What would be the diameter of a sphere of iron with the mass of our sun? Use standard density of metalic iron. I realize it would then collapse into a star in its own right and become much smaller as it’s density increases. How does this relate to the large diameter that our sun will have when at its red dwarf stage? Thank you.

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    $\begingroup$ The last part of your question is unclear. The Sun is in its "red dwarf" stage now. Do you mean its final white dwarf phase? $\endgroup$ – Rob Jeffries Mar 20 '19 at 19:50
  • $\begingroup$ The first part of your question is simple arithmetic, not physics. What's the relevance of iron? FWIW, it'd take roughly 1000 stellar systems like ours to gather that much iron. I guess you could cheat & grab it from a large star in the silicon burning phase, but you have to be quick. That phase only lasts a few days, and generally ends in a type II supernova explosion. $\endgroup$ – PM 2Ring Mar 20 '19 at 21:31
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    $\begingroup$ @PM2Ring The centre of a massive star is roughly a 1 solar mass ball of iron, shortly before core collapse... $\endgroup$ – Rob Jeffries Mar 20 '19 at 22:36

The object would shrink to become an "iron white dwarf", supported by electron degeneracy pressure.

If you want to assume that you have a cold sphere of iron supported by electron degeneracy pressure, then you can use the Nauenberg (1972) approximation for the radius of a cold$^*$ white dwarf with $\mu_e$ mass units per electron and a Chandrasekhar mass $M_C = 5.816 M_{\odot}/\mu_e^2$. $$ R = \frac{2.45354}{\mu_e} R_{\rm Earth} \left(\frac{M}{M_{C}}\right)^{-1/3} \left[ 1 - \left(\frac{M}{M_c}\right)^{4/3}\right]^{1/2}$$

Thus for $^{56}$Fe, $\mu_e = 2.15$, $M_c = 1.26M_{\odot}$, and if the mass of your star is $M=1M_{\odot}$, we have $R= 0.635R_{\rm Earth} \simeq 4045$ km.

The Sun will end its life as a white dwarf with a mass of about $0.5M_{\odot}$ and will be made of a mixture of carbon and oxygen with $\mu_e =2$ and $M_c = 1.45M_{\odot}$. Putting this into the Nauenberg approximation we have an estimated radius for the end-of-life Sun as a cold white dwarf of $R = 9700$ km.

$^* $ "Cold" in this context means that the Fermi kinetic energy is much greater than $k_B T$. In practice this means $T<10^9$ K. A contracting iron ball roughly in hydrostatic equilibrium will never reach such temperatures because of highly efficient neutrino emission above $10^8$ K.

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  • $\begingroup$ FWIW, the radius for room temperature iron under no compression ($\rho=7.874 g/cm^3$) is just over 392,150 km. If you assembled this sphere from cold iron it would get very hot as it compresses into degenerate matter. (And even if there were no compression it could get hot from the heat of formation, unless you built it very carefully). But how hot could it get? I guess the total thermal energy can be calculated from the difference in gravitational potential of the initial & final spheres, but how long would it take to compress itself, and how much heat would be radiated away while it shrank? $\endgroup$ – PM 2Ring Mar 20 '19 at 21:19
  • $\begingroup$ @PM2Ring a 1 solar mass ball of iron would shrink and heat up, but of course it would not become hot enough to avoid degeneracy - it would start emitting neutrinos long before that. It is no different at all in that respect from a 1 solar mass ball of carbon. $\endgroup$ – Rob Jeffries Mar 20 '19 at 22:25
  • $\begingroup$ Ok, and from your edit I see the temperature is under a billion K. (BTW, I wasn't suggesting the heat would be sufficient to prevent degeneracy). $\endgroup$ – PM 2Ring Mar 20 '19 at 22:39
  • $\begingroup$ @PM2Ring Cold and degenerate are synonyms in the study of compact objects. $\endgroup$ – Rob Jeffries Mar 20 '19 at 22:58

Hint: I'm assuming you know (or can find) the mass of the Sun $m_\odot$ and the density of solid iron on Earth $\rho_{Fe}$. Now, you just need to know the relation between density, mass and volume.

Regarding what happens after someone created such a sphere, I don't know for sure. The mass is below the Chandrasekhar limit of $1.4 M_\odot$, so it will not become a neutron star. (If it was the case, the diameter would be significantly reduced because its density would become around $10^{14}$ larger than the density of the Sun.) Nevertheless, the extra pressure due to the mass would decrease the diameter of the iron sphere since it's density would rougly be $10^6$ the density of the Sun.

If it had enough energy (e.g, from approaching the iron to create a sphere), we may get a white dwarf. In time it would cool down and it's colour would go to yellow, red, and brown, since the colour depends on the surface temperature (hotter means whiter and brighter). However, red dwarfs and brown dwarfs are different types of object.

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  • $\begingroup$ The answer has nothing to do with the density of solid iron on Earth. The object would not become a neutron star because it is below the Chandrasekhar mass for an iron white dwarf. $\endgroup$ – Rob Jeffries Mar 20 '19 at 19:47
  • $\begingroup$ @RobJeffries: you're right, I was thinking about neutron degeneracy pressure. I'll edit my answer. $\endgroup$ – Ertxiem - reinstate Monica Mar 20 '19 at 19:51

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