# The Moon’s eccentric orbit

As far as I can tell no long term study of lunar eccentricity has been done, so  I decided to begin investigating this myself, even though I am a complete novice in astronomical calculations. The key resource for all planetary motions in the solar system is the JPL ephemeris, and the easiest way to access it is through their Horizon web interface. Initially I ran a simulation of the lunar orbit relative to the earth-moon barycentre over a 10 year period beginning in december 2000.

I was amazed  to discover that the moon’s orbit around the earth is anything but simple and the quoted parameters are just approximations. The eccentricity of the orbit is changing almost on a daily basis due to complex variations in gravitational  effects depending on the relative positions of the sun and the earth, and also Jupiter and Venus. Figure 1 plots the eccentricity from Dec 12 2000 until October 12 2010 on a daily basis. The eccentricity varies in total between the extreme values of  0.026 and 0.077. This is a huge range which can alone  change the strength of lunar tides up to a maximum of 20%.

Figure 1: Variations in eccentricity of the moon’s orbit around the Earth-Moon barycentre. The blue curve is a fit to two oscillations with time periods 31.8 days and 205.9 days. Click to expand. Note Start Date should have been 18th November 2000 !

There are at least 2 regular resonances which at first sight seems odd because neither coincide with the orbital period of the moon (27.32days) nor that of the earth (365.25 days). There are also beats in the amplitude. Following this german article, I made a least squares fit shown as the blue curve which reproduces almost perfectly the signal .

eccentricity(d) = 0.55 + 0.014cos(0.198*d + 2.148) + 0.0085cos(0.0305*d +10.565)

eccentricity(d) = 0.o55 + 0.014cos(0.198*d + 2.148) + 0.0142cos(0.0305*d +10.565) where d is the number of days after 18/11/2000

This variation in eccentricity changes the perihelion distance from the earth significantly causing  large variations in the strength of spring tides on a yearly basis. The eccentricity becomes a maximum when the semi-major axis of the orbit lines up with the sun. This happens every 205.9 days – more than half a year due to the precession of the orbit every 18.6 years.  The 31.8 day variation is I think  the regular orbital change in distance from the sun.

Horizons only goes back as far as 8000 BC, so in order to investigate paleoclimate effects of eccentricity of lunar orbit around the time of the last interglacial we now need to find another tool.

PhD High Energy Physics Worked at CERN, Rutherford Lab, JET, JRC, OSVision
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### 58 Responses to The Moon’s eccentric orbit

1. Euan Mearns says:

Clive, the Moon is far to easy 😉 What I’d be interested to see is the eccentricity of Venus, Jupiter and Saturn combined compared with Earth – do they rhyme? Or even hum. And by way of education, what causes Earth’s obliquity to change. And since Earth is a gyroscope, does changing obliquity have the same effect as rotating a gyroscope about its axis? What is that force called?

• Euan Mearns says:

And thought flow… Earth is a magnetic gyroscope, wobbling about its axis inside the magnetic field of The Sun. I’m not sure anyone understands what causes Earth’s dipole to flip “randomly”. Does the polarity have any impact on Earth’s dipole ability to deflect cosmic rays? I’m writing a post on 10Be, and while I have a PhD in isotope geochemistry, I’m way out of my depth on certain things.

• Clive Best says:

The JPL ephemeris models all planets, moons, satellites and asteroids. NASA use it to plan their space probes. So yes it can give you changes in eccentricity of Jupiter, Venus etc. The numerical integration of all gravitational interactions results I the Milankovitch cycles. I have only just started to look into this – I hope I can get out the other end. Much of the pioneering work was done by French matehmaticians.

I have a steep learning curve here but intend to keep my feet on terra firma.

2. Clive Best says:

The flip of the earth’s magnetic field is another mystery which no-one has explained. I think the last inversion was a couple of million years ago and another is also due.

The origin must be astronomical as well. It would cause complete havoc if the field switched over tomorrow morning !

3. Pete Russell says:

I’m sure that at school sometime in the middle of the last century, I was told that the Earth’s orbital period was about 365.25 days. When did that change, or am I missing something?

• Clive Best says:

Oops – you are right. That was a typo ! thanks.

