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Gas giants can generate heat via the Kelvin-Helmholtz mechanism. It's oft-repeated that Jupiter actually generates more heat internally via this method than it receives from the Sun. Scale this mechanism up enough and you get into brown dwarf territory, with something that is effectively a star to a nearby observer (whether it is a star or not seems to be contentious).

A giant in "star mode" obviously emits vast amounts of heat and is agreed to have a very narrow, but technically possible, habitable zone in its own right, enough to be independent of any larger star it may be orbiting. How does this zone change in relevance as the giant decreases and it moves towards "planet mode"? Is it possible for an object to have a gentle enough "heat gradient" (?) that despite being cool enough that it doesn't visibly glow, it still noticeably warms its moons?

Example scenario: a superjovian/sub-brown-dwarf at, or just beyond, the outer edge of a parent star's main habitable zone. The giant has its own planetary moon system; on these worlds, the brightest object and dominant feature in the sky is the giant, which is cool enough to have visible stripes, no internal "glow" (the light would mainly be reflected sunlight), and to not vaporize any equipment sent into its upper atmosphere. I guess 300-500K.

How would the heat given off by such an object affect its satellites? Would it be able to raise their temperature by any significant amount - does a low central temperature on a massive object still cause a reasonably large zone of warming? How would/could this mini-zone interact with the zone of the primary star? (i.e. the sliding scale of "how far out can I move this before I have to make the giant brighter?" vs. "how far in can I move this before the giant's heat is completely overpowered?")

Allowing an extremely generous definition of "habitable" (e.g. a world that would have had a surface temperature of 150K is now merely Antarctic at the equator thanks to the giant).

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  • $\begingroup$ Please note white dwarfs typically radiate more heat. $\endgroup$ – user2284570 Sep 16 '16 at 20:47
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It's possible, but heat generated by the Kelvin-Hemlholz mechanism may be too variable to complex life to develop solely as a result of this source of heat. This paper suggests that the temperature of Jupiter, when it first finished an initial phase of contraction, was quite high, at around 25000K. At this temperature, it would have a small habitable zone around it, but then began to cool as it radiated this initial heat off into space. It continues to contract, but at a slower rate, leading to less heat production.

However, it's possible that you could define a habitable zone based on the level of tidal heating the moons experience, rather than based on heat radiated by the primary planet. The heat given off can be calculated by the equation $q = 36\rho n^5r^4e^2/38\mu Q$, where $r$ is the orbital radius, $e$ is eccentricity, $n$ is mean orbital motion, $\rho$ is density, $\mu$ is shear modulus, and $Q$ is a dimensionless constant.

This won't give you a habitable zone based purely on distance from the primary planet, but one which is strongly influenced by distance. Too close to the primary and the planets will end up like Io, too far and they will freeze.

Note that, with moons heated internally instead of externally, you will end up with some environments that are very different from earth, but still habitable. Europa, for example, may be habitable, but life will exist in lightless oceans under miles of ice.

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    $\begingroup$ To go with that I think I read that Jupiter sized planets to Brown dwarfs tend to be similar in size, but can very a lot in Mass, I think it said about 20 Jupiter masses is where the Brown dwarfs start in classification. $\endgroup$ – bowlturner Dec 23 '14 at 21:28
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    $\begingroup$ 36/38 = 18/19 . These 2's are in short supply: let's not waste them! Also, so rare to get a nice 19 in physics. $\endgroup$ – Dan Sheppard Dec 21 '15 at 11:56
  • $\begingroup$ @ckersch, "...Jupiter, when it first finished an initial phase of contraction, was quite high, at around 25000K. At this temperature, it would have a small habitable zone around it,..." Was that period of time long enough for the OP's story to take place then, in their gas giant? $\endgroup$ – Len Jan 30 '18 at 20:30
  • $\begingroup$ @Len Depends on the story. It's a very long period of time compared to the time spans of civilizations, but probably not a long enough time for complex, multicellular life to develop on its own. $\endgroup$ – ckersch Jan 30 '18 at 21:32
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Really awesome answer by ckersch. I want to add in some math to get an idea of how large this kind of habitable zone would be. Formulas are from here and here, if you want to investigate them further, though I'll try to explain them here.

We may assume that the source of energy is gravitational potential energy, defined as $$U_g = -G \frac{M_p m_s}{r}$$ where $U_g$ is potential energy, $G$ is the gravitational constant, $m_1$ is the mass of object one, $m_2$ is the mass of object two at a given distance, and $r$ is that distance. We'll treat these masses as shells going outward from the center.

