I am writing a blog post and while I am interested in astronomy I havent done any physics like this since high school. My question is: what would the mass and density of Sirius A be if the mass of Sirius B is roughly the same as the Earth? If Sirius B is actually only as heavy as the Earth then Sirius A would need to be far less dense. How dense would it need to be to be the same size it currently is and what would its mass be?

The orbit of Sirius A and Sirius B has been observed. But the true mass and density of Sirius A and Sirius B cannot be known with certainty like with the orbit. So, it is possible that Sirius B actually has the mass and density roughly equal to the Earth. The light we see when we look at Sirius B is some kind of alien technology they are using to block out cosmic radiation.

If this fictional scenario were the case then what would the mass and density of Sirius A need to be in order for the orbits to be the same given that Sirius B has the mass and density roughly equal to the Earth?

Additionally, how dense would Sirius A be compared to Betelgeuse in this scenario?

The orbital motion reveals that Sirius A, the brighter star, has 2.02 ± 0.03 solar masses and Sirius B, the white dwarf, has 1.00 ± 0.02 solar masses.

Betelgeuse has a radius 1,100 times that of the Sun but its mass is only about 8 – 20 times the Sun. This means the density of Betelgeuse is much, much lower than the Sun. The average density of Betelgeuse is about a million times less dense than Earth’s atmosphere at sea level. That’s about the same as a vacuum found in an insulating Thermos bottle.

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    $\begingroup$ You may find this helpful: en.wikipedia.org/wiki/Gravitational_two-body_problem $\endgroup$ – PM 2Ring Nov 18 '20 at 5:35
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    $\begingroup$ While painful, it would be great if you took a look at that link and made at least a start at a calculation, the reason being that for whatever reason, questions often get better and more thorough answers if "what have you tried" is addressed in some way. Of course someone could prove me wrong and post an answer soon just as easily :-) $\endgroup$ – uhoh Nov 18 '20 at 5:40

The masses of the two objects are completely determined by a careful measurement of their relative positions on the sky over decades, combined with an accurately known distance.

There is no degeneracy in this solution. For a given orbital period and separation of the two objects, the masses are entirely determined. If Sirius B were not a massive white dwarf then the orbit would be different. Specifically, if you made the total system mass smaller, but with the same orbital period, then the size of the orbit would be smaller. If you tried to compensate by adding mass to Sirius A, then it would almost coincide with the centre of mass of the system and would show almost no orbital motion.

The density is then found by combining these masses with radii estimated from the measured luminosities and temperatures of the components and is therefore also fixed (to about $\pm 10$% because of the observational uncertainties in the mass and radius determinations).

  • $\begingroup$ Thank you for the thoughtful and professional comment but I am going to have to attempt to speak against your conclusion. Rotation has a non-zero effect on the orbit of a large body. Your instincts are to deny this. You might say that bodies have rotational invariance. But the effect that the tides have on the Moon is observed proof that the rotations of the bodies have an effect on the orbit. We have never been to Sirius so you just dont know that rotation isnt playing a factor. $\endgroup$ – Thomas Blobaum Nov 18 '20 at 8:36
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    $\begingroup$ @ThomasBiobaum For a given orbital separation and orbital period, the rotation has no effect on the mass determination. The transfer of angular momentum between the rotation of individual bodies and the orbit changes the orbital period and separation, but in such a way that it has has zero effect on the masses you would estimate from observing the orbit. $\endgroup$ – ProfRob Nov 18 '20 at 12:05

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