4. John F. Hultquist says:

I was amazed to discover that the moon’s orbit around the earth is anything but simple …”

The Moon’s path about the Sun has nearly the same look as does the Earth’s.
Said another way: The often seen diagram of the Moon literally going around Earth is a false representation.
Picture a blue car traveling in the center lane of a 3-lane highway with a white car behind. The white car can accelerate and pull into the right hand lane. Going faster than the car in the center lane, the white car will eventually pull ahead. Then it can cross to the left in front of the blue car and when, having achieved a position in the left lane it can slow down. Now the blue car can pass the white car. As the blue car pulls ahead the white car can return to the center lane. Repeat. Neither a circle nor an ellipse is anywhere to be found . . .

. . . except as a mathematical construct.

http://www.wired.com/wiredscience/2012/12/does-the-moon-orbit-the-sun-or-the-earth/

• Clive Best says:

This is very true. The moon really orbits the sun but is “locked into” the earth’s orbit round the sun. The gravitational attraction of the sun on the moon is about twice that of the earth on the moon. It is only relative to the earth that the moon appears to be in orbit around the earth. However as far as tidal effects of the moon on the earth is concerned the distance of the moon from the earth determines what the tidal force is and this goes as 1/R3. That is why the lunar tide dominates on earth rather than the solar tide, even though the gravitational force of the moon on earth is much less than the sun.
“>
One could argue that the solar system is also in orbit around the centre of the galaxy, although I wouldn’t know how to plat that one.

• brianrlcatt says:

Um, I thought the reason the actual tidal effect of the Sun on the Earth was much smaller than the Moon’s, although the gravitational force of the Moon on the Earth is much less than Sun’s, is because the Earth is in an equilibrium state created by centrifugal force bing in equilibrium with the centripetal force of gravity, so, in my mind anyway, the main effect of the Sun is on the Moon’s orbit, acceleration and deceleration as it orbits, which then affects the Earth indirectly, etc. Obs.

The Moon’s effect is larger because there is no centrifugal force to counter its effect, the Earth is not orbiting the Moon, etc….

PS I suppose that means the planets create plasma tides in the Sun. Whatever.

5. A C Osborn says:

Clive you might like to take a look at E M Smith’s Forum, he posted on the Moon’s orbit almost at the same time as you, he has collected a lot of data links.
See
http://chiefio.wordpress.com/2014/01/24/the-moons-orbit-is-wrong-it-can-change-a-lot-and-tides-will-too/
and
http://chiefio.wordpress.com/2014/01/25/a-remarkable-lunar-paper-and-numbers-on-major-standstill/

• John F. Hultquist says:

When Chiefio moved from the left coast to the other left coast followed by only episodic postings I failed to keep up. Recently, I noticed his comments starting to reappear on WUWT. I guess I should get back in the habit of checking in there.
The top post you link to there (2014/01/24) has many links to his previous discussions of the moon. One needs some popcorn before starting.

• clivebest says:

Thanks for the pointer.

Indeed E M Smith has some important insights new to me.

great.

6. A C Osborn says:

He is the the thinking man’s Willis Eschenbach.

• Clive Best says:

However, E M Smith seems to suffer from the same ego problem as Willis – my two comments have now been in moderation for well over 24 hours !

7. John F. Hultquist says:

Clive,
I don’t think E. M. will object to this but in dealing with him it helps to know that he is sort-of living out of a suitcase for awhile. Not really — but he is from the CA Bay Area but has some unique computer skills and drove to Florida with a car load of things for a short duration job. He moved from one motel to another for a time while finding a way to settle in. As I mentioned above, I haven’t kept up because for a time his posts were rare. The second thing to know is that he is on the very functional end of the Asperger syndrome. If the work he went to Florida to do isn’t what is occupying his time, he could be on to a project or “find.” If that is the case, whatever he comes up with will be interesting.
For myself, I’m here in comfort and warmth (hydro/electric power) watching most of the US and Canada freeze solid. We have just at freezing temps, total cloud cover, rime ice on everything, no wind, and very little reason to go outside except to exercise dogs and feed horses. I’ve been doing a lot of reading. Cheers.

8. A C Osborn says:

Clive, I have never had any problem with him.
It might be an idea to have a look at his Archives, he has posted some excellent analysis in the past.

