1. Nov 4, 2005

### WarrenPlatts

I think I might have a possible explanation for the peculiar, west-to-east rotation of Venus, where a day is longer than a year.

Venus "wants" to have one side always facing the sun, so that one rotation would equal one revolution (like the Moon). However, once Venus achieved this state, every time it was in opposition to Earth (at which point it is only .3 AU's away), it would also "want" to have one side always pointing at Earth. This represents a glancing, gravitational, inelastic collision that would induce a tiny westward "kick" to Venus everytime it speeds past the Earth that might be enough to cause the slow retrograde rotation. It's sort of like when a cue ball strikes a glancing blow on the left side of another pool ball, a left-ward spin is induced.:tongue:

I've completely forgot how to do calculus, so I don't have much of an idea as to how to quantitatively attack this problem to see if I am crazy or not. :uhh: Any ideas?

Last edited: Nov 4, 2005
2. Nov 4, 2005

### Andre

I do wonder what will happen to this thread.

3. Nov 4, 2005

### WarrenPlatts

Why? Granted, I probably should have put it in the celestial mechanics forum. . . . Also, I just found out the idea is not original to me, and that it is still being debated by astronomers so it's not entirely out of the mainstream. The main problem is that one would expect the tidal effects of the sun to predominate.

BTW, I did read your interesting paper on Venus, but it didn't mention possible tidal effects of Earth.

4. Nov 4, 2005

5. Nov 4, 2005

### SpaceTiger

Staff Emeritus
Do you have a reference for this?

6. Nov 4, 2005

### Phobos

Staff Emeritus
Hmm. I suppose I could be hard-nosed about this and direct you to the IR forum (probably a dead end without any math to present), but perhaps we can keep this going in the CM forum if we keep to the following (or similar):
(1) what are the main current scientific hypotheses/theories on this matter?
(2) how would we begin to quantitatively address Warren's hypothesis?

A more thorough working out of this idea would need to go to the IR forum.

7. Nov 4, 2005

### WarrenPlatts

http://astro.uchicago.edu/home/web/lucia/a100/lectures/venus.html#retro

(my emphasis)

Last edited: Nov 4, 2005
8. Nov 4, 2005

### Andre

There is absolutely no textbook consensus about the retrograde rotation of Venus. Consequently, anything goes or the first one who does not get discredited.

To introduce a tiny bit of math for Warren, what would be order of magnitude gravity difference for Venus to the Sun at 0,7 AU versus Venus to the Earth at a pseudo harmonic varying range between 1,7 and 0,3 AU?

See how Correia and Laskar disregard any notion about resonance with Earth:

http://www.imcce.fr/Equipes/ASD/preprints/prep.2002/venus2.2002.pdf

There may be wild reasons for the resonance, coincidence probably being one of them.

9. Nov 4, 2005

### Andre

10. Nov 4, 2005

### WarrenPlatts

I don't have much math, but some simple back of the envelope geometry allows me to make a prediction.

Every two years, the same side of Venus basically points toward Earth. Thus, consider the following table:

Earth Venus
Year Position

0 12 O'clock
2 9
4 6
6 3
8 12
Repeat cycle. . .

Assume that you're looking down on the solar system and the planets are moving counterclockwise.

Now consider this map of Venus:

http://www.solarviews.com/raw/venus/venmap.gif

One immediately sees that one side of Venus consists of dark lowlands and the other side is has lighter colored highlands; in other words its sort of like the Moon, which always has its mare basalts pointed at Earth.

Hence my prediction:

Every 4 years, when the Earth is at perigee (July 4), Venus will be in conjunction relative to the Earth (i.e., lined up with the Sun), and Venus's dark side will be facing Earth.

So, the question is: What time is it on Venus?

No, the night before last, I couldn't fall asleep until after sunrise because I was thinking about this. I know--I need to get a life. . . .

11. Nov 4, 2005

### SpaceTiger

Staff Emeritus
It would seem that astronomers are debating the connection between the earth and venus' rotation, but not for quite the reasons your model would suggest. First of all, the reason we suspect a connection is observational; that is, we observe a resonance between the earth's orbital period and venus' rotation period. From a theoretical point of view, it makes very little sense that the earth should be the dominant factor here. The tidal force is given by:

$$F_t=\frac{2GM_{perturber}M_{perturbee}\Delta r}{r^3}$$

where $\Delta r$ is the size of the perturbee and r is its distance from the perturber. To find the relative magnitudes of the tidal force from the sun and earth, you just do:

$$\frac{F_{t,earth}}{F_{t,sun}}=\frac{M_{earth}}{M_{sun}}(\frac{r_{sun}}{r_{earth}})^3$$

At the minimum earth-venus separation, this gives

$$(3 \times 10^{-6})(18) \simeq 5 \times 10^{-5}$$

That's pretty tiny. Any "kick" provided by the earth should be quickly corrected by the sun's tidal field. Also, it's not obvious to me why your argument should apply only to Venus. There are other planets in the solar system that could be given similar types of kicks by their neighbors.

