The Semi-Major Axis of Binary Stars

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  • #1
Tom MS
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Wikipedia seems to think that a binary system is defined by a single semi-major axis, but I've seen other sources such as hyperphysics that define it using two semi-major axes. Is the semi-major axis of the system simply the average of the two?
 

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  • #2
jim mcnamara
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I do not know what the standard approach is to this problem. I, too, have seen both. Never for the same system, AFAIK. My take is:

Mass difference of the two stars makes a big difference in the axes of the orbit of each companion. Take this binary as an example:
O type star, mass ~90M☉ and a type G star with mass 1M☉. It is likely to have an orbit , described simply -- the smaller star orbits the bigger one. One semi-major axis.

When companions are closely matched mass-wise a simple one-axis description does not work well. Two semi-major axes, one for each orbital component more closely matches the data. Since the variety of binaries' mass differences is large there must be a more or less standard approach to the problem. Or it may relate to available data for the system. Don't know.

Also note that Wikipedia is never the best final arbiter on scientific questions.
 
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Vanadium 50
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Since two points determine a line, how can you have multiple semi-major axes?
 
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Janus
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Since two points determine a line, how can you have multiple semi-major axes?

There is the semi-major axis of the distance between the centers of the two objects, and then there are the semi-major axis for the barycentric orbits of each body.
 
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  • #5
Ken G
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Since the center of mass of the whole system won't move, the orbits of the two stars independently (with two semi-major axes) are simply scaled-down versions of the "orbit" of the displacement between the stars (one semi-major axis that is the sum, not the average, of the other two). The scale factor is set by the mass ratio, where it is 1/2 if the two stars have the same mass, and 1 for the lower-mass star in the limit that the ratio goes to 0. So the scale factor for the individual orbit of star 1 is $m_2/(m_1+m_2)$, and its orbit is just a miniature copy of the orbit of the total displacement, shrunk by that factor (and of course, the two stars' independent orbits are in phase with each other but mirror reflected).
 

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