Can a gas giant have its own habitable zone?
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).
This post was sourced from https://worldbuilding.stackexchange.com/q/6909. It is licensed under CC BY-SA 3.0.
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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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