I've just read this article: http://www.sci-news.com/astronomy/science-kepler-432b-new-super-jupiter-exoplanet-02490.html

And I wondered how this could be possible?

Maybe it's because this gas giant doesn't have the same composition as Jupiter. But then why isn't it just almost entirely composed of hydrogen and helium like most gas giants?

Or maybe it's because the stronger gravity resulting from the additional mass tends to try to shrink the planet. But could this effect really completely counterbalance the fact that more mass should lead to an increase in the size of the planet?

  • $\begingroup$ Note that Saturn has less than a third of the mass of Jupiter, and its radius isn't that different... $\endgroup$
    – Micah
    Commented Feb 16, 2015 at 0:59
  • $\begingroup$ @Micah according to en.wikipedia.org/wiki/List_of_Solar_System_objects_by_size Jupiter has 3.34 times the mass of Saturn, so we we'd expect 1.49 times the radius assuming equal density. It is interesting that Saturn does have about half the density of Jupiter, which in turn has a lower density than the Sun. In g/cm3: Sun 1.4 Jupiter 1.3 Saturn 0.7 Uranus 1.27 Neptune 1.6 with the inner rocky planets all being significantly higher. Both Jupiter and the Sun are denser than water! Also Venus, Earth, Uranus, Neptune & Saturn have about the same surface gravity, with Jupiter about double. $\endgroup$ Commented Feb 16, 2015 at 10:08

2 Answers 2


The reason is electron degeneracy pressure.

The cores of giant planets are dense enough that the electrons in the gas occupy about $h^3$ of phase space each. The Pauli exclusion principle means that they cannot all occupy low energy/momentum states. This means that even at relatively cool temperatures the gas can still exert considerable pressure due to the momenta of the electrons.

A degenerate gas behaves in an anti-intuitive way when it supports a star or planet. A simple argument is the following.

The gravitational potential $\Omega$ and internal pressure $P$ of a planet in equilibrium are related by the virial theorem. $$ \Omega = -3 \int P\ dV,$$ The pressure of a completely degenerate electron gas is proportional to density $\rho$ to the power of 5/3; i.e. $P \propto \rho^{5/3}$ and does not depend on temperature. This is quite a "hard equation of state - the planet becomes difficult to compress.

If we assume the planet has constant density - a terrible approximation, but good enough for a dimensional analysis, then $$ -\frac{3GM^2}{5R} = - 3 \int \frac{P}{\rho}\ dM \propto -3 \rho^{2/3} \int dM,$$ where $M$ is the mass of the star and $\int dm = M$. Substituting $\rho =3M/4\pi R^3$ for the average density, we can easily see that $$ R \propto M^{-1/3}$$ i.e. a more massive star supported by degeneracy pressure is actually smaller, though the dependence on mass is weak.

Now the centres of giant (exo)planets are not completely degenerate, and their outer layers are not really degenerate at all, so this strange behaviour is somewhat moderated. But nevertheless there exists a broad range of planetary masses, from below a Jupiter mass up to tens of Jupiter masses where we expect the radius of the planets to be roughly similar.

The plot below shows some theoretical models compared with some observations from Chabrier et al. (2008). This covers both stars and planets. Notice how the radii of low-mass stars basically decrease (proportional to mass) as the mass decreases and hence $\rho \propto M^{-2}$. But these are supported by perfect gas pressure. As we approach the brown dwarf regime and higher internal densities the electrons become (partially) degenerate and the character of the curves changes and flattens.

Data for transiting exoplanets is also shown. They show a diversity of radii at a given mass that is not completely explained at the current time. Some of it is almost certainly due to irradiation by the parent star (these are almost all "hot Jupiters"). But there may also be composition effects.

Mass-radius from Chabrier et al. (2008)

EDIT: In response to Steve Everill's points

Note that the $R \propto M^{-1/3}$ behaviour applies roughly between about a few Jupiter masses and 70 Jupiter masses. At lower masses there are various interactions with the ions, Thomas-Fermi corrections etc. that change the ideal degenerate gas behaviour and flatten the relation. This means that when we plot density versus mass for exoplanets, we find that density is proportional to mass (i.e. that the radius is roughly constant). See below - data extracted from exoplanets.org. Below a tenth of a Jupiter mass, the equation of state become much more incompressible and the behaviour changes again.

For normal low-mass stars, the central temperature does not vary a great deal. It is set by the ignition of the pp-chain. Thus the central pressure is $\propto \rho$ for a perfect gas. If you insert this into the treatment I gave above for degenerate stars you find that $R \propto M$ and indeed, the average density of low-mass stars higher.

Density vs mass from exoplanets.org

  • $\begingroup$ I can't help but feel that when we get up to Electron degeneracy, we would a) have surpassed fusion requirements b) be in the regime of Chandrasekhar masses... $\endgroup$
    – Aron
    Commented Feb 16, 2015 at 4:04
  • 1
    $\begingroup$ @Aron Fusion requires high temperatures, and the Chandrasekhar mass is 1.4 solar masses. Neither is applicable in the case of anything reasonably called a planet, whereas degeneracy in fact is important starting around a Jupiter mass. $\endgroup$
    – user10851
    Commented Feb 16, 2015 at 6:28
  • $\begingroup$ @Aron the density in the centres of stars increases with decreasing mass. It is the core temperature that determines if and when fusion can take place. It never gets high enough in brown dwarfs and planets. $\endgroup$
    – ProfRob
    Commented Feb 16, 2015 at 8:41
  • $\begingroup$ drawing a line through the lowest and rightmost purple points of your graph, I get a gradient of about 0.25/0.8=1/3.2. That isn't so different from the gradient of 1/3 that would be expected on purely geometrical grounds assuming identical density. $\endgroup$ Commented Feb 16, 2015 at 9:50
  • $\begingroup$ @steveverrill Wishful thinking I'm afraid. Density goes roughly as $M$. Degeneracy is not the only thing going on - the electrons are not non-interacting. $\endgroup$
    – ProfRob
    Commented Feb 16, 2015 at 10:03

It's not really related to your question but I've read that heavier white dwarfs are smaller than lighter white dwarfs and heavier neutron stars are thought to be smaller than lighter ones. When you get that much mass together, gravity tends to win.

Even on the scale of the earth or Mercury, the planet's cores are crushed into greater density. I'm not sure the exact numbers, but the earth's core might be as much as 50% denser than the same materials would be on the surface. The Earth's core has a density of about 13 G/Cm^3 where Iron's density is about 8 G/Cm^3 and the core is about 80% iron. The percentage of heavier elements could throw my estimate off a bit, but that's in the range of 50% more dense than it would be at standard pressure.

  • 1
    $\begingroup$ It is related! Similar physics governs the sizes of White dwarf stars and neutron stars; electron degeneracy pressure and neutron degeneracy pressure respectively. $\endgroup$
    – ProfRob
    Commented Feb 16, 2015 at 8:44

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