point. Your thought exercise is for two unattached bodies, unconstrained in motion relative to one another. When there is a constraint on how they move with respect to one another, that requires constraint forces, and those constraint forces depend upon the relative centripetal acceleration, as well as gravity.
You have simplified the problem too far. A distributed body is not in free-fall. Only its center of gravity is in free-fall. Every other part of the body is in constrained motion with respect to that point. Otherwise, the component trajectories would diverge based on their relative positions and velocities within the gravity well.
Willis Eschenbach says:
February 23, 2014 at 6:40 pm
“Until you can show the math is wrong, you’ve done nothing.”
Willis, you appear to be operating under a misapprehension of what I have been telling you.
E.g., your link here:
I am in no way suggesting that the second tidal bulge is created by centrifugal “force”. If you look back upthread to my discussion with Steve Fitzpatrick, you will see that I clearly and unequivocally disagree with that statement. I thought that should have been pretty clear by now, but it appears perhaps you missed that discussion.
Since I am not claiming this, your links denouncing those who do are inapplicable.
What I am telling you is that the centripetal acceleration further elongates the bulges. Look again at my comment here, which touched off this argument.The centripetal accelerations add additional terms which create greater tension, resulting in greater bulging on both ends.
Leif claims that this additional term is due to “spin” of the Earth, and therefore should not create any differential bulging. But, what he calls “spin” is measured with respect to a non-inertial frame, so that is an incorrect interpretation. The inertially fixed point is the barycenter of the two body system, and it is with respect to this that you must measure the dynamically important variables. The fact that, in steady state, the differential accelerations at equal radii from the Earth are equivalent from either point of view does not make them equally valid.
For example, suppose we had a space vehicle with four tensile cables with probes mounted at the ends. Something like, maybe, this beast. If the spin rate is greater than the orbit rate, then the probes will extend out due to the centrifugal “force” of the spin. However, at spin rate equal to the orbit rate, we have a resonance condition. If the configuration were initially with one probe extended down, one up, and the other two side to side, the two side to side ones would not stay there, held in place by the putative spin. They would start oscillating and eventually, due to energy dissipation, they would either flop upwards or downwards to lie alongside the radially distributed ones. The CG of the system would continue in orbit without change, but the location of that CG amoung all the bodies would shift. These kinds of dynamics are well known and extensively studied in space physics.
Just so, the centripetal acceleration of the Earth also creates two tidal bulges. Not a uniform displacement, because displacements at the sides would be unstable. These bulges add to the two bulges present due to gravitational forces. I would go further, but detailed explanation of how this happens is rather involved, and I am obviously having enough trouble getting my point across as it is.
Bart:
Do you accept that there is approximately 21 km of bulge created in the rock by Earth’s axial rotation alone?
Do you accept that there is approximately 0.3 m of bulge created in the rock by the Sun and Earth’s axial rotation and orbit?
Do you accept that there is approximately 0.6 m of bulge created in the rock by the Moon and Earth’s axial rotation and orbit?
Do you accept that the bulges caused by Earth’s axial rotation and orbit in the oceans are lower in magnitude than those that occur in the rock?
Do you accept that the orbital changes that modulate the above produce even smaller variations still?
RichardLH says:
February 24, 2014 at 3:56 am
“Do you accept that the orbital changes that modulate the above produce even smaller variations still?”
The contribution of the centripetal acceleration about the Sun to the stress produced along the Sun-Earth line is 50% of that induced by gravity alone. That is non-negligible.
Something which has been missed in all this is that much of this is convention. You can solve for the equipotential surface in either an inertial frame, or in the rotating frame of reference. The latter is, IMO, conceptually simpler, but if done properly, the results should be the same. What I call the effects of centripetal acceleration in the rotating frame can be designated the relative acceleration of the Earth in the inertial frame. Since the centripetal acceleration is a function of the gravitational pull of the Sun, one can make the claim that it is all gravity. All roads lead to Rome, or at least, all good roads lead to Rome.
This tempest in a teapot started when Leif suggested that my equations above were wrong, because I had committed the sin of speaking of centripetal acceleration. It is a sin, because the usual connotation of incorporating fictitious forces into the solution produces the erroneous conclusion that the bulge is two-sided because of the centripetal acceleration. Willis, I think, also believed that was my position, and would have been correct in calling me down upon it if that were, indeed, what I was claiming.
