# Earth-Moon without Sun

hello there

let's imagine that the Earth-Moon system became isolated (ie, not subjected to Sun's gravity anymore), would Moon run around the Earth at a lesser speed than now? (if it would, in which proportions then?); and does Moon's orbit would change in circumference?

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Matterwave
Gold Member
To first approximation, there would be no difference. The orbit of the Moon can be computed as if it were in isolation with the Earth, and the Sun were not there. To higher degrees of precision, there may be a difference, but I'm not so sure what those would be.

Janus
Staff Emeritus
Gold Member
One effect the Sun has on the Moon's orbit is to alter its eccentricity. It increases it when the orbit's semi-major axis lines up with the Earth-Sun radial and decreases it when it is at a right angle. Another is that it causes the Lunar orbit to precess.

I post this question because i read the following:

As we can see, a revolution of Earth-Moon system around the Sun for 1 synodic month is equivalent to an arc of 27° of the terrestrial orbit around the Sun. This angle (27°) is the same as the one between Moon-position at 1 sideral month and at 1 synodic month according to Earth's frame of reference.

The vector "E" (beginning from the Moon) perpendicular to the segment Earth-Sun points to a different direction every synodic month.

Thus from this "triangle" of vectors, we can calculate mean Moon's speed E' around Earth without Sun's gravitionnal influence.

ll E ll=square root [ (E sin phi)² + (E')² ]

thus ll E' ll= ll E ll * cosine phi
ll E' ll= current Moon's speed * cosine phi
ll E' ll= 3680km/hour * cosine 27°

ll E' ll=3279 km/hour

THUS: In an isolated Earth-Moon system, Moon would run at a mean speed of 3279 km/hour around our planet.

I agree with the value of phi, but is this method to calculate Moon's speed in an isolated Earth correct?

thank you

Staff Emeritus
2019 Award
I took a look at that website, and wouldn't consider anything on it to be reliable.

Hey, Janus!

A website states that if Sun's gravitationnal effect were "put off", Moon's orbit around the Earth would be almost a perfect circle (whereas with the Sun it's an ellipse).

In that condition, would the mean moon's speed become equal to V= 3680*(1-2e) where "e" is the actual excentricity of Moon (about 0.05) and "2e" the variation ratio, as they suggest?

[PLAIN]http://img40.imageshack.us/img40/9568/1dtp.jpg [Broken]

thank you

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Hello everyone. I've been discussing this with Termina for a while now on youtube and as we could not reach an agreement I suggested he asked on a physics forum.

Please don't let the source of the claim put you off discussing this topic. Ultimately the purpose of this discussion is to determine

A: The average velocity of the Moon
B: The average velocity of the Moon if the Sun were not there

I'd appreciate it if you would join in, thanks!

tony873004
Gold Member
This is easy enough to simulate. Here are some graphs showing the the Moon's velocity, eccentricity, and inclination over the course of 1 year, first with the Sun present, then repeated after deleting the Sun
http://orbitsimulator.com/BA/moonwithoutsun.GIF

This is easy enough to simulate. Here are some graphs showing the the Moon's velocity, eccentricity, and inclination over the course of 1 year, first with the Sun present, then repeated after deleting the Sun
http://orbitsimulator.com/BA/moonwithoutsun.GIF
Thanks for the data tony. What we are trying to establish is the distance the Moon would travel in a complete orbit of the Earth vs the distance it would travel in the same time if the Sun were not present.

Any help would be much appreciated.

tony873004
Gold Member
Thanks for the data tony. What we are trying to establish is the distance the Moon would travel in a complete orbit of the Earth vs the distance it would travel in the same time if the Sun were not present.

Any help would be much appreciated.
In the graphs, you're given time and velocity. So you can compute distance. d=vt. You can also see how many orbits were completed in each graph by counting the cycles. So you have everything you need to compute distance per orbit.

Hey, Janus!

A website states that if Sun's gravitationnal effect were "put off", Moon's orbit around the Earth would be almost a perfect circle (whereas with the Sun it's an ellipse).
Termina, that is not true. First of all, a two-body system acting via gravity NATURALLY is elliptical (or hyperbolic). The sun is not responsible for the Moon's elliptical orbit.

In the graphs, you're given time and velocity. So you can compute distance. d=vt. You can also see how many orbits were completed in each graph by counting the cycles. So you have everything you need to compute distance per orbit.
Hi Tony

Please don't think I am being lazy. There are two reasons I cannot do what you say. The first is that Termina and I are here to get an objective answer from more experienced people (experienced in physics) to a disagreement we have been having elsewhere so if I come up with the answer then Termina will think I have concocted it; and the other is much more simple, if I tried to determine the result from those graphs I would get it wrong :)

I spent some time calculating how far the Moon would travel in X amount of time based on its average velocity and assuming that any effect on this velocity by the gravity of the Sun as the Moon heads toward it will be equally negated as the Moon travels away from it.

