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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.

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2 Answers 2

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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.

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    $\begingroup$ 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? $\endgroup$
    – Ray61
    Commented Oct 26, 2022 at 14:03
  • $\begingroup$ @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. $\endgroup$
    – ProfRob
    Commented Oct 26, 2022 at 16:24
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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

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

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    $\begingroup$ 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! $\endgroup$
    – Ray61
    Commented Oct 25, 2022 at 16:36
  • $\begingroup$ @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! $\endgroup$
    – Connor Garcia
    Commented Oct 25, 2022 at 17:16
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    $\begingroup$ So separation is double 0.3 AU = 0.6 AU. $\endgroup$
    – Ray61
    Commented Oct 25, 2022 at 22:54
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    $\begingroup$ 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...? $\endgroup$
    – Ray61
    Commented Oct 25, 2022 at 22:58
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    $\begingroup$ Let us continue this discussion in chat. $\endgroup$
    – Ray61
    Commented Oct 26, 2022 at 13:56

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