<p>The Moon is moving around the Earth in the same direction it is rotating. This means that it would be moving across the sky in the opposite direction the stars are, in case the Earth stopped rotating, however, the Moon is moving in the same direction as the stars due to diurnal motion (only slower, because it also has velocity in the opposite direction). Its angular speed across the celestial sphere will be <br></p>

<h2><span class="math-container">$\omega = \frac{2*\pi}{T_{stars}} - \frac{2*\pi}{T_M} = \frac{2*\pi}{T}$</span></h2>

<p><br>where <span class="math-container">$T_M$</span> and <span class="math-container">$T_{stars}$</span> are the sidereal period of the Moon and the sidereal day length. Substituting the values above, we get that the Moon takes about 51 minute over 24h to go around the celestial sphere, crossing the same meridian. Note that this is a rough approximation because the Moon's orbit is not circular, but it explains the value you mentioned.</p>

CC BY-SA 4.0
2019-12-16T13:18:42.070Z
2019-12-16T13:18:42.070Z
Tosic
<p>On the website <a href="https://epod.usra.edu/blog/2018/07/lunar-synodic-curve.html" rel="nofollow noreferrer">Lunar Synodic Curve</a> a nice image of an Moon analemma is shown together with an explantion on the reason why for the different phases of the Moon the images have to be taken with 24h 51min apart. It is explained that this difference is due to the longer synodic month of 29days 12hrs 44mins (compared to the sidereal month of 27days 7h 43mins). Whereas the explanation that due to the additional rotation time for the moon to reach the same angular relation to the sun an additional angle of 29.1 degs has to be travelled, the last step to come to the additional 51mins each day is not clear to me. Any clues on how to calculate this from the sidereal and synodic month duration or the associated mean motions of moon and sun would be very welcome.</p>

<the-moon><solar-system>
Delay of Moon passing a defined meridian each day
Rainer
# Delay of Moon passing a defined meridian each day

(Asked by: username_0 on 2019-12-15)

On the website [Lunar Synodic Curve](https://epod.usra.edu/blog/2018/07/lunar-synodic-curve.html) a nice image of an Moon analemma is shown together with an explantion on the reason why for the different phases of the Moon the images have to be taken with 24h 51min apart. It is explained that this difference is due to the longer synodic month of 29days 12hrs 44mins (compared to the sidereal month of 27days 7h 43mins). Whereas the explanation that due to the additional rotation time for the moon to reach the same angular relation to the sun an additional angle of 29.1 degs has to be travelled, the last step to come to the additional 51mins each day is not clear to me. Any clues on how to calculate this from the sidereal and synodic month duration or the associated mean motions of moon and sun would be very welcome.

**Tags:** the-moon, solar-system


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## 1 answers

### Answer by username_1 on 2019-12-16. Score: 1 | Accepted answer

The Moon is moving around the Earth in the same direction it is rotating. This means that it would be moving across the sky in the opposite direction the stars are, in case the Earth stopped rotating, however, the Moon is moving in the same direction as the stars due to diurnal motion (only slower, because it also has velocity in the opposite direction). Its angular speed across the celestial sphere will be

## $\omega = \frac{2\*\pi}{T\_{stars}} - \frac{2\*\pi}{T\_M} = \frac{2\*\pi}{T}$

where $T\_M$ and $T\_{stars}$ are the sidereal period of the Moon and the sidereal day length. Substituting the values above, we get that the Moon takes about 51 minute over 24h to go around the celestial sphere, crossing the same meridian. Note that this is a rough approximation because the Moon's orbit is not circular, but it explains the value you mentioned.


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