<p>If you have the period then you also know the <em>angular</em> velocity. This in turn can be expressed in terms of the separation (be careful about where the centre of mass is) and the linear velocity (which you are told). This gives you the separation.</p>
<p>If you have the period, then Kepler's third law can be used to estimate the separation in terms of the total system mass (which is double the mass of one component). This gives you the component masses.</p>

CC BY-SA 4.0
2022-10-26T14:03:33.610Z
Thank you... I have now found the separation of the binary star components but I'm unclear how to find to masses of the binary star components compared to the Sun. I've plugged in all the values I know in the formula (Period P, separation a, gravitational constant G) from which I can find a value for the sum of the masses of the binary star components, but how do I get the masses relative to the Sun from this?
Ray61
CC BY-SA 4.0
2022-10-26T16:24:05.630Z
@Ray61  If you know the sum of the masses and that each component has the same mass, then what's the problem? Divide the mass by 2e30 kg to put the masses in units of a solar mass.
ProfRob
CC BY-SA 4.0
2022-10-24T16:24:21.510Z
2022-10-24T16:24:21.510Z
ProfRob
<p>Let's rephrase:</p>
<p>Suppose a tiny satellite is moving at 19 AU per year and takes 0.1 years to complete a circular orbit around a star.  What is the circumference of the orbital path?  What is the distance from the star to the satellite? (Hint: What is the equation relating the radius to the circumference of a circle?)</p>
<p>Suppose two stars of equal mass are orbiting their mutual barycenter, each with speed 19 AU per year with an orbital period of 0.1 years.  What is their separation?</p>
<p>Kepler's third law for circular binary systems can be written as: <span class="math-container">$$\frac{P^2}{a^3}=\frac{4\pi^2}{G(m_1+m_2)}$$</span><br />
Here, <span class="math-container">$P$</span> is the orbital period, <span class="math-container">$a$</span> is the separation distance betwen the stars, <span class="math-container">$G$</span> is the gravitational constant, and <span class="math-container">$m_1$</span> and <span class="math-container">$m_2$</span> are the masses of the respective stars.  Can you solve for the sum of the masses?  What is the mass of each star?</p>
<p><strong>Edit:</strong> What value should we use for <span class="math-container">$G$</span>?  You might be tempted to use:
<span class="math-container">$$G = 6.674\times 10^{−11} \frac{\text{m}^3}{\text{kg}* \text{s}^2}$$</span></p>
<p>If you use this value, your units won't cancel.  Instead, we can plug in the values for Earth's orbit into Kepler's third law above with <span class="math-container">$P=1 \text{year}$</span>, <span class="math-container">$a= 1 \text{AU}$</span>, and <span class="math-container">$m_1+m_2 \approx M_\odot$</span> (since the mass of the Earth is negligible compared to the mass of the Sun).  Then solve for <span class="math-container">$G$</span> to get:</p>
<p><span class="math-container">$$ G = 4\pi^2 \frac{\text{AU}^3}{\text{year}^2M_\odot} $$</span></p>
<p>This value of <span class="math-container">$G$</span> will allow nice cancelation of units so that your final solution will be in terms of Solar Masses (<span class="math-container">$M_\odot$</span>) as required by the question.</p>
<p>If you want another source for the same info, check out <a href="https://en.wikipedia.org/wiki/Gravitational_constant#Orbital_mechanics" rel="nofollow noreferrer">wikipedia</a></p>
<blockquote>
<p><span class="math-container">$$ G=4\pi ^{2}\mathrm {\ AU^{3}{\cdot }yr^{-2}} \ M^{-1}\approx
&gt; 39.478\mathrm {\ AU^{3}{\cdot }yr^{-2}} \ M_{\odot }^{-1}$$</span></p>
</blockquote>