But, my equations are correct, and it is a fully valid way of looking at the problem. It is the standard approach for calculating the tension in the tether for one of these animals. It was, in fact, Leif who was wrong, because he was looking at the Earth as a static reference frame, and his calculation for the delta-specific force was off by a factor of 2/3. I was put off by it because the equations I presented above are so direct and simple, and obviously correct, that if anyone actually took the time to look at it, one should immediately see that they are correct.
In want to post here and will do ASAP but gravity is causing me an issue. Seems CAGW snow has caused a collapse of the roof on the barn.
FWIW the rocky belt would have been caused by a planet/moon system approaching\g from the rear (generally) of the direction of the sun’s travel. change the direction of the force causing a change of the gravitational orbit and BOOM, possibly.
Above comment done in haste. I should have used spell check before posting but long day and another unwanted project due to collapsed roof has my needed attention and will take considerable effort. Lesson learned, if to know you might need a series of tension cables (structurally speaking) in the event of extremely heavy snow but it’s July while you’re building, well don’t let it slip your mind. That heavier than usual winter snow will come even if it takes a few years.
Gravity is a powerful force. Don’t take it lightly.
This thread has really opened my mind to much thought and I look forward to some discussion sometime in the future. Terrestrial gravitational forces are the most pressing need from my view point at this time.
Bart:
“The contribution of the centripetal acceleration about the Sun to the stress produced along the Sun-Earth line is 50% of that induced by gravity alone. That is non-negligible.”
Fact checking the link I have already provided (as have many others)
Tidal trivia.
•Amplitude of gravitational tides in deep mid-ocean: about 1 meter.
•Shoreline tides can be more than 10 times as large as in mid-ocean.
•Amplitude of tides in the Earth’s crust: about 20 cm.
•Ratio of sun/moon tidal forces on Earth is 0.465.
•Angular velocity of Earth’s axial rotation: 7.29 x 10-5 rad/s.
•Angular velocity of moon’s revolution around Earth: 2.67 x 10-6 rad/s.
•Earth polar diameter: 12710 km.
•Earth equatorial diameter: 12756 km.
•Difference between these diameters: 46 km.
•Difference between these radii: 23 km, or 0.4 %.
•Centripetal acceleration at Earth’s surface due to Earth’s axial rotation: 0.034 m/s2
•Size of centripetal acceleration at Earth’s surface due to Earth’s circular motion around the Earth-moon barycenter: 3.3 x 10-5 m/s2.
Add to those
6,378.14 kilometres of rock (average).
3.8 kilometres of water (average)
Centripetal forces
Earth rotation 0.034 m/s^2
Earth orbital round Sun ~0.0000952 m/s^2
Angular velocity of Earth’s axial rotation in radians per second = 7.29 x 10^-5 rad/s
Angular velocity of Earth around the Sun in radians per second = 1.99 × 10^-7 rad/s
Earth’s average orbit of 149.59787 * 10^6 kilometres
Earth’s equatorial radius of 6,378.14 kilometres
“Earth orbital round Sun ~0.0000952 m/s^2”
You’ll want to check that calculation again. Make sure you are using consistent units.
Earth rotation 0.034 m/s^2
Earth orbital round Sun ~0.0000952 m/s^2
Angular velocity of Earth’s axial rotation in radians per second = 7.29 x 10^-5 rad/s
Angular velocity of Earth around the Sun in radians per second = 1.99 × 10^-7 rad/s
As there is a 10^-2 between the two angular velocities I would have thought it was correct. Do you have a different value, and, if so, how did you calculate it?
But, the radius about the Sun is much greater than the radius of the Earth.
Of course, but that makes my point not yours. It is, after all, the difference between a straight line and the curve that is the reason for the Centripetal force. Smaller difference, smaller force.
The next centripetal acceleration down in size is the one I listed above.
Size of centripetal acceleration at Earth’s surface due to Earth’s circular motion around the Earth-moon barycenter: 3.3 x 10-5 m/s2.
Then we get the orbital round the Sun.