I've read on NASA's page on neutron star that one teaspoonful of that star would weigh over 20 billion tonnes on Earth. If it was somehow possible to bring it to earth would it:

  1. Burn and disappear during Earth atmosphere entry?

  2. Assuming we have 20 billion tonnes of mass occupying the volume of a teaspoon here on Earth, would it fall through the ground under its own weight?


2 Answers 2


If we take neutron star material at say a density of $\sim 10^{17}$ kg/m$^{3}$ the neutrons have an internal kinetic energy density of $3 \times 10^{32}$ J/m$^{3}$. This is calculated by multiplying the number density of the neutrons $n_n$ by, $3p_{f}^2/(10m_n)$, the average KE per fermion in a non-relativistically degenerate gas and where $p_f =(3h^3n_n/8\pi)^{1/3}$ is the Fermi momentum.

So even in a teaspoonful (say 5 ml), there is $1.5\times10^{27}$ J of kinetic energy (more than the Sun emits in a second, or a billion or so atom bombs) and this will be released instantaneously.

The energy is in the form of around $10^{38}$ neutrons travelling at around 0.1-0.2$c$. So roughly speaking it is like half the neutrons (about 250 million tonnes) travelling at 0.1$c$ ploughing into the Earth. If I have done my Maths right, that is roughly equivalent to a 40km radius near-earth asteroid hitting the Earth at 30 km/s.

So, falling through the Earth is not the issue - vapourising a significant chunk of it is.

Note that the beta decay of the free neutrons that dominate the neutron material is also energetic, but a slow process. On these 10 minute timescales, the neutrons could have exploded to a radius of a tenth of an au.

  • 9
    $\begingroup$ Wikipedia tells me that the Chicxulub impactor let slip around $4.2\times 10^{23}{\rm J}$, so this is 3500 times bigger than the dinosaur death dealer. Flabbergasting. $\endgroup$ May 26, 2015 at 12:04
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    $\begingroup$ @WetSavannaAnimalakaRodVance Yes. I am not often surprised by numbers in astrophysics - but was on this occasion. $\endgroup$
    – ProfRob
    May 26, 2015 at 12:25
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    $\begingroup$ This should be the accepted answer. $\endgroup$
    – fig
    Jul 16, 2021 at 16:18
  • $\begingroup$ Can we use the virial theorem to say that 1 kg of neutronium has $3×10^{15}$ J of internal energy so it took double that amount of energy to produce it, with $3×10^{15}$ J lost to radiation? $\endgroup$
    – PM 2Ring
    Dec 21, 2021 at 21:20

The reason that the density is so high is because the pressures are so immense. If we somehow teleported a teaspoonful of neutron star material to earth, it would very rapidly inflate because the pressures aren't high enough to crush it into its dense form. This would effectively be an enormous explosion.

It is difficult to describe what it would inflate out into - the neutron star material can be imagined as an incredibly dense soup of neutrons with some protons and leptons in small numbers. The protons and leptons would make neutron-rich elements like deuterium, but most of the matter would consist of free neutrons. These free neutrons would undergo beta decay to produce neutrinos, protons, and electrons, which would likely recombine to make very large amounts of hydrogen, some helium, and a few heavier atoms. In all of these cases, the atoms would be neutron-rich isotopes, though.

The behavior would look most like a very rapidly expanding gas. It would explode with such force that it wouldn't even need to "fall through the ground" - it would obliterate the floor entirely.

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    $\begingroup$ just want to highlight that we don't know this for sure: its possible that after reaching nuclear densities, matter is essentially stable entirely due to nuclear strong force. It is believed that islands of nuclear stability become more strong as $Z$ and $N$ grow above 150. However finding such super-dense pieces of matter will be almost impossible in the surface of the earth, since this will likely sink inmediately until reaching the center of the earth $\endgroup$
    – lurscher
    May 19, 2011 at 5:50
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    $\begingroup$ @lurscher: If it's all neutrons, it cannot be stable after an appropriate amount of beta decay, which happens enormously quickly as the mass of a bound proton in a proton free environment is much less than a neutron squeezed by other neutrons through the exclusion principle. Once it decays into proton-neutron material, it will fall apart like any charged droplet into nuclei of various sizes. The only way nuclear stuff can be stabilized is if it somehow becomes net-neutral strange matter of some kind, and then it would slowly gobble up normal matter. Such strange matter probably doesn't exist. $\endgroup$
    – Ron Maimon
    Nov 5, 2012 at 17:43
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    $\begingroup$ The timescales for beta processes are slow. Much slower than the expansion timescales of the neutrons which would be moving at a significant fraction of $c$. The ultimate fate of the gas (that which didn't "interact" with the earth) would be an expanding cloud of protons, electrons and (anti) neutrinos. $\endgroup$
    – ProfRob
    Feb 17, 2015 at 12:48
  • $\begingroup$ The neutrons are going to be thrown out at very high speed. I would expect an awful lot of them to smack nuclei and get absorbed by nearby atoms, it would make the dirtiest nukes look clean by comparison. Much nastier than simply staying put and undergoing beta decay. $\endgroup$ Apr 19, 2019 at 20:16
  • $\begingroup$ This can really benefit from the note that by "rapidly expanding gas", it means "nuclear explosion". $\endgroup$
    – ikrase
    Feb 19 at 7:25

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