<p><a href="https://arxiv.org/abs/1703.10625" rel="noreferrer">Bond et al. (2017)</a> measure the orbital period of the Sirius system to be <span class="math-container">$50.1284 \pm 0.0043$</span> years. I believe this is the most precise and accurate value (I cannot find any more recent papers, with new determinations, that cite this paper).</p>
<p>An earlier, comprehensive study by <a href="https://ui.adsabs.harvard.edu/abs/1978ApJ...225..191G/abstract" rel="noreferrer">Gatewood &amp; Gatewood (1978)</a>  gave <span class="math-container">$50.090\pm 0.056$</span> years; consistent with the later measurement, but less precise.</p>
<p>I am unclear where 49.9 years would come from and it is inconsistent even with measurements from the 1970s.</p>

CC BY-SA 4.0
2021-07-07T17:33:26.653Z
2021-07-07T17:41:08.207Z
2021-07-07T17:41:08.207Z
ProfRob
<p>Calculations like this one for the orbital period of a binary system are hardly ever done analytically.
Astronomers usually take some known approximations and extend it to system like Sirius, and get an estimate for their period numerically (and a good one, by the looks of it).</p>
<p>If you're curious, we can derive this version of the 3rd law by considering the acceleration of the center of mass of the two stars:</p>
<p><span class="math-container">\begin{equation}
m_{1}r_{1}=m_{2}r_{2}
\end{equation}</span>
<span class="math-container">\begin{equation}
r = r_1+r_2
\end{equation}</span></p>
<p>The latter being the total separation between the two stars.</p>
<p>Additionally we <em>approximate</em> the orbits to be circular, we can say that:</p>
<p><span class="math-container">\begin{equation}
Pv_i = 2\pi r_i
\end{equation}</span></p>
<p>Where P is the period of orbit.
Which gives <strong>r</strong> as:</p>
<p><span class="math-container">\begin{equation}
r = \frac{P}{2\pi}(v_1+v_2)
\end{equation}</span></p>
<p>Now you just substitute this into the standard Kepler's 3rd Law and do a bit of rearranging:</p>
<p><span class="math-container">\begin{equation}
\frac{m_2^3}{(m_1+m_2)^3}=\frac{Pv_1^3}{2\pi G}
\end{equation}</span></p>
<p>Now we have an approximate relationship between the mass of the stars, their orbital period and the observed orbital velocity (not taking into account the angle of observation).</p>
<p>In conclusion, you shouldn't really be looking at decimal places when it comes to calculations like this one; we're not proving theorems here, so an error below 0.1 of a year is honestly amazing to me. Especially since we have been looking at Sirius for thousands of years, but only realised it's a binary system in the last 150.</p>

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2021-07-07T16:50:07.947Z
*Welcome to Stack Exchange!* It's necessary to address the question as asked: What exactly is the orbital period value of Sirius binary star system?* One value given is 49.9, the other is 50.1284±0.0043 Considering that there are photographic plates with positional data for two full orbits now using a variety of techniques with great results, I think it is just fine to believe those error bars. The arXiv paper is linked in a comment below the question. Why not skim it and see if you might want to adjust your conclusions a bit.
uhoh
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2021-07-07T16:55:44.033Z
In science the decimal places given, especially when also with an error are not something to disregard lightly, but they indicate (in astronomy) usually the 95% or 99% confidentiality of the authors in that result. Thus a general statement like "disregard" does not honour the given and achieved precision - nor does it answer the question. An answer will need to dwell on why different sources have different values which might not be in agreement within their errors. E.g.: are there systematic or other errors not accounted for? Different data sets? Etc
planetmaker
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2021-07-08T00:19:01.913Z
Also, observation of double stars, where one object gravitates around the other, are the only way of knowing a star’s mass, so in this case, we *cannot* know the period from the mass, but we derive the mass from the period. This kind of observation is the only *direct* way to know a star’s mass.
Pierre Paquette
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2021-07-08T09:21:55.157Z
@planetmaker Error bars/confidence intervals in astronomy almost always indicate 68% (“1-sigma”) limits, not “95% or 99%”.
Peter Erwin
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2021-07-07T16:18:39.533Z
2021-07-07T16:18:39.533Z
Marco Leonardi
<p>My Textbook <a href="https://www.cambridge.org/academic/subjects/physics/astrophysics/introduction-modern-astrophysics-2nd-edition" rel="nofollow noreferrer">An Introduction to Modern Astrophysics 2nd edition (2017)</a>, page 557 mentions that the modern value is 49.9 years. But <a href="https://en.wikipedia.org/wiki/Sirius" rel="nofollow noreferrer">Wikipedia mentions</a> 50.1284±0.0043 years.</p>
<p>Which is correct? Or, has the orbital period value changed and what Wikipedia mentions is the updated value?</p>

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2021-07-07T15:47:42.937Z
Did you check the reference wikipedia links to its data source?
planetmaker
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2021-07-07T15:51:28.227Z
Is it the 1st or [2nd edition (2017)](https://www.cambridge.org/academic/subjects/physics/astrophysics/introduction-modern-astrophysics-2nd-edition)? Actually, the Wikipedia page lists reference #11 for the orbital data, which is also from 2017! [The Sirius System and its Astrophysical Puzzles: Hubble Space Telescope and Ground-Based Astrometry](https://arxiv.org/abs/1703.10625) So this turns out to be an interesting question!
uhoh
CC BY-SA 4.0
2021-07-07T16:58:15.567Z
@uhoh It is the 2nd edition (2017).
James Smith
CC BY-SA 4.0
2021-07-07T17:16:15.073Z
@planetmaker yes, did just now.
James Smith
<observational-astronomy><astrophysics><binary-star><astrometry>
What exactly is the orbital period value of Sirius binary star system?
James Smith
# What exactly is the orbital period value of Sirius binary star system?

