<p>The formula you gave is to find the hour angle of the star while setting, not the setting time itself. Suppose the hour angle of star is $HA\star$, and $RA=\alpha$,then the Local Sidereel Time is given by $LST=HA\star+\alpha$. At the time of setting, $HA\star=10^h1^m30.2^s$. Thus, $LST=13^h16^m30.2^s$at the time of setting.</p>

<p>Now, $RA\odot=6^h$ on $\text{June} 21^{st}$ and $HA_\odot =\arccos(\tan(23.43^{\circ}))=4^h17^m17^s$ at sunset. Thus the star sets with the sun after $\text{ June} 21^{st}$. Since $RA_\odot$ increases approximately by $2^h$ every month, we can assume that the star sets with the sen near $\text{July }21^{st}$. Let it set after $n$ days. At $\text{July }21^{st}$, $\delta_\odot = 20.09^{\circ}$ which gives us $HA_\odot=4^h34^m11^s$, and $RA_{\odot}=8^h9^m44^s$. The $HA_\odot$ at sunset is approximately the same for nearby dates. Since $\delta_\odot$ is also near the maximum, we can approximate $\Delta RA_\odot= \omega_\odot n$. Thus,</p>

<p>$$13^h16^m30.2^s=8^h9^m44^s + \omega_\odot n +  4^h34^m11^s$$
$$ \omega_\odot n =0.5433^h$$
$$n=8$$</p>

<p>At  $\text{July } 29^{th}$, $\delta_\odot=18.23^{\circ}$, $HA_\odot= 4^h43^m6^s$, $RA_\odot=8^h42^m12^s$ which gives us $LST=13^h25^m18^s$, which is a bit more than expected. Trial and error for few days before gives us that on $\text{July } 26^{th}$, the sun sets at LST $13^h15^m9^s$ which would be the most appropriate answer.</p>

<p>Thus, the sun sets along with the star at $\text{July } 26^{th}$</p>

CC BY-SA 3.0
2018-03-29T15:51:07.633Z
2018-03-29T19:04:27.713Z
2018-03-29T19:04:27.713Z
Jim Haddocc
<blockquote>
<p>Problem: Suppose an observer's latitude is 45 S and a star's RA/DEC is
3h15min/41S. On what date will the sun set with the star.</p>
</blockquote>
<p>I have been stuck on this problem for a while.</p>
<p>I found out that the time of setting of a star is found using the formula:</p>
<blockquote>
<p><span class="math-container">$$\alpha+\arccos(-\tan(\delta).\tan(\phi))$$</span>
<span class="math-container">$\implies3\frac{1}{4}^h+\arccos(-\tan(-41)\tan(-45)) = \alpha_{sun}+\arccos(-\tan(\delta_{sun})\tan(-45))$</span></p>
<p><span class="math-container">$\implies 47.8 = \alpha_{sun}+arccos(tan(\delta_{sun}))$</span></p>
</blockquote>
<p>I am not able to solve this any further.</p>
<p>Is there something wrong in my approach? Something I am missing?</p>
<p>If not how do I solve the above equation further?</p>
<p>Thanks in anticipation!</p>

CC BY-SA 3.0
2018-03-29T13:42:38.997Z
It means that the star is on the horizon at the same time as the sun is on the horizion in the West. Its related to the old problem of estimating the Heliacal rising of stars.
James K
CC BY-SA 3.0
2018-03-29T14:36:12.677Z
You sure you're not doing it the hard way?  Here's the standard definition: "0 hours right ascension is by convention the right ascension of the sun on the vernal equinox, March 21."  the DEC is not involved in this set/rise calculation.  So maybe if you can calculate the sun's set time, just invert until you get the desired RA value.
Carl Witthoft
CC BY-SA 3.0
2018-03-29T19:04:58.967Z
I've edited my answer to make it more accurate
Jim Haddocc
<star><the-sun><declination><right-ascension>
When does a star rise with the sun?
Arhaan Ahmad
# When does a star rise with the sun?

(Asked by: username_0 on 2018-03-29)

> Problem: Suppose an observer's latitude is 45 S and a star's RA/DEC is
> 3h15min/41S. On what date will the sun set with the star.

I have been stuck on this problem for a while.

I found out that the time of setting of a star is found using the formula:

> $$\alpha+\arccos(-\tan(\delta).\tan(\phi))$$
> $\implies3\frac{1}{4}^h+\arccos(-\tan(-41)\tan(-45)) = \alpha\_{sun}+\arccos(-\tan(\delta\_{sun})\tan(-45))$
>
> $\implies 47.8 = \alpha\_{sun}+arccos(tan(\delta\_{sun}))$

I am not able to solve this any further.

Is there something wrong in my approach? Something I am missing?

If not how do I solve the above equation further?

Thanks in anticipation!

**Tags:** star, the-sun, declination, right-ascension
> **Comments:**
> **username_1**: It means that the star is on the horizon at the same time as the sun is on the horizion in the West. Its related to the old problem of estimating the Heliacal rising of stars.  
> **username_2**: You sure you're not doing it the hard way?  Here's the standard definition: "0 hours right ascension is by convention the right ascension of the sun on the vernal equinox, March 21."  the DEC is not involved in this set/rise calculation.  So maybe if you can calculate the sun's set time, just invert until you get the desired RA value.  
> **username_3**: I've edited my answer to make it more accurate  

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## 1 answers

### Answer by username_3 on 2018-03-29. Score: 3 | Accepted answer

The formula you gave is to find the hour angle of the star while setting, not the setting time itself. Suppose the hour angle of star is $HA\star$, and $RA=\alpha$,then the Local Sidereel Time is given by $LST=HA\star+\alpha$. At the time of setting, $HA\star=10^h1^m30.2^s$. Thus, $LST=13^h16^m30.2^s$at the time of setting.

Now, $RA\odot=6^h$ on $\text{June} 21^{st}$ and $HA\_\odot =\arccos(\tan(23.43^{\circ}))=4^h17^m17^s$ at sunset. Thus the star sets with the sun after $\text{ June} 21^{st}$. Since $RA\_\odot$ increases approximately by $2^h$ every month, we can assume that the star sets with the sen near $\text{July }21^{st}$. Let it set after $n$ days. At $\text{July }21^{st}$, $\delta\_\odot = 20.09^{\circ}$ which gives us $HA\_\odot=4^h34^m11^s$, and $RA\_{\odot}=8^h9^m44^s$. The $HA\_\odot$ at sunset is approximately the same for nearby dates. Since $\delta\_\odot$ is also near the maximum, we can approximate $\Delta RA\_\odot= \omega\_\odot n$. Thus,

$$13^h16^m30.2^s=8^h9^m44^s + \omega\_\odot n + 4^h34^m11^s$$
$$ \omega\_\odot n =0.5433^h$$
$$n=8$$

At $\text{July } 29^{th}$, $\delta\_\odot=18.23^{\circ}$, $HA\_\odot= 4^h43^m6^s$, $RA\_\odot=8^h42^m12^s$ which gives us $LST=13^h25^m18^s$, which is a bit more than expected. Trial and error for few days before gives us that on $\text{July } 26^{th}$, the sun sets at LST $13^h15^m9^s$ which would be the most appropriate answer.

Thus, the sun sets along with the star at $\text{July } 26^{th}$


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