Is it possible that all dark matter is made of rogue planets (free-floating planet)? (and other stuff like asteroids or meteoroids)

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    $\begingroup$ it has been envisaged in the 90s that some of the dark matter in halos was made of brown dwarfs; this was later ruled out through lack of indirect detection via light deflection (known as micro lensing). $\endgroup$
    – chris
    Mar 28 '14 at 20:57
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    $\begingroup$ To expand on @chris' comment about dense, cold planet like object in interstellar space of our Galaxy add up to not more than 3% of the excess mass density needed to explain the galactic rotation curves of similar galaxies. These things are out there, but they aren't the answer to the puzzle. $\endgroup$ Mar 31 '14 at 4:08

First of all I'll start with a few ideas:

  1. Baryonic Matter: Baryons are elementary particles made up of 3 quarks. This includes protons and neutrons, and the term baryonic matter refers to matter made of baryons, such as atoms. Examples of non-baryonic matter includes neutrinos, free electrons and other exotic matter.
  2. Things like planets, stars, dust, etc. are all made of atoms, and so are classified as baryonic matter.

Now, how do we know that dark matter is present in the universe?

Astronomers measure the gravitational pull of galaxies and galaxy groups/clusters based on how objects behave when interacting with these objects. Some examples of this include tidal gas/dust stripping, the orbit of stars in a galaxy and gravitational lensing of distant light from a large cluster. Using this they determine the mass of the galaxy (or galaxy group). We can also determine the mass of a galaxy or group by looking at it and adding up the mass of all the objects (like stars, dust, gas, black holes, and other baryonic matter). While these methods both give us approximations, it is clear that the gravitational mass of galaxies and groups exceeds the baryonic mass by a factor of 10-100.

When astrophysicists first found this phenomenon they had to come up with a plausible explanation, so they suggested that there is some new, invisible matter called dark matter. (Aside: some astrophysicists also came up with other explanations like modified gravity, but so far dark matter does the best job at explaining observations).

Okay, so now how do we know dark matter is not any sort of baryonic matter?

There are a few reasons astrophysicists know that it is extremely unlikely that dark matter is baryonic. First of all if all the stars in a galaxy shine on an object it heats up, this heat causes the release of radiation, called thermal radiation, and every (baryonic) object above zero kelvin (or -273.14 deg celcius) emits this radiation. However, dark matter does not emit any radiation at all (hence the name dark!)

If dark matter were baryonic it would also mean that it could become light emitting. If we got a clump of baryonic matter* and put it in space it would gravitationally contract, and would eventually form a star or black hole** - both of which we would be able to see.

So, because of these reasons the dark matter in galaxies and in galaxy groups/clusters cannot be baryonic, and so cannot be planets, dead stars, asteroids, etc. It would definetely not be planets as there is no way 10-100 times the mass of the stars in a galaxy would be planets, as the mechanism for making planets relies on supernovae, and the number of supernovae needed for the that many planets would be far too high to match our observations. I hope that this answered your question!

*provided the clump of baryonic matter was large, and the amount there is in galaxies definitely is!

** we don't observe black holes directly, but can see radiation from their accretion disks.

  • $\begingroup$ Thanks for taking the time to write a clear explanation. I have a question on one bit I am trying to follow. Am I right to understand that in the thought experiment on a "clump of dark matter" that equally if it is not baryonic it should also gravitationally contract too? This follows from its existance from your first paragraph, unless it is so exotic that it can affect the orbits of stars around galaxies but does not interact with its own kind. Does this make sense? $\endgroup$
    – Puffin
    Feb 4 '16 at 21:41
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    $\begingroup$ @Puffin I'm not sure I completely understand what you are asking, but dark matter does interact with other dark matter - however this interaction is purely gravitational. Baryonic and dark matter also only interact through gravity, but baryonic matter interacts with other baryonic matter through gravity, electromagnetism, nuclear forces, etc. Since baryonic matter interacts in this way it can "lose energy" through radiation and other means to contract, but as dark matter has no way to "lose energy" it can't contract as efficiently. Does this answer your question? $\endgroup$
    – Robbie
    Feb 5 '16 at 1:11
  • $\begingroup$ Thank you. Your answer covers it well, I think. My knowledge is rather flaky here and am having to take a large leap. Are you saying that, for example, gravitational waves would provide a means for matter to lose orbital energy and thus between the two types of matter, baryonic and dark, and would thus allow conventional matter to form stars and galaxies whilst dark matter remained more distributed? $\endgroup$
    – Puffin
    Feb 5 '16 at 12:20
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    $\begingroup$ Yes, gravitational waves are one way for matter to lose energy (although it is very minor). As a gas cloud contracts it heats up, and this heat radiates away, cooling the gas and allowing it to contract more and more. This is why matter can form stars and planets and cool stuff like that but dark matter can't. $\endgroup$
    – Robbie
    Feb 6 '16 at 10:57
  • $\begingroup$ OK, thank you, its much clearer with the heat example. $\endgroup$
    – Puffin
    Feb 6 '16 at 11:01

