<p>The semi-major axis is 4.3 <em>arcseconds</em>. At a distance of 3.51 parsecs, this corresponds to 15.09 au. Using your formula I get 40.3 years, which is close enough given the 2 significant figure precision of the semi-major axis.</p>

CC BY-SA 4.0
2018-11-11T17:23:08.333Z
2018-11-11T17:23:08.333Z
ProfRob
<p>I've been experimenting with Binary Star System calculations but came across some issues.</p>

<p>I can correctly calculate the Semi-Major Axis of the Earth's orbit around the Sun by doing the equation:</p>

<p><a href="https://i.sstatic.net/acbQf.png" rel="noreferrer"><img src="https://i.sstatic.net/acbQf.png" alt="T=2pi*SQRT(a^3/GM)"></a></p>

<p>Using data from Wikipedia. </p>

<pre><code>a = (1.521e11 + 1.47095e11) / 2.   &lt;- average radius of Earth's orbit (meters)
μ = G * M = 6.67408e(-11) * 1.98847e30 gravitational constant * Mass of Sun (kg)
</code></pre>

<p>I got ≈ 31557884 seconds which equates to ≈ a year</p>

<p>Now my issue is that I've been searching around for a formula that calculates the period and etc of a Binary System where two masses (such as stars) orbit around each other.</p>

<p>So then I try essentially the same formula, <a href="https://en.wikipedia.org/wiki/Orbital_period" rel="noreferrer">this</a>:</p>

<p><a href="https://i.sstatic.net/dIaZ6.png" rel="noreferrer"><img src="https://i.sstatic.net/dIaZ6.png" alt="T = 2pi * sqrt(a^3 / (G(M1 + M2)))"></a></p>

<p>I used data from the <a href="https://en.wikipedia.org/wiki/Procyon" rel="noreferrer">Procyon A and B</a> (binary system)</p>

<pre><code>a = Semi-Major Axis = 4.3 AU
M1 = Procyon A Mass = 1.499 Solar Masses
M2 = Procyon B Mass = 0.602 Solar Masses
</code></pre>

<p>these give me a T (period) of 4731227. It says 40.82 years on Wiki.</p>

<p>What am I doing wrong? Is it the data? is it the formula? My comprehension of the question? Thank you for the help!</p>

CC BY-SA 4.0
2018-11-11T09:52:05.853Z
Welcome to Astronomy! For starters, in case you are in a hurry: https://en.wikipedia.org/wiki/Reduced_mass but I think someone will write a full answer fairly soon.
uhoh
CC BY-SA 4.0
2018-11-11T10:47:11.460Z
or [Two-body problem](https://en.wikipedia.org/wiki/Two-body_problem), *also or* [Gravitational two-body problem](https://en.wikipedia.org/wiki/Gravitational_two-body_problem).
uhoh
CC BY-SA 4.0
2020-04-06T12:36:15.247Z
T[s], a [m], M [kg]
Andrew Łukasik
<orbit><binary-star>
Doing Correct Calculations with Binary Star Systems
chriscrutt
# Doing Correct Calculations with Binary Star Systems

(Asked by: username_0 on 2018-11-11)

I've been experimenting with Binary Star System calculations but came across some issues.
I can correctly calculate the Semi-Major Axis of the Earth's orbit around the Sun by doing the equation:

Using data from Wikipedia. 
a = (1.521e11 + 1.47095e11) / 2.   <- average radius of Earth's orbit (meters)
μ = G * M = 6.67408e(-11) * 1.98847e30 gravitational constant * Mass of Sun (kg)

I got ≈ 31557884 seconds which equates to ≈ a year
Now my issue is that I've been searching around for a formula that calculates the period and etc of a Binary System where two masses (such as stars) orbit around each other.
So then I try essentially the same formula, this:

I used data from the Procyon A and B (binary system)
a = Semi-Major Axis = 4.3 AU
M1 = Procyon A Mass = 1.499 Solar Masses
M2 = Procyon B Mass = 0.602 Solar Masses

these give me a T (period) of 4731227. It says 40.82 years on Wiki.
What am I doing wrong? Is it the data? is it the formula? My comprehension of the question? Thank you for the help!

**Tags:** orbit, binary-star
> **Comments:**
> **username_1**: Welcome to Astronomy! For starters, in case you are in a hurry: https://en.wikipedia.org/wiki/Reduced_mass but I think someone will write a full answer fairly soon.  
> **username_1**: or [Two-body problem](https://en.wikipedia.org/wiki/Two-body_problem), *also or* [Gravitational two-body problem](https://en.wikipedia.org/wiki/Gravitational_two-body_problem).  
> **username_2**: T[s], a [m], M [kg]  

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## 1 answers

### Answer by username_3 on 2018-11-11. Score: 5 | Accepted answer

The semi-major axis is 4.3 *arcseconds*. At a distance of 3.51 parsecs, this corresponds to 15.09 au. Using your formula I get 40.3 years, which is close enough given the 2 significant figure precision of the semi-major axis.


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