(Asked by: username_0 on 2021-07-07)

My Textbook [An Introduction to Modern Astrophysics 2nd edition (2017)](https://www.cambridge.org/academic/subjects/physics/astrophysics/introduction-modern-astrophysics-2nd-edition), page 557 mentions that the modern value is 49.9 years. But [Wikipedia mentions](https://en.wikipedia.org/wiki/Sirius) 50.1284±0.0043 years.

Which is correct? Or, has the orbital period value changed and what Wikipedia mentions is the updated value?

**Tags:** observational-astronomy, astrophysics, binary-star, astrometry

> **Comments:**
> **username_1**: Did you check the reference wikipedia links to its data source?  
> **username_2**: Is it the 1st or [2nd edition (2017)](https://www.cambridge.org/academic/subjects/physics/astrophysics/introduction-modern-astrophysics-2nd-edition)? Actually, the Wikipedia page lists reference #11 for the orbital data, which is also from 2017! [The Sirius System and its Astrophysical Puzzles: Hubble Space Telescope and Ground-Based Astrometry](https://arxiv.org/abs/1703.10625) So this turns out to be an interesting question!  
> **username_0**: @username_2 It is the 2nd edition (2017).  
> **username_0**: @username_1 yes, did just now.  

---

## 2 answers

### Answer by username_3 on 2021-07-07. Score: 8

[Bond et al. (2017)](https://arxiv.org/abs/1703.10625) measure the orbital period of the Sirius system to be $50.1284 \pm 0.0043$ years. I believe this is the most precise and accurate value (I cannot find any more recent papers, with new determinations, that cite this paper).

An earlier, comprehensive study by [Gatewood & Gatewood (1978)](https://ui.adsabs.harvard.edu/abs/1978ApJ...225..191G/abstract) gave $50.090\pm 0.056$ years; consistent with the later measurement, but less precise.

I am unclear where 49.9 years would come from and it is inconsistent even with measurements from the 1970s.


---

### Answer by username_4 on 2021-07-07. Score: -1

Calculations like this one for the orbital period of a binary system are hardly ever done analytically.
Astronomers usually take some known approximations and extend it to system like Sirius, and get an estimate for their period numerically (and a good one, by the looks of it).

If you're curious, we can derive this version of the 3rd law by considering the acceleration of the center of mass of the two stars:

\begin{equation}
m\_{1}r\_{1}=m\_{2}r\_{2}
\end{equation}
\begin{equation}
r = r\_1+r\_2
\end{equation}

The latter being the total separation between the two stars.

Additionally we *approximate* the orbits to be circular, we can say that:

\begin{equation}
Pv\_i = 2\pi r\_i
\end{equation}

Where P is the period of orbit.
Which gives **r** as:

\begin{equation}
r = \frac{P}{2\pi}(v\_1+v\_2)
\end{equation}

Now you just substitute this into the standard Kepler's 3rd Law and do a bit of rearranging:

\begin{equation}
\frac{m\_2^3}{(m\_1+m\_2)^3}=\frac{Pv\_1^3}{2\pi G}
\end{equation}

Now we have an approximate relationship between the mass of the stars, their orbital period and the observed orbital velocity (not taking into account the angle of observation).

In conclusion, you shouldn't really be looking at decimal places when it comes to calculations like this one; we're not proving theorems here, so an error below 0.1 of a year is honestly amazing to me. Especially since we have been looking at Sirius for thousands of years, but only realised it's a binary system in the last 150.

> **Comments:**
> **username_2**: *Welcome to Stack Exchange!* It's necessary to address the question as asked: What exactly is the orbital period value of Sirius binary star system?* One value given is 49.9, the other is 50.1284±0.0043 Considering that there are photographic plates with positional data for two full orbits now using a variety of techniques with great results, I think it is just fine to believe those error bars. The arXiv paper is linked in a comment below the question. Why not skim it and see if you might want to adjust your conclusions a bit.  
> **username_1**: In science the decimal places given, especially when also with an error are not something to disregard lightly, but they indicate (in astronomy) usually the 95% or 99% confidentiality of the authors in that result. Thus a general statement like "disregard" does not honour the given and achieved precision - nor does it answer the question. An answer will need to dwell on why different sources have different values which might not be in agreement within their errors. E.g.: are there systematic or other errors not accounted for? Different data sets? Etc  
> **username_5**: Also, observation of double stars, where one object gravitates around the other, are the only way of knowing a star’s mass, so in this case, we *cannot* know the period from the mass, but we derive the mass from the period. This kind of observation is the only *direct* way to know a star’s mass.  
> **username_6**: @username_1 Error bars/confidence intervals in astronomy almost always indicate 68% (“1-sigma”) limits, not “95% or 99%”.  

---

