My Textbook An Introduction to Modern Astrophysics 2nd edition (2017), page 557 mentions that the modern value is 49.9 years. But Wikipedia mentions 50.1284±0.0043 years.

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


2 Answers 2


Bond et al. (2017) 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) 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.


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.

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    $\begingroup$ 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. $\endgroup$
    – uhoh
    Commented Jul 7, 2021 at 16:50
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    $\begingroup$ 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 $\endgroup$ Commented Jul 7, 2021 at 16:55
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    $\begingroup$ 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. $\endgroup$ Commented Jul 8, 2021 at 0:19
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    $\begingroup$ @planetmaker Error bars/confidence intervals in astronomy almost always indicate 68% (“1-sigma”) limits, not “95% or 99%”. $\endgroup$ Commented Jul 8, 2021 at 9:21

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