My calculation for the distance the Moon travels in 1 Moon sidereal day is as follows.
Figures from http://nssdc.gsfc.nasa.gov/planetary/factsheet/moonfact.html
Note: It is the technique that is important, not the exact answer

Moon sidereal day = 655.728 hours
Moon sidereal day = 2360620.8 seconds
(multiply hours by 60^2 to get seconds)
Moon average velocity = 1023km/s
(multiply by seconds in sidereal day to get average distance travelled in 1 Moon sidereal day)
Distance travelled by Moon in 1 sidereal Moon day = 2414915078.4km

Now whether or not this figure is exactly right is not the source of our disagreement. Termina thinks that if there were no effect of the Sun's gravity on the Moon this distance would be approximately 10% lower - which to me is frankly ridiculous but I am willing to accept I am wrong.

I shan't go into the details of how Termina calculated this figure at this point unless Termina wishes to later. At the moment all we really want to know is the answer to the following question...

If the Moon travels 2414915078.4km in 655.728 hours in the presence of the Sun's gravity. How far would it travel in the same time if the Sun's gravity were not there?

I'm sorry if this is all very basic for you guys, but I cannot express how grateful I am for your time on this!

Cheers!

TheRationaliz, i've never thought Sun would affect Moon movement in such extend, rather after hearing such claims i'm here to check whether they are right or not.

tony873004, nice graphs! can you tell me where you found them?

TheRationaliz, i've never thought Sun would affect Moon movement in such extend, rather after hearing such claims i'm here to check whether they are right or not.
Okay, good. Then I hope someone here will be kind enough as to tell us how much (if at all) the gravity of the Sun will affect the distance travelled by the Moon during an Earth sidereal month (Moon sidereal day.)

tony873004
Gold Member
...tony873004, nice graphs! can you tell me where you found them?
I made them in Excel using data generated from an n-body simulation. I let the solar system run for 1 year, recording data. I then started the simulation over again, but this time I deleted the Sun.

Going back to the spreadsheet data used to make the graphs, I had the spreadsheet compute the average velocities. Over the course of the year plotted, the Moon's average velocities are:
without the Sun:1.02071 km/s.
With the Sun:1.021342 km/s

So the Moon with the Sun travels slightly faster.

So the distances travelled in 1 lunar sidereal day are:
without sun: d=vt = (1.02071 km/s)(2360620.8 s/moon sidereal day) = 2409509 km
with sun : d=vt = (1.021342 km/s )(2360620.8 s/moon sidereal day) = 2411011 km

So the moon with the sun included travels 1492 km farther during the amount of time, a difference of about 0.06%

However, this experiment is sensitive to when I magically make the Sun disappear. In the above experiment, the Sun disappeared when the Moon was at a waning crescent phase, nearly a new moon. Repeating the 1-year experiment 10 days later, when the Moon is at 1st quarter yields.

Moon without Sun: 1.031003 km/s
Moon with Sun: 1.023383 km/s

This time, the Moon without the Sun is slightly faster.

Tony

Hopefully with this objective analysis and the fact that it was calculated using a solar system simulator Termina will now accept that the Moon will not travel 10% slower without the gravity of the Sun and will accept that the calculation we were looking at is merely a numerology trick.

Thank you so much for taking the time and effort to calculate these effects for us! I appreciate your time greatly!

Thanks!

I suspected their initial claim were pseudoscientific "gymnastics", but i wasn't certain enough since i don't know celestial mechanics like the back of my hand.
now, i'm certain!

The website where i get those doubtful claims even asserts it's Einstein who said one must make 3682km/h*cosine26,9° to know Moon speed value without Sun!!!!!

D H
Staff Emeritus
The Earth-Moon system by itself would almost exactly follow an elliptical orbit. (the only deviation is due to general relativity, which in the case of the slow-moving Moon around the not-so massive Earth is negligible). Call this period T', to distinguish it from the period where the Sun is present.

$$T' = 2\pi\,\sqrt{\frac{a^3}{G(M_e+M_m)}$$

where a is the semi-major axis of the orbit of the Earth and Moon about their center of mass, G is the universal gravitational constant, and Me and M[/sub]m[/sub] are the masses of the Earth and Moon. Rather than using G and the masses of the Earth and Moon, all of which suffer from a lack of accuracy, it is far better to use the Earth gravitational coefficient and the Moon/Earth mass ratio:

$$G(M_e+M_m) = GM_e(1+M_m/M_e) \approx 1.0123 \mu_e$$

where $\mu_e = 398600.4418\pm0.0009\,\text{km}^3/\text{s}^2$. The Moon/Earth mass ratio os 0.0123, to many places of accuracy. What to use for a? The French Ephemerides of the Earth, Moon, and Sun specifies a value of 384,7481 km. The resulting value for T' is 0.4 minutes longer than the sidereal month; essentially no change.