9. A C Osborn says:

10. Euan Mearns says:

“Fascinating and thought provoking!
I have been trying to use the JPL ephemeris to calculate the moon’s orbital parameters 17,000 years ago. I beginning to conclude that the Milankovitch cycles really only apply to the earth-moon barycentre. The effective moon-earth orbit is far more complex as you have described. Therefore it seems likely that larger excursions in the gravitational coupling between the two must result as the overall eccentricity increases.

How could we prove this hypothesis ?”

Clive, you posted this at Chiefio. I have been asking Rog what the origin of Milankovitch is and he never answers – so I guess he doesn’t know. Does anyone know? Or did Milankovitch simply observe the cycles and conclude they were orbital without mathematical evidence?

I’ve posted my 10Be post, a bit nervous about a complex post like this not being reviewed before hand, and so would appreciate some friendly informed commentary before I start to spread it to the bigger blogs.

Solar influence of glaciation in Greenland

I think there is strong evidence that D-O cycles and Bond cycles are linked to solar variability. I will be doing a second post on Bond cycles in the Holocene. I have kind of assumed then that the big changes at glacial terminations are also linked to the Sun, but am not wedded to that idea.

• clivebest says:

As far as I know Milankovitch spent many years calculating how the orbits of the other planets vary the gravitational pull on the earth. This cannot be done analytically – even with 3 bodies. He had to do laborious integrations over long periods to get the 100,000 year ellipticity cycle.

Obliquity and precession are gyroscopic effects and I think this involves also the flattened spherical shape of the earth and the moon ! I don’t know whether Milankovitch was the first to propose them or whether he just put them all together. His actuak “theory” was that summer insolation above the arctic circle determined whether ice sheets wax or wane depending on the amount of solar insolation. If it was low then then the ice sheets slowly grew in size each winter.

However the data doesn’t really support this. There is a correlation with the differential of ice volume so there is more melting when the obliquity is high and precession is right but it doesn’t explain interglacials at all.

11. John F. Hultquist says:

Just for reference because I don’t know without reading more and don’t have the time tonight: Milankovitch issues are discussed by Luboš Motl with link to a paper by Roe:

http://motls.blogspot.com/2010/07/in-defense-of-milankovitch-by-gerard.html

http://earthweb.ess.washington.edu/roe/GerardWeb/Publications_files/Roe_Milankovitch_GRL06.pdf

Also note the comment by Nigel Calder and link there.

• Clive Best says:

Yes. This paper gets is correct. The increase in insolation at northern latitudes increases the rate of ice melting – NOT the total volume of ice. Roe’s result explains the details within the ice ages but it doesn’t explain why the ice sheets collapse every 100,000 years into an interglacial. I looked into the Roe paper here http://clivebest.com/blog/?p=4585 There is a clear correlation but it isn’t quite perfect.

So Roe proved that Milankovitch cycles are at work. In fact before 900,000 years ago glacial cycles were perfect 41,000 y obliquity cycles. However no-one has been able to explain the 100,000y cycle since then.

12. Roger Andrews says:

Clive: On the general question of what causes ice ages here’s another theory you can probably discard 😉

http://malagabay.wordpress.com/2013/12/28/the-drake-passage-impact-event/

• Clive Best says:

For sure the isolation of Antarctica over the South Pole caused the drift to an the series of ice ages starting 3-5 million years ago. Whatever caused the Drake Passage it resulted in S America splitting from Antarctice and seems to have caused the AMOC. Circular winds drive eckman currents kick starting the global MOC.

13. E.M.Smith says:

Clive,
Not ego, just limited time. New commenters go to moderation until white listed, then are instant up for future comments. You are now unmoderated and instant up. As ponted out above, I am in new housing at a new job away from everything and everyone where I lived for 40+ years trying to cope. Picking up wifi at starbucks Iin a slow and hard way to run a blog… I appologized for the delay at my place when your comments went up.

Per Milankovitch, he did his calculations with pen and paper while in a Nazi prison for several years. Fascinating story in a book an the history of the discovery of the ice ages titled something like: The Ice Ages

FWIW; I think you are on the right track with the lunar tides and moving faster than I can, so I’ll be playing catch up with you for a while 🙂

14. Doug Cotton says:

The IPCC et al are wrong because their basic assumptions are wrong. They need a paradigm shift in your thinking, because planetary temperatures are not controlled by this imaginary radiative forcing concept which I debunked two years ago in my peer-reviewed paper “Radiated Energy and the Second Law of Thermodynamics.” Temperatures are set by the gravito-thermal gradient (modified by inter-molecular radiation) both above and below any planetary surface.