It should also be noted that the theories suggested by the links you provided are qualitatively different from yours. Rather than spinning venus down and then up in the opposite direction, they're suggesting that it was simply flipped over. That is, the torque provided by the earth was parallel to the plane of the solar system rather than perpendicular to it.

I don't want to discourage you from thinking about these things, but your model is oversimplifying the issue a bit. In particular, your prediction:

implies not only that venus' spin period should be in resonance with the earth, but also its orbital period. This is known not to be the case.

12. Nov 5, 2005

### WarrenPlatts

Actually, there is a remarkable 8:13 resonance between the orbital periods of Earth and Venus, respectively, that is even tighter than the rotation resonance.

cf.: http://www.exo.net/~pauld/activities/astronomy/transitvenus/venustransit2004.htm

13. Nov 5, 2005

### SpaceTiger

Staff Emeritus
The existence of a small, approximately natural number ratio in the periods does not necessarily mean that there is a resonance. In this case, it's probably a coincidence. To my knowledge, the only recognized orbit-orbit resonance of the planets is between Neptune and Pluto (3:2). Really, we probably shouldn't even be calling the spin-orbit relationship between earth and venus a "resonance", more like a "near-resonance."

Anyway, it would still be inconsistent with your prediction, which implies a 2:1 orbit-orbit resonance between Venus and Earth (or 4:1 or 1:1, but I'll assume you weren't suggesting these).

14. Nov 5, 2005

### Andre

let's check that resonance.

Factsheet: http://nssdc.gsfc.nasa.gov/planetary/factsheet/venusfact.html

So with sidereal years (224.701) and days (243.0208) basically after every two Earth years (730.512 Earth days) Venus has completed 3.005965 Venus days and consequently is 1,44 hrs beyond the same orientation.

After eight Earth years (2922.048 days) Venus has just completed it's 13th year (2921.113 earth days, 12.02001 Venus days) So they have approximately the same orientation again reference each other like the closest distance to each other. This is after 12.02386 Venus days, so Venus is about 5 hrs 45 minutes beyond perfect resonance.

In this position Venus always shows it's dark side of course(shadow)

Nice map, but forget about the colors. It's radar imagery just reflecting radar reflectivity. Smooth surface is dark, rough surface is bright. Brightness also increases with higher metal contends of the rocks. The strong reflectivity of the higher terrain is also a formal Venus mystery.

15. Nov 5, 2005

### tony873004

There's a big difference between ratios that are almost integers and ratios that are integers. And ratios that are integers aren't necessarily integers at any given moment, but over long periods of time, they average into integers.

Take a look at the two illustrations. The first one shows the Sun, Venus and Earth in what last month's Sky & Telescope magazine called a "Kepler's Pretzel". Notice how the top loop is a little thicker than the rest. That's because I let it run one loop beyond a complete cycle. After Venus traced its 5 loops, it began to almost repeat the pattern. But it is not repeating the pattern, and ultimately each loop will advance upon and pass the loop ahead of it.

The second illustration is an animated GIF of Neptune and Pluto in a rotating frame. Neptune is the almost-stationary blue dot. Pluto's orbit is the purple path. It shows a completely different story for Neptune and Pluto. At any given moment, their periods are not exactly 3:2. So Pluto’s perihelion advanced upon Neptune’s position. But when Pluto's perihelion advances too close to Neptune, orbital energy is exchanged which ever so slightly changes Pluto's period. This prevents it from coming any closer to Neptune. Pluto's perihelion then spends thousands of years advancing in the opposite direction until it encounters Neptune from the other side. Then the exchange of energy happens again and repels Pluto's perihelion once again, preventing it from ever coming close to Neptune.

In the Earth/Venus situation, no such mechanism exists. So their periods are just coincidentally a ratio of approximate integers. But the Neptune/Pluto situation corrects itself when it tries to go astray, ensuring that its ratio averages Exactly 3:2 over long periods of time.

As an analogy, picture 2 cars driving down the freeway at almost the same speed. They appear to be in a 1:1 ratio with each other. For every 1 mile the first car drives, the second car also drives 1 mile. But the drivers are strangers to each other. They are just coincidentally driving at about the exact same speed, and in the same direction. After 100's of miles they are likely to be quite a distance from each other, especially if one stops for gas, or a rest, etc. It becomes obvious that their 1:1 resonance isn't exactly 1:1.

Now imagine 2 drivers that know each other. They are driving together to the same destination. They also appear to be in a 1:1 resonance. But not only do they have similar speeds, but when their distances get too great, they correct for that to maintain an acceptable spacing. If one has to stop for gas, and the other one doesn't need gas, he pulls over anyway. After thousands miles, they are still within a few hundred yards of each other. Although at any given moment their speeds might not be exactly the same, their speeds, and therefore their resonance averages exactly 1:1 no matter how far they drive.

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16. Nov 6, 2005

### WarrenPlatts

Eurekaaaaaaa!!!!!!!!!!!!!!!!!!

I figured it out at 2AM Saturday morning, but after that I slept good.

It makes predictions and is surprising in ways even to me. :surprised However, I'm afraid to post the solution here quite yet. But believe me, it works! :!!)

Meanwhile. . . .

SpaceTiger: using your formula, I ran the numbers for Jupiter, and Earth exerts 10 times more tidal force than Jupiter, the only other possible contender. Granted, the Sun exerts 25,000 times more force than Earth. But think of a heavy pendulum. A small impulse timed just right will still move it. What I need to do is integrate the total impulse exerted by Earth over over 144 Earth days from the maximum eastern elongation to the maximum western elongation (this is just a line from Earth tangent to the orbit of Venus), taking into account the motion of the planets, and then see what the net force direction is and how much of a torque this could apply to the rotational angular momentum of Venus. This should only require basic calculus, and Newton's law of gravitation.

No, no, my friend. At the risk of giving away too much, what I am suggesting is that in addition to the 2:3 and 8:13 resonances, there are also 5:8, 5:13 and 12:13 resonances at work, and probably others that I haven't discovered yet! :!!)
Believe me, I have an elegant solution to this problem.

However, I wonder about the accuracy of the measurement of the length of the sidereal rotation. I understand that this has been determined using Earth-based radar. Thus, I need to know what the expected error is of these measurements. Most figures I've seen just say 243, I saw one that said 243.1, my CRC handbook says 243.01, yet based on the NASA website, you say 243.0208--7 significant figures. My model suggests that these estimates are off by a couple of one hundreths of one percent--about an hour. To put this figure in perspective .0002 of 24 hours is about 17 seconds--small, but significant. So, this could be a fatal flaw for the resonance model; on the other hand, if future revisions of the day length become closer to that predicted by the resonance model, that would be strong confirmation that the model is true.

Also, I haven't yet taken into account the eccentricity of Earth's orbit, and certain precessions, recessions or librations might account for the difference.

My point exactly! This is evidence that the mass distribution within Venus is anisotropic with respect to the volume of Venus--just like the Moon.

Like I said, if one fudges the Venus day length a tiny bit, the resonance model is extremely precise. This model does imply that Venus is in dynamic equilibrium, and so there must be some sort of mechanism that has to do with the Earth that maintains this equilibrium. Just because we don't know exactly how this mechanism works does not imply that such a mechanism does not exist.

Regarding the Kepler's pretzel, are the loops supposed to represent the rotation of Venus, or the crazy orbital path of some small object? Also, I downloaded gravity simulator, but couldn't get the time step command to work. Any suggestions?

Bottom line is that there are way too many cooincidences--and this requires a special explanation. It's like in biology. A watch implies a watchmaker, or at least natural selection. Rather than trying to derive Venus's rotation from first principles, a better approach is to assume that the rotation is there because it is a good design for the intrinsic value of least action. One's task is then to reverse engineer this design--a top down approach. Physical science can benefit from a healthy dose of biology-style teleology.

Last edited: Nov 6, 2005
17. Nov 6, 2005

### tony873004

Sorry, I didn't explain the Kepler's Pretzel well. This shows the motion of Venus in an Earth-centered universe. It doesn't represent Venus' rotation. The dot in the middle represents Earth, the yellow circle represents the path of the Sun around Earth, and the white pretzel represents the path of Venus around the Earth. No wonder this model drove the ancient people crazy! In a day and age when we know that Earth is not in the center, Kepler's Pretzel still helps to illustrate things.

