The previous generation of stars famously are the origin of all the heavier elements (up until iron?) in the solar system. So a big portion of the solar system mass actually is made up of Carbon, Silicon, Iron and the like because of that. But in the center, and only in the center, there is a star with presumably almost no heavy elements inside. How can that be? Am I wrong about the actual mass concentrations or is there really an imbalance, i.e. is the element distribution really lighter towards the center of the solar system? I would presume that the previous generation of stars just ended in a more or less uniform cloud of debris, from which the solar system formed. But if so, why aren’t there star systems where the star has a very different composition and is kind of a spluttering, dirty fusion machine (metaphorically, I mean)?


2 Answers 2


The solar system contains very little of elements heavier than Helium - less than 2% by mass.

This is reflected in the chemical abundances measured in the photosphere of the Sun. i.e. The Sun does contain heavier elements.

Your question is the wrong way around; it is not that the heavier elements have not sunk into the middle, it is that the vast majority of hydrogen and helium that was in the same place as the planets when they formed, did not end up as part of the planets. In fact, even this is only partially true. The mass of planetary material in the solar system is also dominated by the hydrogen and helium in the gas giants.

So the conundrum is only why the smaller planets don't have a similar composition to the Sun. The answer to that is temperature and gravity. A small, hot planet just doesn't have the gravity to retain fast moving hydrogen and helium atoms, unless they are trapped in some compound (like water!).

Thus, the small planets close to the Sun are depleted of light elements.

  • 2
    $\begingroup$ Oh I see, thanks! So does that effectively mean that really only a slight portion of matter was fused in the early stars before they became unstable? $\endgroup$
    – lthz
    Commented Oct 7, 2017 at 13:10
  • 6
    $\begingroup$ @lthz Yes, that is certainly true, but more importantly, the debris from those stars is mixed with a much greater quantity of material that has never been part of a star. $\endgroup$
    – ProfRob
    Commented Oct 7, 2017 at 14:09
  • 7
    $\begingroup$ @lthz - Another way to look at it: The vast majority of the hydrogen in the universe has never been in a star. See What percentage of the hydrogen today has never been in a star. $\endgroup$ Commented Oct 7, 2017 at 14:11

Toward the end of a stars lifecycle the star looses the ability to continue using hydrogen ect). To power itself, it begins to, (in a desperate attempt to live.) Create heavier elements such as iron. Now iron can't certainly not sustain a star.

As such the iron destroys the star, thus marking the end of its life cycle.

Small amounts of iron probably exist in stars, (from the universes point of view.) But it's not enough to actually effect the star it's in.

Iron is of course not the only things starsform that ultimately kill the star, I'm just using it as an example here.

None the less stars are quite interesting, just as anything else in space.

P.S. This is just my vary basic understanding of space, and I'm learning a lot on this website.

Have a nice day/night.

  • $\begingroup$ This doesn't really address the OP's question, which is about the spatial distribution of heavy elements. $\endgroup$
    – HDE 226868
    Commented Oct 9, 2017 at 18:02
  • 1
    $\begingroup$ I understand your enthusiasm but it really doesn’t add anything relevant. $\endgroup$
    – lthz
    Commented Oct 9, 2017 at 19:04
  • $\begingroup$ Welcome to StackExchange. In answers, we're generally looking for more complete answers informed by a deeper understanding of the subject matter. If you're not familiar with this field but you still want to contribute, you might have better luck asking questions about things you find hard to grasp as you learn. You can also contribute to other Stack Exchange sites when you are familiar enough with the subject matter to provide an in depth answer. Good luck. $\endgroup$
    – jpmc26
    Commented Oct 9, 2017 at 22:45

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .