<p>The earth’s orbit eccentricity affects its angular velocity around the Sun. Reversely, the sun velocity in the sky would be effected observing from the Earth. The Moon phase is determined by the relative position of the Sun in the sky, and therefore when the velocity of Sun varies, the phase change of the moon would be varied, too.</p>
<p><strong>Edit:</strong>
Sorry for misunderstanding the question, I roughly calculated the synodic month when the Earth is at perihelion and aphelion.
The aphelion distance is
<span class="math-container">$R_{aphelion}= 1.52098 \times 10^8 km$</span>  and perihelion distance is  <span class="math-container">$R_{perihelion} = 1.47098 \times 10^8 km$</span>
I then obtained the angular velocity <span class="math-container">$\omega_{perihelion}$</span> and <span class="math-container">$\omega_{aphelion}$</span>with simple Newtonian physics</p>
<p><span class="math-container">$$\omega = \sqrt{\frac{GM}{R^3}}$$</span> which comes from
<span class="math-container">$$\frac{GMm}{R^2} = m\omega^2R$$</span></p>
<p>Then, taking the period of siderial month
<span class="math-container">$T_{siderial} = 27.3215 \ day$</span> , we may obtain the angular velocity of the moon relative to the Earth <span class="math-container">$ \omega_{moon} $</span></p>
<p>With these parameters, we can <strong>roughly</strong> calculate the synodic month when the Earth is at perihelion and aphelion.</p>
<p><span class="math-container">$$
T_{perihelion} = \frac{2 \pi}{\omega_{moon} - \omega_{perihelion}} = 29.5909 \ day
$$</span>
<span class="math-container">$$
T_{aphelion} = \frac{2 \pi}{\omega_{moon} - \omega_{aphelion}} = 29.4713 \ day
$$</span></p>
<p>which differs by around 0.4%</p>

CC BY-SA 4.0
2023-10-18T05:03:18.223Z
Good answer. But the distance to the Sun also affects the size of the Moon's orbit. See  https://astronomy.stackexchange.com/a/49267/16685
PM 2Ring
CC BY-SA 4.0
2023-10-18T11:02:48.300Z
@PM2Ring Thanks! That’s inspiring
Greeddeer
CC BY-SA 4.0
2023-10-18T13:09:47.690Z
Thanks for better explaining my question, but that's not an answer to it (it should be posted as an edit to my question, or as a comment). My question was "How much does it actually contribute?".
George Lee
CC BY-SA 4.0
2023-10-18T17:29:29.100Z
@GeorgeLee I edited my answer. Since the contribution of eccentricity isn't independent of other factors and further derivation is beyond my ability, I only roughly calculated it. Hope it helps.
Greeddeer
CC BY-SA 4.0
2023-10-18T17:32:02.983Z
@George As I said in my linked answer, lunar theory is complicated. However, Professor Richard Fitzpatrick has a simplified model of lunar motion: https://farside.ph.utexas.edu/books/Syntaxis/Almagest/node39.html
PM 2Ring
CC BY-SA 4.0
2023-10-18T17:41:32.657Z
@George OTOH, it's ok to model the Earth-Moon system's motion around the Sun as a Kepler ellipse. So the ratio of the Sun's angular speeds at perhelion & aphelion is given by $$\frac{\omega_p}{\omega_a}=\left(\frac{1+e}{1-e}\right)^2$$
PM 2Ring
CC BY-SA 4.0
2023-10-18T20:30:28.197Z
Can you please explain the formula of $\omega = \sqrt{ \frac {GM}{R^3}}$, or give a link to an explanation?
George Lee
CC BY-SA 4.0
2023-10-18T20:36:01.650Z
I can't view most images in your answer. I tried reloading multiple times, and it didn't help. I don't have this problem with other answers, so it's probably fixable from your end.
George Lee
CC BY-SA 4.0
2023-10-19T01:38:58.230Z
@GeorgeLee Sure, I edited the images to latex, and briefly explained how $\omega$ was derived
Greeddeer
CC BY-SA 4.0
2023-10-19T20:31:08.413Z
Thanks. I'm just wondering if that formula accounts for Kepler's 3rd law (which would cause Earth to go faster than regular orbit speed at perihelion, and vice versa at aphelion).
George Lee
CC BY-SA 4.0
2023-10-19T20:33:13.280Z
We have to take in account that Earth can't be a full month at perihelion/aphelion, so the extremes are actually a bit less extreme.
George Lee
CC BY-SA 4.0
2023-10-21T03:30:04.293Z
@George $\omega = \sqrt{\frac{GM}{R^3}}$ is for a circular orbit, so it totally ignores eccentricity.
PM 2Ring
CC BY-SA 4.0
2023-10-18T04:18:27.067Z
2023-10-19T01:36:07.073Z
2023-10-19T01:36:07.073Z
Greeddeer
<p>Usually, only the eccentricity of Moon's orbit is used as an explanation for the varying length of the lunar month. But Earth's orbit's eccentricity should also contribute to that. How much does it actually contribute?</p>