When a photon from a cooler source strikes a warmer target, that target “recognises” that this photon has the same characteristics of photons which it can emit. It has exactly the right frequency (thus energy) that is required to cause an electron in the target to move up between two quantum energy states. But the process is immediately reversed, and a new identical photon is emitted as part of the target’s “quota” as per its Planck function. So the target does not need to convert some of its own thermal (kinetic) energy to electron energy (then to a photon) and so it cools more slowly as a result of the back radiation, as we all know.

But non-radiative processes can increase their rate of cooling to compensate for slower radiative cooling.

Furthermore, most slowing of surface cooling (and cooling of the 2 metre high surface layer of the troposphere where we measure climate) is caused by slowing of conduction into nitrogen and oxygen molecules.

Have you ever wondered why the temperatures don’t keep falling at a rapid rate all through the night when upward convection almost ceases? It’s due to the fact that the “environmental lapse rate” is a state of thermodynamic equilibrium in which the non-radiative processes have a propensity to form a -g/Cp thermal gradient, but this is reduced (usually by no more than about a third) by the temperature levelling effect of inter-molecular radiation. For example, on Uranus the -g/Cp gradient of about 0.76K/Km is reduced to about 0.72K/Km by radiation between just a small percentage of methane molecules, whereas on Venus it is reduced more like 25% by carbon dioxide, which thus leads to a significantly lower surface temperature on Venus. On Earth it is reduced mostly by water vapour which reduces the insulating effect of the atmosphere by inter-molecular radiation, just as it reduces the insulating effect between the panes of double glazed windows as it helps energy leap-frog across the gap at the speed of light, overtaking the far slower diffusion heat transfer.

And it is because of all this that we actually have evidence that the gravito-thermal gradient exists, and thus the greenhouse conjecture is demolished and there is zero (warming) sensitivity to carbon dioxide. It actually cools by a mere 0.1 degree at most.

15. Greg says:

This happens every 205.9 days – more than half a year due to the precession of the orbit every 18.6 years.

this not due the nodal precession, it is the alignment of the precession of the lunar apsides:

1/(2/365.25-2/pApsides) = 205.888171586386

Since it does not matter which end of the line of apsides is towards the sun it’s 365.25/2 ; similarly once the line of apsides has gone 180 deg. the pattern repeats.

Welcome to the fun and games: you will soon find out why madmen are referred to as loonies 😉

• Greg says:

If you dig out the theory of the torques involved, the component causing precession is a sin2 and the other is a cos.sin , both have a period twice that of the roughly annual alignment, this results in a frequency doubling:

sin(2*x) = 2*sin(x+pi/4.)**2-1

• Greg says:

A few years back I did this graph of high-pass filtered LOD:

Knowing what I now do I should have plotted sin2 rather than the abs() for the apsides.

The high-res part of the data since 1962 is a bit short for detecting 18y cycles by spectral analysis.

• Greg says:

I found this paper linked on cheifeo’s pages above. It ties to explain anomalous variations in lunar eccentricity.

http://arxiv.org/abs/1102.0212

Actually, the values for its mass and dis-
tance needed to explain the empirically determined increase
of the lunar eccentricity would be highly unrealistic and in
contrast with the most recent viable theoretical scenarios for
the existence of such a body. For example, a terrestrial-sized
body should be located at just 30 au, while an object with
the mass of Jupiter should be at 200 au.

So the “highly unrealistic ” possibility of binary system does not even get considered, though we may imply it from the above. A binary star of similar mass to Sol ie 200*mJ would presumably have to be located at about sqrt(200) time the Jupiter like object ie 14*200 au.

At least there is an observed anomaly from lunar range finder and Lorenzo Iorio has provided some numbers as a starting point. I wonder if this anomaly could be expressed as a rotation of the solar system.

• Clive Best says:

Amazing – thanks.

16. Greg says:

Clive, what was the residual of the fit you did? Could you post waht that looks like?

