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I am confused on how to calculate the distances for each star in a binary system to their centre of mass. I am trying to solve it from a book I am reading which shows the solution but I do not understand the solution.

So the info I am given is:

A system is located at a distance D (in parsecs). The two stars orbit around their centre of mass. Star A has an angular separation of a (in arcsecs) and star B has an angular separation of b (arcsecs).

So now I am trying to find the physical separation of the stars to their centre of mass.

The book's solution (once converting to correct units) is:

Distance for star A = D tan(a)
Distance for star B = D tan(b)

I don't fully understand why they use tangent here because when I draw out a diagram of this situation i seem to need to use sine not tangent.

Here is how I drew it:

enter image description here

Why do they use tangent function and not the sine function? Have I drawn my diagram correctly?

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2 Answers 2

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It actually doesn't really matter: the angular separation is sufficiently small that you can use the small angle approximations $\sin \theta \approx \theta$ and $\tan \theta \approx \theta$ (where $\theta$ is in radians). In both cases the next term in the Taylor series is $\mathcal{O}(\theta^3)$ and the effect on the answer of dropping this and all subsequent terms of the Taylor series is in most realistic cases going to be utterly dwarfed by the uncertainty in the distance.

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  • $\begingroup$ Oh so i was correct anyway and doubted myself for no reason? One thing i did wonder though is - does this not assume that the system is face on to the Earth, what if it was slanted, wouldn't this equation fall apart? $\endgroup$
    – WDUK
    Sep 22, 2020 at 7:34
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There's one mistake in your drawing:

The center of mass of a two-body system lies on the straight line between their centers. So, the 90° angles should be placed between D and distA (and D and distB). Then the tangent applies instead of the sine.

But, as Antispinwards said, it doesn't matter for small angles.

What matters is, whether our viewing direction is really perpendicular to their orbiting plane. Even circular orbits will look elliptical if not viewed exactly from "above". But that might be beyond the scope of exercise from your book.

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  • $\begingroup$ If it wasn't perpendicular to viewing direction, how would you be able to calculate it ? Yeah the book doesn't go beyond basics of it. $\endgroup$
    – WDUK
    Sep 23, 2020 at 19:59
  • $\begingroup$ @WDUK As long as you assume circular orbits, at their maximum separation you will have a perpendicular view. $\endgroup$
    – Ralf Kleberhoff
    Sep 23, 2020 at 20:03
  • $\begingroup$ I thought circular orbits were pretty uncommon unless you mean "close enough to perfect circles" ? $\endgroup$
    – WDUK
    Sep 23, 2020 at 20:44

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