This is a question which spun of this Can the Sun / Earth have a dark matter core?

I have argued that it is possible to pretty accurately estimate how much mass of Sun is in the regular matter, so comparing this estimate to the gravitational effect of Sun could rule out any significant esoteric mass sources.

A couple of points brought in there:

What do you mean by "visible mass"? We can only see the first ~100km of the Sun. The gravitational effects of dark matter are identical to those of normal matter. There is no debate about what the mass of the Sun is; only what it consists of. - Rob Jeffries

While we may know the Sun's total mass pretty accurately from the masses and velocities of orbiting objects, I think we cannot know how much of that mass may be dark matter. (We may be able to give an upper bound above which the Sun wouldn't work as it does now. But apart from that, I'm confident that a couple of percent of DM inside the Sun wouldn't be observable.) – JimmyB

By visible mass I mean attributed to "regular" matter as opposed to dark matter. I agree with Jimmy. The gravitational effects are the same, precisely, but other's are not, so judging from its radius, color, dynamics (solar seismology is also a thing) etc. I would bet that we could calculate how much regular matter there is, and how well it fits the total mass as observed by gravitational effects.

Are there such estimates and how precise they are, a couple of %? Could there be a discrepancy between Sun's gravitational effect and mass attributed to it's regular matter?

Edit: To avoid confusion I guess the answer I'm looking for is x% of Hydrogene, y% of Helium, z% of heavier elements, w% neutrinos etc. to fit Standard model and star physics we know and a% of possibly dark matter, because it seems to pull planets stronger than all the regular matter constituants that we know of.

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    $\begingroup$ The effects you talk about do not yield the total mass of the Sun in ordinary matter. They principally depend on how much gravitating mass there is (dark or ordinary), but there are second-order effects that tell you how that mass is distributed and which are expected to be different for dark and ordinary matter. In that respect your question, though more specifically about the Sun, is a duplicate of the one you refer to. $\endgroup$
    – ProfRob
    Apr 16, 2018 at 16:57
  • $\begingroup$ I have the same impression as Rob Jeffries. The Sun's mass is unambiguous. My question was whether we know the entire $2x10^{30}$ kg is comprised of ordinary matter, not whether there's an extra dark matter halo around the Sun or anything like that. Therefore this question feels like a duplicate of mine. If you are asking something else, please explain. $\endgroup$
    – Allure
    Apr 16, 2018 at 20:05
  • $\begingroup$ @RobJeffries I state that you can measure Sun's mass in two-way. Gravitational effect and using fundamental physics star dynamics etc. If one can with high certainty attribute 100% of Sun's mass to Standard Model there is no room for any form of Dark Matter. I am asking how high is the upper bound Jimmy is talking about and what is the certainty of it. In this respect I believe it's not a dupe, though some of the edits to your answer in Allure's question might answer mine. I wasn't aware of them. $\endgroup$
    – luk32
    Apr 16, 2018 at 23:54
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    $\begingroup$ There's a useful citation in this answer that I wrote to a much less speculative question, which matches observed solar neutrino emissions with 0% dark matter. $\endgroup$
    – rob
    Apr 17, 2018 at 1:16
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    $\begingroup$ please read this hyperphysics.phy-astr.gsu.edu/hbase/Tables/suncomp.html . It is not simple to assign elements in the sun 's core, (it is nuclear physics, not the standard model)where dark matter would end up by gravitational attraction. Lots of modeling goes in the dynamo model (en.wikipedia.org/wiki/Solar_dynamo) that would be affected by too many neutrals , so the magnetic field and the sun spot numbers and distributions might be affected, but this needs a model. It cannot be answered in a question here. $\endgroup$
    – anna v
    Apr 17, 2018 at 6:32


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