<p>Since a solar year, i.e. the time over which the Earth completes an orbit about the Sun, is about 365.24 days, I would change 365 to a more precise value of 365.24</p>
<p>Then the quantity <span class="math-container">$n\equiv2\pi/365.24$</span> will be the so-called <em>mean motion</em> of the Earth, i.e., the mean orbital angular velocity.</p>
<p>Multiplying it by the number of day, we obtain a discretised version of the <em>mean anomaly</em>, as Greg Miller rightly pointed out in his comment above.</p>
<p>Whether you wish to define your <span class="math-container">$\Gamma$</span> as <span class="math-container">$n\,(n_{day} -1)$</span> or as <span class="math-container">$n\,n_{day} $</span> is a matter of choice. In the former case, your <span class="math-container">$\Gamma$</span> will be the mean anomaly value in the beginning of the day number <span class="math-container">$n_{day}$</span>. In the latter case, in the end of this day. For example, in <a href="https://hal-mines-paristech.archives-ouvertes.fr/hal-01676634v1/preview/2018_basics_solaire_wald_v1.pdf" rel="nofollow noreferrer">this lecture course</a> the author preferred the latter option -- but then, again, it was a matter of his choice.</p>

CC BY-SA 4.0
2023-06-15T17:39:41.360Z
2023-06-15T17:39:41.360Z
Michael_1812
<p>I am searching for a <em>proper definition</em> for day angle.
The day angle  <span class="math-container">$\Gamma$</span> is calculated as follows:<span class="math-container">$$\Gamma = 2\pi\cdot \frac{n_{day} - 1}{365}$$</span></p>

CC BY-SA 4.0
2023-06-15T16:05:17.033Z
Some context and the source might help.  But it looks like what Kepler calls the "mean annomoly".
Greg Miller
<declination>
Definition for Day Angle
Sadira Khan
# Definition for Day Angle

(Asked by: username_0 on 2023-06-15)

I am searching for a *proper definition* for day angle.
The day angle $\Gamma$ is calculated as follows:$$\Gamma = 2\pi\cdot \frac{n\_{day} - 1}{365}$$

**Tags:** declination
> **Comments:**
> **username_1**: Some context and the source might help.  But it looks like what Kepler calls the "mean annomoly".  

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## 1 answers

### Answer by username_2 on 2023-06-15. Score: 1

Since a solar year, i.e. the time over which the Earth completes an orbit about the Sun, is about 365.24 days, I would change 365 to a more precise value of 365.24

Then the quantity $n\equiv2\pi/365.24$ will be the so-called *mean motion* of the Earth, i.e., the mean orbital angular velocity.

Multiplying it by the number of day, we obtain a discretised version of the *mean anomaly*, as Greg Miller rightly pointed out in his comment above.

Whether you wish to define your $\Gamma$ as $n\,(n\_{day} -1)$ or as $n\,n\_{day} $ is a matter of choice. In the former case, your $\Gamma$ will be the mean anomaly value in the beginning of the day number $n\_{day}$. In the latter case, in the end of this day. For example, in [this lecture course](https://hal-mines-paristech.archives-ouvertes.fr/hal-01676634v1/preview/2018_basics_solaire_wald_v1.pdf) the author preferred the latter option -- but then, again, it was a matter of his choice.


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