This is consistent with lunar theory. What you are talking about is the variation of the Moon. This has a period term and a secular term. The secular term is exceeding small, measured in arcseconds/century. The affect of the Sun on the period of the Moon, and hence on the distance traveled, is essentially null.

--------------------------

1A widely-used value for the Moon's semi-major axis is 384,399 km. This is not the semi-major axis. It is instead the inverse of the mean of the inverse of the distance. Using that value yields a period that is about 53 minutes shorter than the observed sidereal month. This says the Moon would, on average, move slightly faster than it does with the Sun present. This is counter to lunar orbit theory (essentially no effect), and in any case, it does not support termina's position.

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The website where i get those doubtful claims even asserts it's Einstein who said one must make 3682km/h*cosine26,9° to know Moon speed value without Sun!!!!!
Yes, they are using an old debating trick known as an "appeal to authority". They throw in some complex numbers that most people won't be able to follow (especially illiterate people in 3rd world countries) and say "Look! It's not US claiming this, it's Einstein. Are you saying Einstein was wrong? Do you think you are more intelligent than Einstein?"

Then they fail to actually provide evidence that Einstein every said why they claim he said, and even if Einstein had said such a thing it doesn't mean he was automatically right just because he was clever. Einstein apparently once wrote the foreword in a book in which he stated that plate tectonics theory was implausible...he was wrong :-)

Still, I am more than just a little bit pleased to see we are now in agreement that this claim is merely a BS numerology trick.

TONY

hehe, the duplicity of this guy :-)

After 2 weeks arguing with me that the claim is a fact he came on here pretending to be sceptical. Then when it was proven to him beyond a doubt that it is a false claim and he accepted that proof he back to the speed-info.com forum and claimed that it is true after all....the Moon would slow down even more because the absence of the Sun would cause the Earth to rotate more slowly due to missing gravity effect on the Earth's oceans.

Of course he also believes that multiplying the orbital distance of the Moon by cosine(26.928) will magically compensate for both this orbital friction and the general relativity factor in one go. I never realised COSINE was so powerful :-)

Does multiplying the circumference of an ellipse by COS actually mean anything anyway?

Boy, I can see why you guys wouldn't want to get into discussions about claims of religious miracles :-D

hehe, the duplicity of this guy :-)

After 2 weeks arguing with me that the claim is a fact he came on here pretending to be sceptical. Then when it was proven to him beyond a doubt that it is a false claim and he accepted that proof he back to the speed-info.com forum and claimed that it is true after all....the Moon would slow down even more because the absence of the Sun would cause the Earth to rotate more slowly due to missing gravity effect on the Earth's oceans.

Of course he also believes that multiplying the orbital distance of the Moon by cosine(26.928) will magically compensate for both this orbital friction and the general relativity factor in one go. I never realised COSINE was so powerful :-)

Does multiplying the circumference of an ellipse by COS actually mean anything anyway?

Boy, I can see why you guys wouldn't want to get into discussions about claims of religious miracles :-D
that's not me!:rofl:

Arguing on youtube that this claim is accurate, and then posting on here that you thought all along it was false. Leaving here exclaiming how you now know for a fact it is rubbish, straight back onto youtube saying that the people on this physics forum don't understand physics properly.

Please do explain to the people on this forum how taking the distance travelled by the Moon in a sidereal month and multiplying it by cosine(26.928) removed the gravitational affect of the Sun *and* cancels out the increase in Moon velocity due to the Earth's oceans.

I'm sure there are people here who might find it amusing :)

Arguing on youtube that this claim is accurate, and then posting on here that you thought all along it was false. Leaving here exclaiming how you now know for a fact it is rubbish, straight back onto youtube saying that the people on this physics forum don't understand physics properly.

Please do explain to the people on this forum how taking the distance travelled by the Moon in a sidereal month and multiplying it by cosine(26.928) removed the gravitational affect of the Sun *and* cancels out the increase in Moon velocity due to the Earth's oceans.

I'm sure there are people here who might find it amusing :)

You're far from the truth! as far i'm concerned, i've never stated on youtube that such far-fetched equation like Vmoon*cosine 26 was right. I'm not wormhole199.

Well, it's a non physics matter so that part of our discussion should be held elsewhere to save annoying others :-)