Just above I posted the LOD plot but it seems guests links don’t show. You may like to make it visible.

• Greg says:

BTW what you show as the sum of two cosines in not what is usually called a modulation, though any such sum can also be expressed as an amplitude modulation. It’s a trig identity.

You could see it as two frequencies close to that of the central period 205.888 days
Their mean frequency would match that. Half the difference would give 8.85y

Since the amplitudes are not equal you will have a residual 8.85 which underlies the modulated semi-annual oscillation.

In fact I think you are missing the modulation of the circa 6mo part.

You could try a sum of two periods either side of 6mo and the longer period. Or multiply the 6mo harmonic by the longer one *as well as * adding the longer term.

That way I think you will have a much closer fit.

• Clive Best says:

Greg,

I just installed a plugin which allows comments to include an image. Just add a line with the url on it – like this

17. Greg says:

BTW I had an article on Climate Etc this week showing a common lunar signal in SST and cyclone energy.

http://judithcurry.com/2016/01/11/ace-in-the-hole/

I point out that the famous circa 60y periodicity is probably a lunar+solar beat :

p1=9.1;p2=10.8;
cos(2*pi*x/p1)+cos(2*pi*x/p2)

• clivebest says:

That is remarkable observation. The change in latitude of the horizontal tidal drag with lunar precession must affect ocean heat transport.

You mean cosA + cos B = 2cos(A+B)/2.cos(A-B)/c ?

• Greg says:

cosA + cos B = 2 * cos((A+B)/2) * cos((A-B)/2)

In essence yes but I’m not sure that applies to your data in the way I thought. I’m trying to fit to some eccentricity data from a gravity sim and gnuplot is having difficultly converging to a stable solution.

• Greg says:

The sim is claiming to output eccentricity. I think there’s phase modulation going on the pseudo monthly part.

I gather you were getting barycentre distance. Is that right? I may try JPL so I have the same thing.

That is not quite the same but should be similar. I’m not sure what the sim is doing or how it can get instantaneous values of eccentricity.

• Greg says:

Here is the interference pattern of 9.1 and 10.8

The envelop matches the 60 y climate cycle quite well.

• Greg says:

The strongest solar peak was in the 1960s which was not hot. but if it was getting countered by the lunar cycle as above it fits. It’s not that simple and solar is far more complex than jst 10.8 and is not harmonic. However, I think there is a good indication of the origin of the mysterious 60y pseudo-periodicity there.

Also note two adjacent beats are not the same, the modulation freq is 118y not 59. There is a phase inversion at each node. SST shows an odd quirk in 1950 that is suggestive of such a phase inversion ( could also me “adjustment” artefacts ).

The 1975 PDO “phase change” is often interpreted as just meaning it goes from neg to positive but if you look at the detail it is just this kind of phase reversal too.

• Greg says:

PDO 1975 looks more like the flip in 1850 on this graph. Like it should have gone down but forgot and started going up again instead.

• Greg says:

OK, just did a spectrum on the grav sim output and it shows two very clear peaks as you fitted.

I also see some minor peaks at 27.5 34.7 37.5 and interestingly at 9.5 days.

These may relate to something I found in Arctic sea ice:
http://climategrog.wordpress.com/nh_seaice_spd_27d_detail/

• Greg says:

• Greg says:

Here is something you may like to try to explain since you are interested in these lunar periods.

I have dug deeply and got the most accurate figures I can find for the apsidal and nodal periods. It comes from top level astronomers in Paris Bureau des Measures and are defined as 4th order polynomials based on j2000. They take account of lunar laser range finder data. and are give with 12 sig figs. !?

from Jean Chapront [2002]
pNodal= 6793.476501
6793.476501/365.25 = 18.5995249856263

The time series (3rd order) for Lunar apse cycle comes from (Chapront [2002], page 704)
(3232.60542496 + T_1000 * ( 0.0168939 + T_1000 * ( 0.000029833 – T_1000 * 0.00000018809 ) )) / 365.25

## evaluate at y2k:
pApsides=3232.60542496
3232.60542496/ 365.25 = 8.85039130721

1/(1/(2*pApsides)-1/pNodal)/365.25 = 366.318712740139

So the two cycles are slipping by one earth roatatin per day.