CC BY-SA 4.0
2022-10-25T16:36:18.783Z
Thank you for your advice, I tried reworking the problem but I am still hitting a blockage. I managed to deduce that the radius of the orbiting binary star is 0.3 AU, but I have no solution for the masses of the stars in the binary system, only that they are equal to each other. Any further advice appreciated!
Ray61
CC BY-SA 4.0
2022-10-25T17:16:26.737Z
@Ray61 Excellent work.  If the radius of each star's orbit to the barycenter is 0.3 AU, what is $a$, the separation between the 2 stars?  Now you should know $a$, and $P$=0.1 years, $G = (2\pi)^2$ (for AU and solar masses).  Now, plug all these values into Kepler's 3rd law as written above.  Note that since the masses are the same, you can substitute $m_1+m_2$ by $2m$.  Solve for $m$!  Your result will be in the units of Solar Masses, which is the units asked for in the original question!
Connor Garcia
CC BY-SA 4.0
2022-10-25T17:24:37.360Z
@Ray61 If you get an answer, please let us know what it is! (Hint: I got a mass for each star between 7 and 11 solar masses.)
Connor Garcia
CC BY-SA 4.0
2022-10-25T22:54:42.173Z
So separation is double 0.3 AU = 0.6 AU.
Ray61
CC BY-SA 4.0
2022-10-25T22:58:05.147Z
Plugging into Kepler's 3rd law gives me: 2m = [(0.6)^3 x 4(pi)^2]/[(0.1)^2 x 6.672x10^-11] = 1.278x10^11.  Can't see how you get between 7 and 11 solar masses...?
Ray61
CC BY-SA 4.0
2022-10-26T00:09:53.057Z
@Ray61 Great progress!  You got the separation!  I wonder if our disconnect is the units value of $G$. Instead of 6.672x10^-11, try $4\pi^2$
Connor Garcia
CC BY-SA 4.0
2022-10-26T13:56:28.187Z
Let us [continue this discussion in chat](https://chat.stackexchange.com/rooms/140155/discussion-between-ray61-and-connor-garcia).
Ray61
CC BY-SA 4.0
2022-10-24T17:27:40.357Z
2022-10-26T16:00:29.887Z
2022-10-26T16:00:29.887Z
Connor Garcia
<p>I'm struggling with the following problem:
An eclipsing binary has period 0.10 years and each component has speed in its orbit of 19AU per year (or 90km/s) Assuming both stars have the same mass and that their orbits are circular, calculate their separation in AU and their masses relative to the Sun.</p>
<p>I've tried solving using Newton's laws of motion and gravity and Kepler's 3rd law, but can't seem to get the expressions I need for the separation and relative masses.</p>

<binary-star><keplers-laws>
Calculating the separation of binary stars
Ray61
# Calculating the separation of binary stars

(Asked by: username_0 on 2022-10-24)

I'm struggling with the following problem:
An eclipsing binary has period 0.10 years and each component has speed in its orbit of 19AU per year (or 90km/s) Assuming both stars have the same mass and that their orbits are circular, calculate their separation in AU and their masses relative to the Sun.

I've tried solving using Newton's laws of motion and gravity and Kepler's 3rd law, but can't seem to get the expressions I need for the separation and relative masses.

**Tags:** binary-star, keplers-laws

---

## 2 answers

### Answer by username_2 on 2022-10-24. Score: 5 | Accepted answer

Let's rephrase:

Suppose a tiny satellite is moving at 19 AU per year and takes 0.1 years to complete a circular orbit around a star. What is the circumference of the orbital path? What is the distance from the star to the satellite? (Hint: What is the equation relating the radius to the circumference of a circle?)

Suppose two stars of equal mass are orbiting their mutual barycenter, each with speed 19 AU per year with an orbital period of 0.1 years. What is their separation?

Kepler's third law for circular binary systems can be written as: $$\frac{P^2}{a^3}=\frac{4\pi^2}{G(m\_1+m\_2)}$$  
Here, $P$ is the orbital period, $a$ is the separation distance betwen the stars, $G$ is the gravitational constant, and $m\_1$ and $m\_2$ are the masses of the respective stars. Can you solve for the sum of the masses? What is the mass of each star?