The fact that there is not enough luminous matter to explain the gravitational properties of galaxies and clusters of galaxies does not inexorably lead to the conclusion of weird dark matter that does not emit or absorb light.

For a long time in the 70s, 80s and 90s, there was very much a search for small, dark objects that could make up this missing mass. This included things like cold white dwarfs, low mass brown dwarfs and "rogue planets". None of these searches - and specifically the microlensing surveys, which continue today - turned up anything like the numbers of objects required to solve the "dark matter" problem.

A more fundamental problem for the "rogue planets" idea is that we are now quite sure that whatever dark matter is, it is not in the form of everyday "baryonic matter" that emits and absorbs electromagnetic radiation (a k.a. light).

Measurements of the primordial abundances of helium and deuterium, created in the big bang, combined with measurements of the universal expansion rate, give a direct estimate of the amount of baryonic matter in the universe. Whilst the inferred amount is greater than can be counted up in the luminous matter of the universe, it is insufficient by a factor of 6 compared with what is required to explain the dynamics of galaxies, clusters of galaxies and the universe as a whole.

Further indirect evidence of the non-baryonic nature of this matter comes from simulations of structure formation in the universe. The "dissipationless", slow-moving nature of non-baryonic matter is what is required to explain (i) the relatively tiny fluctuations in the cosmic microwave background, emitted when the universe was 380,000 years old; and (ii) how this then develops into the rich set of galaxies and galaxy clusters that we seen in the universe today.

In other words, most of the "dark matter" must be non-baryonic and so cannot be made up of rogue planets.


Yes, it's definitely possible. For our solar system, you'd need about 18,000 Saturns of "rogue planets". That seems pretty doable for that amount of matter to be between us and the closest star, and for us to not observe it.

Just to put it in perspective, if earth was the size of a dime, a Saturn rogue planet would be a bit smaller than a volleyball. With that same scale, the nearest planet would be over 50,000 km away; that would make a 2D area (3D makes it even bigger) about the size of 60 Pacific Oceans. Seems fairly easy to have 18,000 volleyballs floating unobserved in 60 Pacific Oceans.

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    $\begingroup$ The question is asking about dark matter, not just "things we can't see". While it's certainly possible (likely, in fact) that there are many rogue planets we can't see, they probably don't make up the dark matter portion of the Universe, because planets are made of baryonic matter, and dark matter almost certainly isn't. $\endgroup$
    – Jim421616
    Nov 17 '20 at 18:55
  • $\begingroup$ +1 I like your comparison. But like @Jim421616 said, dark matter is something that we can't see. $\endgroup$ Nov 17 '20 at 22:13

… I am confused. “Dark Matter”, by definition, is “matter we cannot see”. You are assuming that we would be able to “see” a planet-sized chunk of baryonic matter at stellar distances. I am not referring to nebulas and such — those are relatively low mass, highly diffuse, clouds of light elements with very free electrons with which to produce high amounts of photons/em radiation. Not everything behaves as such. Photons are photons — ever take a picture of a face with the light behind the subject? The light behind will cause any foreground light to be washed out; it is also why we can’t see into our own galactic center all that well.

That “volleyballs in an ocean” comment above really puts that into context. Thank you, kind individual. That would be, what, one volleyball per over 200,000 square miles? I have lost soccer balls in public parks before, let alone losing one within a city, but those are state sizes/scales.

Come on astronomy community, you are not that arrogant, are you?

  • $\begingroup$ Please add further details to expand on your answer, such as working code or documentation citations. $\endgroup$
    – Community
    Aug 30 at 16:45
  • $\begingroup$ It is incorrect to say that the reason that "rogue planets" are ruled out as a source of dark matter is simply because they cannot be seen. $\endgroup$
    – ProfRob
    Sep 1 at 7:04

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