The body has a total radius of $R$, and a density $\rho$. The mass contained within an arbitrary radius $r_{\text{arbitrary}}$ is a function of this radius, denoted as $m(r_{\text{arbitrary}})$. Each shell has a surface area of $4 \pi r^2$, which is simply the formula for the surface area of a sphere - those these shells are essentially hollow spheres. To find the total gravitational potential energy, we have to integeate over the entire radius of the body: $$\Sigma U_g=-G\int_0^R \frac{m(r_{\text{arbitrary}})4 \pi r^2 \rho}{r}dr$$ The mass contained within the radius $r_{\text{arbitrary}}$ can be reduced to the product of density and volume ($m=v \cdot \rho$). Volume, however, can fortunately be expressed in terms of the radius as $V=\frac{4}{3} \pi r^3$, so we have $$m=\frac{4}{3} \pi r^3 \rho$$ and then $$\Sigma U_g=-G\int_0^R \frac{\frac{4}{3} \pi r^3 \rho 4 \pi r^2 \rho}{r}dr$$ This becomes $$\Sigma U_g=-G \left(\frac{16}{3} \pi ^2 \rho ^2 \right)\int_0^R r^4 dr$$ Integrating, we get $$\Sigma U_g= \left(-G\frac{16}{3} \pi ^2 \rho ^2 \right) \left[\frac{r^5}{5} \right]_0^R$$ $$\Sigma U_g= \left(-G\frac{16}{3} \pi ^2 \rho ^2 \right) \left[\frac{R^5}{5} - \frac{0^5}{5} \right]$$ and finally $$\Sigma U_g=-\frac{16}{15}G \pi ^2 \rho ^2 R^5$$ We go back to our definition of mass as a function of volume and density, and find $$\Sigma U_g=-\frac{3M^2G}{5R}$$ However, half of the available energy is turned into kinetic energy, so we divide that by two to find that $$\Sigma U_g=-\frac{3M^2G}{10R}$$ If we can find the time $t$ over which the energy is radiated, we have our power $P$. The intensity over a given surface area is $$I=\frac{P}{4 \pi r^2}$$ so we have $$I=\frac{3M^2G}{4 \pi r^2 10R t}$$ Let's say you have a time $t$ for how long you want the body to emit the energy. Choose the mass $M$ and radius $R$ of the body, and take the intensity $I$ from an orbit of the planet around a star in the star's habitable zone - in other words, take the solar intensity from a spot in Earth's orbit. You can solve for $r$ to figure out what that distance would be: $$r=\sqrt{\frac{3M^2G}{4 \pi I 10R t}}$$ You can then do some guess-and-check to figure out the inner and outer boundaries of the habitable zone. I don't have the time to do this at the moment, but I may be able to later. For now, I encourage you to play around with the equations a bit and see what kind of setup you can come up with.

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Tidal effects between a gas giant and its moons can heat up the moons to various degrees. This in turn causes volcanoes on the moons if strong enough. Io, a moon of Jupiter, is so overheated and has so many volcanoes that it would be uninhabitable even if it had an oxygen nitrogen atmosphere.

So there is a minimum distance that a moon has to be from its gas giant in order to avoid too much vulcanism. Lesser amounts of Vulcanism could contribute enough heat to compensate for insufficient solar radiation on an Earth sized moon. And if a moon orbits too far from its gas giant planet it will have an unstable orbit.

An Earth like moon of a gas giant planet is likely to be tidally locked to the planet so that it will rotate once every orbit of the planet, keeping one side always facing the planet. Thus the moon's day will equal its monthly orbit of the sun. To be habitable the moon should have a relatively short day length to avoid large temperature differences between night and day. The faster the moon rotates, the more likely it will be to have a strong magnetic field to deflect charged particle radiation.

How fast a tidally locked moon rotates depends on how long it takes to orbit its planet, which depends on the planet's mass and the distance of its orbit.

The habitability of an Earth sized moon of a gas giant planet would depend primarily on how far the planet and the moon orbited from their star and how much the star heated the moon. But it would also depend on how close the moon orbited the gas giant planet. Orbiting too close to the planet or too far from it would make the moon uninhabitable. So gas giant planets do have habitable zones for their moons that depend on various factors including but not limited to how much heat they radiate.

You might find articles I mention in my answers about habitable moons interesting.

Making a slow orbit around a large gas giant1

Captured Earth-Like Moons around Gas Giants2

Here is a link to an article about the habitablity of exomoons:

https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3549631/3

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