If you can see the reason for htat , let me know. 😉

18. climategrog says:

this not due the nodal precession, it is the alignment of the precession of the lunar apsides:

1/(2/365.25-2/pApsides) = 205.888171586386

I have done a detailed spectral analysis of the oeECC for the moon, using SPL data.

It is dominated by the two peaks you found by fitting.

1/(1/31.8122 -1/29.530588853 ) = 411.74 = 1/(1/365.25 -1/3232. )

The RHS of this shows that the circa 6mo bumps in accentricity are due to the apsidal period. The 31.8d term is the temporal coincidence of the new moon with with this maxium in the precessional torque.

All other peaks are much smaller but as your graph shows they are or visible magnitudes.

Of particular interest is a 2.996y period and a 9.61 day peak.

What is interesting in all this is that if you go looking for a lunar signal at the usual periods like 29.5 or 27.545 you may well not find it and erroneously discount the presence of a lunar driver.

• climategrog says:

Excuse all the typos , my eyes are tired 🙁

19. John Greetham says:

I know it’s been a while since there was any activity on this thread but I’ve only found it today whilst looking for a way to calculate the instantaneous eccentricity of the Moon’s orbit.

Seeing the first equation quoted :-

eccentricity(d) = 0.55 + 0.014cos(0.198*d + 2.148) + 0.0085cos(0.0305*d +10.565)

looked very promising so I did the calculations but don’t get the answers I expected. This leads me to believe that I have mis-understood something. I have assumed that ‘d’ is the julian days since epoch (J2000) but at 2nd Jan 2000 (d=1) JPL gives e as 0.0647669 but the formula gives me 0.159215173 — obviously so far out that I must be doing something wrong.

I’ve looked for potential typos but found nothing that makes any sense so I’d be grateful for corrective instruction 🙂

• Clive Best says:

Sorry for the confusion.
I am afraid it is a bit messy

I used Horizon with an arbitrary start time and then counted the days ‘d’ from zero. The start date was 12 December 2000. Therefore to use the formula relative to a julian day ‘j’ with epoch of J2000 you should subtract off the offset first. Try using

d = j-346

and then apply the formula.

• John Greetham says:

Thanks for your quick response – I’ll do just that but not for a couple of days since I’ve just had an urgent request to re-set some music!

I’ll be back before the week-end.

20. John Greetham says:

Sorry Clive, I still can’t rationalize the equation. I’ve downloaded the ephemeris data starting at 12/12/2000 and see that the Eccentricity on that day was 0.066829. Your equation, using d = 0, gives e as 0.572346. Even if I assume that 0.55 should in fact be 0.055 the figure becomes 0.077346, closer but still nearly 16% too high.

Enlarging your graph I read the first plotted figure as 0.0395 which seems to indicate that your calculations differ from mine significantly.

Using d = j-346 ( ie. Jan 1st rather than Dec 12th) I get e = 0.5641138413 (or 0.0691138413) rather than JPL’s figure (for 1/1/2000) 0.06476694.

Looking for ‘reasons’ I downloaded the data for 12th Dec 2000 to 27th Dec 2020 at 30 day intervals – simply to look for long-term variations – solved the equation for each day and compared the results with the JPL figure.

Here I would have added a graph of the result but can’t see a way to attach an image.

I hoped that I might see a straight (or even sinusoidal) line but it doesn’t appear to have any ‘logic’.

Plotting my results I DO see a pattern similar to your graph but since I only looked at 240 figures and they were 30 days apart I didn’t really expect a very close relationship.

If you can point me to the method of adding an image – it must be possible since I see graphs from Greg – I’d be happy to post my results.

21. John Greetham says:

Since you replied to my first comment in just over a day I had expected a response to my follow-up in less than a week Clive. It’s now 8 days, do you have a problem with my logic?

• clivebest says:

Sorry for delay.

I think I have the wrong start date!!! Apologies

I used the Horizon interface on JPL

Current Settings

Ephemeris Type [change] : ELEMENTS
Target Body [change] : Moon [Luna] [301]
Center [change] : Earth-Moon Barycenter [500@3]
Time Span [change] : Start=2000-12-12, Stop=2016-10-28, Step=1 d
Table Settings [change] : defaults
Display/Output [change] : default (formatted HTML)

Using this then I get eccentricity = 0.06828 for December 12 2000 So my start date is wrongly labelled !

