3

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?

George Lee
  • 345
  • 1
  • 10

1 Answers1

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%

Greeddeer
  • 101
  • 3
  • 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 Oct 18 '23 at 05:03
  • @PM2Ring Thanks! That’s inspiring – Greeddeer Oct 18 '23 at 11:02
  • 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 Oct 18 '23 at 13:09
  • @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 Oct 18 '23 at 17:29
  • @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 Oct 18 '23 at 17:32
  • @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 Oct 18 '23 at 17:41
  • Can you please explain the formula of $\omega = \sqrt{ \frac {GM}{R^3}}$, or give a link to an explanation? – George Lee Oct 18 '23 at 20:30
  • 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 Oct 18 '23 at 20:36
  • @GeorgeLee Sure, I edited the images to latex, and briefly explained how $\omega$ was derived – Greeddeer Oct 19 '23 at 01:38
  • 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 Oct 19 '23 at 20:31
  • 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 Oct 19 '23 at 20:33
  • 1
    @George $\omega = \sqrt{\frac{GM}{R^3}}$ is for a circular orbit, so it totally ignores eccentricity. – PM 2Ring Oct 21 '23 at 03:30