Each of those loops represents a Venus inferior conjunction, when Venus makes a close pass to Earth. They happen about every 1.6 years. The next one will happen in January. After every group of 5, the pattern repeats itself. That's why transits of Venus happen in pairs, spaced 8 years apart (1.6 * 5 = 8).

Mars is another interesting object to view in a Kepler's Pretzel. It is currently at a close point to Earth, making a loop. If you run time backwards, you can see that the loop it made 2 years ago was even closer. Hence, the 2003 opposition was the closest in thousands of years. Run time forwards and you will see that it will not loop as close to Earth as it does in 2005 until the year 2018.

To make a Kepler's Pretzel in Gravity Simulator, open the simulation "onlyplanets.gsim". If the user controls are not visible on the screen, press F8 or F9. Then choose Earth as the "Focus Object". Make sure your trails are turned on (The T button), and increase the Time Step. There's a + button on the Time Step control. But don't exceed 16384 or you'll be introducing math errors into the simulation, and your planets will wander from their true positions.

18. Nov 6, 2005

### Andre

Careful, it's illegal to have new ideas here.

Well the mapping of Magelan would probably have been a mess if that rotation wasn't accurate to a good couple of decimals.

I strongly suggest to consult the Correia and Lasker paper I linked to earlier, also part one:

http://www.imcce.fr/Equipes/ASD/preprints/prep.2002/venus1.2002.pdf
http://www.imcce.fr/Equipes/ASD/preprints/prep.2002/venus2.2002.pdf

19. Nov 6, 2005

### SpaceTiger

Staff Emeritus
I think you're missing the point. I'm not saying that the earth has no effect on Venus, I'm saying that it wouldn't make sense for Venus to be "locked" to earth instead of the sun. In your pendulum analogy, any impulse I give it (regardless of timing) will increase its energy, changing the behavior of the pendulum. In the small angle approximation, the period of the pendulum won't increase, but the amplitude of oscillation will.

In the case of "competing tides" between the earth and sun, the forces are behaving quite differently. Imagine that Venus' rotation period is the same as its orbital period (that is, it's tidally locked to the sun). Then imagine that earth comes along and provides a little impulse that tries to lock Venus to its orbital period. What happens? Well, the earth will certainly perturb Venus, such that its rotation period doesn't exactly match that of the sun. However, as the earth and Venus leave conjunction, the impulse from the earth will decrease significantly and the sun will simply "pull" venus back into a lock with its orbit. This is in contrast to the pendulum, where there was nothing to "pull" it back to its original state.

But what about all the angular momentum and energy transfers that are going on? It's certainly true that, when the earth interacts with venus, they exchange energy and angular momentum. When the sun pulls Venus back to its original spin rate, does this mean that the energy and angular momentum are returned to their original state? No, the second interaction was between the sun and venus, not the earth and venus! In fact, the earth's orbit will change as a result of the interaction. Likewise, Venus' orbit will change as a result of its attempts to lock the earth's spin period.

As you can see, the picture is rapidly becoming complicated. In fact, the interactions I've described so far are of even higher order than most of the "perturbations" we consider when calculating orbits. As was already stated, tidal locking is the result of tidal forces, which are much weaker than direct gravitational forces. Not only are all of these forces difficult to compute, but it turns out that the system will actually display chaotic behavior as a result of these perturbations.

I was being generous I said you were oversimplifying things "a bit". You're not far off when you say that the problem requires only "basic calculus, and Newton's law of gravitation", but I think you underestimate the complexity that can arise from a many-body gravitational system.

Your prediction, as it was stated originally, did in fact imply one of the resonances I mentioned earlier, not the ones you give above. The mathematics that go into predicting high-order resonances are extremely complicated, so I suspect you're just doing numerology here.

Last edited: Nov 6, 2005
20. Nov 6, 2005

### WarrenPlatts

:rofl: :rofl: :rofl: OK! OK! OK! :!!) :!!) :!!)

