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If I'm not mistaken, it is believed that the reason for such turbulent weather on the 4 outer gas giant planets is that the internal pressure is so great that it is generating heat, which is causing convection, which causes extreme weather.

Will these planets forever generate heat, or at some point will they freeze up?

What will Jupiter look like a trillion years from now?

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  • $\begingroup$ Please do not use words like "trillion" since they are confusing. It does not convey the same sense in the en.wikipedia.org/wiki/Long_and_short_scales Please use clear scientific words like Gigayears. $\endgroup$ – Envite Dec 4 '14 at 12:59
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    $\begingroup$ 1 trillion years = 1000 Gigayears. $\endgroup$ – Rob Jeffries Aug 10 '16 at 14:12
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The timescale on which Jupiter cools is reasonably well understood and predicted by the current generation of evolutionary models.

Jupiter's luminosity is provided mostly by gravitational contraction. For a planet that only contains gas governed by the perfect gas law, the appropriate timescale for this contraction (or indeed for the luminosity to fall significantly) is given by the Kelvin Helmholtz timescale. $$ \tau = \eta \frac{GM^2}{RL},$$ where $M$ and $R$ are Jupiter's mass and radius and $L$ is its current power output (or luminosity), and the parameter $\eta \sim 1$. This timescale is a few $10^{11}$ years.

However, giant planets like Jupiter are not governed by perfect gas laws. The gas in the centre of Jupiter is dense enough that electrons become degenerate. Degenerate electrons fill the available energy levels up to the Fermi energy. Their consequent non-zero momenta of the electrons exerts a degeneracy pressure that is independent of temperature. As a result, the rate of contraction slows and the release of gravitational potential energy slows; the planet is able to cool and remain in hydrostatic equilibrium without the same degree of contraction.

One can express this change using the $\eta$ parameter. For Jupiter $\eta \simeq 0.03$ (Guillot & Gautier 2014) - i.e. the timescale for the luminosity to fade is 30 times quicker than the naive Kelvin-Helmholtz time and Jupiter's luminosity will scale as the reciprocal of its age and will fall by a factor of a few in $10^{10}$ years. In a trillion years, the luminosity of Jupiter will be lower than it is now by roughly a factor of 250.

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From what I know, the heat was mostly generated when the gas giants are created. Some of this was from the friction caused by internal pressure. However, this heat is not being generated anymore, as it was only generated as matter fell into the planet.

They presumably do generate heat from radioactive elements in the core (though nobody has ever been down and checked if there are any :P), and will also receive a 'boost' from solar heating.

Over time, however, each heat source will diminish. The latent heat from birth will be dissipated into space in the form of radiation, the radioactive elements will decay, and the star it's orbiting will die off.

So I guess that Jupiter will no longer have its dramatic weather in a trillion years.

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  • $\begingroup$ I had always learned that the heat from the Sun was in no way powerful enough to drive the extreme weather on these gas giants, and that there must be some other source of heat. The leading theory was that the extreme pressure at the core was generating enough heat to cause this. But if I understand what you are saying, it's no longer generating heat, but rather just leftover heat from the genesis of the planet? $\endgroup$ – Scottie Nov 14 '14 at 17:43
  • $\begingroup$ I don't think so, as otherwise it would be able to generate heat forever, which would make it an infinite source of energy. However, the residual energy in a gas giant is pretty extreme, and should last for a very long time. It's a similar situation to that of a white dwarf star, which will eventually cool into a dead black dwarf. $\endgroup$ – Superdavo Nov 14 '14 at 18:20
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    $\begingroup$ Jupiter can generate its own heat through the Kelvin–Helmholtz process: as it radiates heat, it contracts, thus converting gravitational potential to thermal. Negative heat capacity is characteristic of gravitationally bound systems. The present Jovian intrinsic luminosity is $L_\text{J} = 8.7\times 10^{-10}$, so the thermal timescale is $$\tau_\text{KH}\sim\frac{GM^2}{RL_\text{J}}\sim 10^{11}\,\mathrm{yr}\text{.}$$ We used the present value, so indications of the long-term future are uncertain. But the point is that it's not obvious that one can dismiss internal heat generation so quickly! $\endgroup$ – Stan Liou Nov 14 '14 at 20:07
  • $\begingroup$ @StanLiou The Kelvin Helmholtz timescale cannot be applied to an object that is supported by (partial) degeneracy pressure. Jupiter's rate of contraction is slowing and it will cease to generate energy via contraction and cool at nearly constant radius, supported by degeneracy pressure. $\endgroup$ – Rob Jeffries Dec 3 '14 at 21:45
  • $\begingroup$ @RobJeffries Yes, that's why I said that it's uncertain to what extent the present conditions are indicative of Jupiter's actual long-term future. However, the point is that dismissing the internal heat generation without deeper analysis as if it were simply obvious is inappropriate. $\endgroup$ – Stan Liou Dec 3 '14 at 23:20
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In "one trillion" years Jupiter's fate will be affected by how violent our Sun becomes when it transforms in Red Giant, 5000 million years (5Gy) from now.

With our Sun being so big and brilliant, it will heat Jupiter quite a lot more than now. But also the mass lose will make Jupiter spiral out towards a bigger orbit, while capturing some extra mass.

So in "one trillion" years Jupiter will be a bigger, colder and denser (as well pointed by Rob Jeffries in his answer), more external planet around a white dwarf.

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  • $\begingroup$ This is an interesting point. But, I think after a trillion (1E12) years, what happened in the first 10 billion will be largely forgotten. ie subsequent evolution depends very weakly on the initial conditions. At most it will "reset the clock to zero" when the Sun becomes a giant. $\endgroup$ – Rob Jeffries Dec 4 '14 at 17:25

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