<the-moon><moon-phases>
How much does the eccentricity of EARTH'S orbit effect the length of the synodical lunar month?
George Lee
# How much does the eccentricity of EARTH'S orbit effect the length of the synodical lunar month?

(Asked by: username_0 on 2023-10-17)

Usually, only the eccentricity of Moon's orbit is used as an explanation for the varying length of the lunar month. But Earth's orbit's eccentricity should also contribute to that. How much does it actually contribute?

**Tags:** the-moon, moon-phases

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## 1 answers

### Answer by username_1 on 2023-10-18. Score: 5

The earth’s orbit eccentricity affects its angular velocity around the Sun. Reversely, the sun velocity in the sky would be effected observing from the Earth. The Moon phase is determined by the relative position of the Sun in the sky, and therefore when the velocity of Sun varies, the phase change of the moon would be varied, too.

**Edit:**
Sorry for misunderstanding the question, I roughly calculated the synodic month when the Earth is at perihelion and aphelion.
The aphelion distance is
$R\_{aphelion}= 1.52098 \times 10^8 km$ and perihelion distance is $R\_{perihelion} = 1.47098 \times 10^8 km$
I then obtained the angular velocity $\omega\_{perihelion}$ and $\omega\_{aphelion}$with simple Newtonian physics

$$\omega = \sqrt{\frac{GM}{R^3}}$$ which comes from
$$\frac{GMm}{R^2} = m\omega^2R$$

Then, taking the period of siderial month
$T\_{siderial} = 27.3215 \ day$ , we may obtain the angular velocity of the moon relative to the Earth $ \omega\_{moon} $

With these parameters, we can **roughly** calculate the synodic month when the Earth is at perihelion and aphelion.

$$
T\_{perihelion} = \frac{2 \pi}{\omega\_{moon} - \omega\_{perihelion}} = 29.5909 \ day
$$
$$
T\_{aphelion} = \frac{2 \pi}{\omega\_{moon} - \omega\_{aphelion}} = 29.4713 \ day
$$

which differs by around 0.4%

> **Comments:**
> **username_2**: Good answer. But the distance to the Sun also affects the size of the Moon's orbit. See  https://astronomy.stackexchange.com/a/49267/16685  
> **username_1**: @PM2Ring Thanks! That’s inspiring  
> **username_0**: Thanks for better explaining my question, but that's not an answer to it (it should be posted as an edit to my question, or as a comment). My question was "How much does it actually contribute?".  
> **username_1**: @GeorgeLee I edited my answer. Since the contribution of eccentricity isn't independent of other factors and further derivation is beyond my ability, I only roughly calculated it. Hope it helps.  
> **username_2**: @George As I said in my linked answer, lunar theory is complicated. However, Professor Richard Fitzpatrick has a simplified model of lunar motion: https://farside.ph.utexas.edu/books/Syntaxis/Almagest/node39.html  
> **username_2**: @George OTOH, it's ok to model the Earth-Moon system's motion around the Sun as a Kepler ellipse. So the ratio of the Sun's angular speeds at perhelion & aphelion is given by $$\frac{\omega_p}{\omega_a}=\left(\frac{1+e}{1-e}\right)^2$$  
> **username_0**: Can you please explain the formula of $\omega = \sqrt{ \frac {GM}{R^3}}$, or give a link to an explanation?  
> **username_0**: I can't view most images in your answer. I tried reloading multiple times, and it didn't help. I don't have this problem with other answers, so it's probably fixable from your end.  
> **username_1**: @GeorgeLee Sure, I edited the images to latex, and briefly explained how $\omega$ was derived  
> **username_0**: Thanks. I'm just wondering if that formula accounts for Kepler's 3rd law (which would cause Earth to go faster than regular orbit speed at perihelion, and vice versa at aphelion).  
> **username_0**: We have to take in account that Earth can't be a full month at perihelion/aphelion, so the extremes are actually a bit less extreme.  
> **username_2**: @George $\omega = \sqrt{\frac{GM}{R^3}}$ is for a circular orbit, so it totally ignores eccentricity.  

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