I can't resist it, so I'll give you my results so far. I guess this place is legit enough to establish my priority if no one else has thought of this (which they probably have, although this problem has only been around for 40 years or so). Maybe if it's original, one of you all can help me coathor a little note to Nature or at least The Journal of the British Interplanetary Society

ANSWER: The real resonance is not between Venus's sidereal rotation and Earth's orbital period--the REAL resonance is between the Venus/Earth synodic period and Venus's diurnal day. :tongue2: :tongue2: :tongue2:

I was struck by the 8 Earth year: 13 Venus year resonance. Yet I realized what's really important are the inferior conjunctions: this is when Earth's gravitational "kick" is going to be the strongest. Yet, I thought it would be odd if the 8th year conjunction happened in exactly the same place every 8 years. So, I ran the numbers: assuming circular orbits and an Earth sidereal year of 365.2564 and a Venus sidereal period of 224.701 days, these planets sweep out .985609 and 1.60213 degrees per day as they orbit. Sure enough, after exactly eight Earth years (2,992 days), the planets are not in conjunction: the actual conjunction occurs 2 days earlier, and 2 degrees to the right (i.e., clockwise looking down). In other words, the 8-year inferior conjuction regresses about 2.8 degrees every 8 years.

So, I thought AHA!: if my theory is right, what I call the "heading" of Venus will be pointing at Earth during the 8-year, rather than 000 as it was the first time. (That is, for my simulation, I arbitrarily defined an inertial coordinate system based on the Sun, using degrees as the unit of measurement. Thus, at the start, both Earth and Venus had an initial bearing from the sun of 000. Then I defined the "heading" of Venus as the local midnight meridian pointed at the Earth at the moment of the first inferior conjunction. Thus, the first heading of Venus was 000.) However, after 8 Earth years (13 Venus years) the inferior conjuction bearing from the Sun was 002. So, I predicted that the Venus heading at the time of conjuction would also be 002; and guess what? The prediction worked out. (Sorry about talk of bearings and headings--I used to be a Navy quartermaster, so that's naturally how I think about these things.)

So then, for the hell of it, I decided to see what Venus's heading would be at the other inferior conjunctions (which occur, of course, every 584 days--which is the Earth-Venus synodic period). And to my surprise: It turns out that the Venus heading matched the Earth/Venus solar bearing. In other words, Venus presents the same side to Earth AT EVERY INFERIOR CONJUNCTION!

Now, I ran my simulation (on an Xcel spreadsheet) out to 64 years (23,257 days), and granted it got a little off. So, I tweaked the Venus sideral day to 243.1675 (versus 243.1 that I had started with), and this made it work out perfectly after 64 years.

And today, after reading Tony's latest post, I found a new resonance: The reason I asked if the loops in his diagram were Venus's rotation is because they match the inferior conjunction period (i.e., 5 conjunctions per 8 Earth years). Tony said they did not, so I checked my simulation again to see about Venus's diurnal days. And again, to my surprise turns out that there are exactly 5 day/night cycles per inferior conjunction! In other words, Venus gets a "kick" out of the Earth every 5 Venus days (diurnal days, that is). :!!):!!):!!):!!) :surprised

Thus, my theory makes three concrete predictions (which is more than Correia and Lasker, et al. can say): (1) the measured rotational period of Venus is too low by about an hour; (2) there is a density assymetry within Venus; (3) one side or the other defined by this axis will always face Earth every inferior conjunction. And prediction (1) might not pan out because of Earth's eccentricity. Tony says the next 8 year conjunction will occur in January--which is during Earth's perihelion. But there are five conjunctions per 8 years, and it is not a circular pattern--rather it goes in a five pointed star or pentacle pattern. This means that the opposite two points are happening at aphelion now. What effect this has, I don't know, but it's bound to throw things off a little over the short term. Considering that the 8-year inferior conjunction recession cycle takes about 1,212 years to complete, and we've only had any idea about Venus's rotation in the last 40 years, an hour off now might be made up 500 years from now. Actually, now that I think about it, the point of the star will be at aphelion in only 121 years from now. So, if it's off by an hour now, it should make itself up within the next century or so (prediction #4).

So, SpaceTiger, how many more coincidences have to pile up before you are willing to consider the alternative hypothesis? Are you telling me that a 25 kg pendulum could not be caused to wobble by a 1 gram impulse precisely timed to its theoretical, small amplitude period?

My hunch is that Venus's orbit is just about perfectly circular for a reason; that is, it is locked into its circular orbit. Therefore, any momentum transfers between Earth and Venus will have to come out of Venus's rotational angular momentum, as opposed to its orbital angular momentum. Probably, the Earth sucks momentum out of Venus, and this is made up by Venus sucking angular momentum from the Sun. But what do I know--I'm a mere graduate student of philosophy.

Last edited: Nov 6, 2005