The first minimum of 0.028528 occurs on April 30 2001 which corresponds to x-axis number 386

So I think the start date really should have beed 18 Nov 2000 !

The formula works but the phase needs to be corrected for any start date.

22. John Greetham says:

Thanks for the update Chris. I’ve now downloaded the JPL data for 18/11/2000 to 1/5/2001 and agree with the figures you quote. However, looking at your graph I see the first minimum at about x = 170 rather than 386. In fact with x = 0 at 18/11/2000 the 30th April 2001 is 163.

That is really of little consequence though. The real issue is that I still can’t get your formula to produce anything like the correct value of ‘e’.

You don’t make it clear whether the components in brackets are in Degrees or Radians so I’ve used both in my calculations.

To assist with understanding I’ve reduced your formula to

e = K + A + B

where K = 0.55,
A = 0.014 cos(0.198d + 2.148)
B = 0.0085 cos(0.0305d + 10.565)

Assuming the bracketed value is in Degrees,
when d = 0 (18/11/2000) – A = 0.01399 & B = 0.9915 so e = 1.5555 which is quite obviously incorrect since (for an ellipse) e must be 0 < 1.
This is also born out by the the JPL data which shows e = 0.03733.

If I assume that the bracketed data is in Radians then A = -0.007639, B = -0.40889 so e = 0.13346, withing the range but still 3½ times the real figure.

Even if I consider that K should really be 0.055 then e is either 1.0605 or -0.3615 – both outside the possible range.

I don't understand your last sentence. What do you mean by 'phase' ? and how do you think it has to be corrected?

• Clive Best says:

John,

Sorry for all the confusion. I checked in the software. I am using

y=0.055+0.014*cos(0.198*x+2.148)+0.0142*cos(0.0305*x+10.565)

where x = day number after 18/11/2000

I clearly screwed up and typed in the wrong value for B !!

Try that and let me know if it is not OK.

23. John Greetham says:

’tis said “To err is human … ” for a real cock up you need a computer !!

I also screwed up and made a silly error only discovered when I came to change the 0.0085 to 0.0142 – – – I’d ADDED rather than multiplied by – – – Duh!!

At least you’ve confirmed that the 0.55 should be 0.055. Now I need to be sure about degrees and Radians.

My re-calculations using the figures in brackets as Radians are within 0.00401 of the JPL figure for 18/11/2000 so I’m assuming that’s correct. Over the 100 days from 18/11 the Max. error is 0.009547 and the Min. error -0.006398. However the trend is increasing at the rate of 0.00047 per day.

By changing the 0.0142 to 0.00995 the trend line is flat (over the 100 days) and the error varies from 0.00587 to – 0.00638.

Looking now at the period from 1/1/2000 to 31/12/2001, my first calculations showed a variation from the JPL figures of between 0.008015 and -0.009698 with the trend decreasing by 0.00000308 per day.

To get the trend to ‘flat line’, the 0.00995 had to change to 0.00725 and this gives a variance of between 0.00883 and -0.00876.

So, not ‘perfect’ and the variation isn’t sinusoidal (nor is it chaotic), it’s primarily periodic following the synodic month but with ‘glitches’. I do think It’s close enough for my purpose so thanks for your work and for making it available.

24. John Greetham says:

It seems that the figure that was originally 0.0085 and that you corrected to 0.0142 and then I modified to 0.0995 and later 0.0725 is VERY sensitive.

Discovering that I’d downloaded the JPL Data for the default TIME of 00:00 – which gives a Julian Day of *******.5, – I downloaded again using a time of 12:00. Now I find that to get a flat line trend it needs to be 0.00685. But that increases the overall error with the Max going to 0.010.

That made me look at the first multiplicand (0.014). By changing that to 0.0145 and putting the second back to 0.00725 I still get a flat line trend but the total error is back down to 0.009244 Max.

• John Greetham says:

Having slept on the results I now see that whilst 0.009 is, of itself, a small figure, when the target figure is quoted to 16 decimal places it is huge! Add to that the fact that it isn’t linear – smaller figures are not necessarily in error by the smallest amount – I’ve now worked out the discrepancy as a %age – – – – +34.05 to -17.81 – – – – far too great to be of any real use 🙁

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