**Edit:** What value should we use for $G$? You might be tempted to use:
$$G = 6.674\times 10^{−11} \frac{\text{m}^3}{\text{kg}\* \text{s}^2}$$

If you use this value, your units won't cancel. Instead, we can plug in the values for Earth's orbit into Kepler's third law above with $P=1 \text{year}$, $a= 1 \text{AU}$, and $m\_1+m\_2 \approx M\_\odot$ (since the mass of the Earth is negligible compared to the mass of the Sun). Then solve for $G$ to get:

$$ G = 4\pi^2 \frac{\text{AU}^3}{\text{year}^2M\_\odot} $$

This value of $G$ will allow nice cancelation of units so that your final solution will be in terms of Solar Masses ($M\_\odot$) as required by the question.

If you want another source for the same info, check out [wikipedia](https://en.wikipedia.org/wiki/Gravitational_constant#Orbital_mechanics)

> $$ G=4\pi ^{2}\mathrm {\ AU^{3}{\cdot }yr^{-2}} \ M^{-1}\approx
> > 39.478\mathrm {\ AU^{3}{\cdot }yr^{-2}} \ M\_{\odot }^{-1}$$

> **Comments:**
> **username_0**: Thank you for your advice, I tried reworking the problem but I am still hitting a blockage. I managed to deduce that the radius of the orbiting binary star is 0.3 AU, but I have no solution for the masses of the stars in the binary system, only that they are equal to each other. Any further advice appreciated!  
> **username_2**: @Ray61 Excellent work.  If the radius of each star's orbit to the barycenter is 0.3 AU, what is $a$, the separation between the 2 stars?  Now you should know $a$, and $P$=0.1 years, $G = (2\pi)^2$ (for AU and solar masses).  Now, plug all these values into Kepler's 3rd law as written above.  Note that since the masses are the same, you can substitute $m_1+m_2$ by $2m$.  Solve for $m$!  Your result will be in the units of Solar Masses, which is the units asked for in the original question!  
> **username_2**: @Ray61 If you get an answer, please let us know what it is! (Hint: I got a mass for each star between 7 and 11 solar masses.)  
> **username_0**: So separation is double 0.3 AU = 0.6 AU.  
> **username_0**: Plugging into Kepler's 3rd law gives me: 2m = [(0.6)^3 x 4(pi)^2]/[(0.1)^2 x 6.672x10^-11] = 1.278x10^11.  Can't see how you get between 7 and 11 solar masses...?  
> **username_2**: @Ray61 Great progress!  You got the separation!  I wonder if our disconnect is the units value of $G$. Instead of 6.672x10^-11, try $4\pi^2$  
> **username_0**: Let us [continue this discussion in chat](https://chat.stackexchange.com/rooms/140155/discussion-between-ray61-and-connor-garcia).  

---

### Answer by username_1 on 2022-10-24. Score: 6

If you have the period then you also know the *angular* velocity. This in turn can be expressed in terms of the separation (be careful about where the centre of mass is) and the linear velocity (which you are told). This gives you the separation.

If you have the period, then Kepler's third law can be used to estimate the separation in terms of the total system mass (which is double the mass of one component). This gives you the component masses.

> **Comments:**
> **username_0**: Thank you... I have now found the separation of the binary star components but I'm unclear how to find to masses of the binary star components compared to the Sun. I've plugged in all the values I know in the formula (Period P, separation a, gravitational constant G) from which I can find a value for the sum of the masses of the binary star components, but how do I get the masses relative to the Sun from this?  
> **username_1**: @Ray61  If you know the sum of the masses and that each component has the same mass, then what's the problem? Divide the mass by 2e30 kg to put the masses in units of a solar mass.  

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