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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?

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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%

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  • $\begingroup$ Good answer. But the distance to the Sun also affects the size of the Moon's orbit. See astronomy.stackexchange.com/a/49267/16685 $\endgroup$
    – PM 2Ring
    Oct 18, 2023 at 5:03
  • $\begingroup$ @PM2Ring Thanks! That’s inspiring $\endgroup$
    – Greeddeer
    Oct 18, 2023 at 11:02
  • $\begingroup$ 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?". $\endgroup$
    – George Lee
    Oct 18, 2023 at 13:09
  • $\begingroup$ @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. $\endgroup$
    – Greeddeer
    Oct 18, 2023 at 17:29
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    $\begingroup$ @George $\omega = \sqrt{\frac{GM}{R^3}}$ is for a circular orbit, so it totally ignores eccentricity. $\endgroup$
    – PM 2Ring
    Oct 21, 2023 at 3:30

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