Habitable moon of a gas giant: working out the sizes and distances

Let's work out some factors.

• Luminosity

  You gave the radius of the inner edge of the habitable zone as 1.976 AU and the outer edge as 2.808 AU. From this, we can calculate the luminosity of the star. There's an explanation of how to do this on Planetary Biology. The formulae are $$r_i=\sqrt{\frac{L_{\text{star}}}{1.1}}$$ $$r_o=\sqrt{\frac{L_{\text{star}}}{0.53}}$$ Plugging in your numbers, I get a luminosity of $$4.295 L_{\odot}\text{ (inner radius)}$$ $$4.179 L_{\odot}\text{ (outer radius)}$$ I'll average those, giving us a luminosity of $4.237$ times the luminosity of the Sun. But a P-type orbit is around two stars, as you said, so we divide by two to get an average luminosity of $2.112$ solar luminosities. We can assume that the two stars are similar because they most likely formed together, and have similar properties.

• Mass

  The mass-luminosity relation can tell us the masses of the stars. It is $$\left(\frac{L}{L_{\odot}} \right)=\left(\frac{M}{M_{\odot}} \right)^a$$ The stars likely have masses similar to the Sun, so we can assume $a \approx 4$. The left side is $4.179$. We write $$4.179^{\frac{1}{4}} \times M_{\odot}=M\approx 1.430M_{\odot}$$ So each star is about $1.430$ solar masses, leaving a combined mass of $2.860$ solar masses.

• Orbital period of the gas giant

  Kepler's Third Law tells us that $$T=\sqrt{\frac{4 \pi ^2}{GM_{\text{star}}}r^3}$$ Here, $M_{\text{Star}}$ is actually the mass of both the stars. If the radius is in the middle of the zone (at about $r=2.392$ AU) $$T=\sqrt{\frac{4 \pi}{6.673 \times 10^{-11} \times 5.689 \times 10^{30}}(3.578 \times 10^{11})^3}=6.902 \times 10^7 \text{ seconds}= 800 \text{ days}$$

• Orbital radius of the moon

  Here we just go in reverse. We do need the mass of the gas giant, though - so going from the graph on TimB's answer here, I'll pick about 5 Jupiter masses, or $9.49 \times 10^{27}$ kilograms. The period will be in between the values you said, so about 52.5 days, which is $4.536 \times 10^6$ seconds. We put this all in and get $$r=\left( \frac{6.673 \times 10^{-11} \times 9.49 \times 10^{27}}{4 \pi ^2}(4.536 \times 10^6)^2 \right)^{\frac{1}{3}}=6.911 \times 10^{6} \text{ kilometers}$$ Obviously, it's still in the habitable zone. But it's far away - although that's because the gas giant is so massive. You may want to opt for a shorter period.

  This system setup appears to be viable if you move the moon closer to the gas giant, giving it a smaller orbital period.

• Tidal Locking

  The formula for the time it takes for a satellite to be tidally locked is $$t \approx \frac{wa^5IQ}{3Gm_{\text{planet}}^2k_2R^5}$$ The factors are described on the Wikipedia page. Here, we can say that $I \approx 0.4m_sR^2$, so $$t \approx \frac{0.4 wQR^2a^6}{3Gm_{\text{planet}}^2k_2r^5}$$ Since $$k_2 \approx \frac{1.5}{1+\frac{19 \mu}{2 \rho gR}}$$ and $g \approx \frac{Gm_s}{R^2}$, $$k_2 \approx \frac{1.5}{1+\frac{19 \mu R}{2 \rho Gm_s}}$$ $$k_2 \approx \frac{3 \rho Gm_s}{2 \rho GM_s+19 \mu R}$$ $$t \approx \frac{0.4 wQR^2a^6(2 \rho GM_s+19 \mu R)}{9G^2m_{\text{planet}}^2 \rho R^5}$$ With $Q \approx 100$, $\mu = 3 \times 10^{10}$, $R \approx R_{\text{Earth}}$ and $\rho = \rho_{\text{Mars}}$, you can figure out the tidal locking time. I'm in a rush, so I don't have time to do the calculation, but I may include it later.

How can I determine how far the moon would need to orbit the gas giant, to not be tidally locked?

Tidal locking will occur at some point in time. You can't get around it.

Tidal forces will also be problematic because moons orbiting gas giants will likely experience tidal forces so strong that tidal heating can render the moon uninhabitable (see Heller & Barnes (2013)).

Capture - Corrections

In my original post, I naively said that there are a bunch of scenarios where capture would be possible. This, as HopDavid pointed out, is blatantly false, because the planet would be traveling in a hyperbolic orbit relative to the gas giant, and so would escape its pull rather easily.

So it has to have its orbit modified somehow.

My suggestion would be an interaction with another body, preferably another gas giant. This could change its orbit such that gravitational capture by the original gas giant is possible. Without this sort of interaction, the planet will just scoot away.

Section on the moon's properties

This may be list-like, but it's the best I can do.

• Mass: You required an atmosphere and a magnetosphere. Both of those require a planet with the right mass and size, as well as composition (which I'll get to). Not many moons have atmospheres complex enough and dense enough to support life. In fact, Mercury can't support an atmosphere. But mass isn't the only thing that plays into this. Titan, one of Saturn's moons, has a mass less than twice that of Mercury, yet it has a rich atmosphere. As Jim2B pointed out, though, such a planet wouldn't be able to hold onto water vapor, as this chart shows, because its escape velocity would be too low:

^{Image courtesy of Wikipedia user Cmglee under the Creative Commons Attribution-Share Alike 3.0 Unported license.}

Also, the maximum mass of the moon is related to the mass of the parent planet, meaning that for a more massive moon, you'll need a much more massive gas giant for it to orbit.

You can attribute this to a few factors:

• The presence of Saturn's magnetosphere
• Low temperatures

• A weak solar wind at that distance from the Sun.

  You've got the distance, the weak stellar wind, and the presence of a gas giant and its magnetosphere. So you really want to aim for a mass similar to that of Titan, at $1.3452 \times 10^{23}$ kilograms.

  • Size: You don't want anything too tiny, because the density of such a body is far greater than expected. Conversely, you don't want anything too big, because the surface gravity would be weaker than you'd like. So go for a surface acceleration of perhaps $0.5g$ - half that of Earth. You can figure out your average radius using $$g=\frac{MG}{r}$$ So you can see why we needed the mass.

  • Composition: You don't want an environment that's hostile to life, so perhaps it would be best to mimic Earth as much as possible. Choose silicate materials for the outer layers, but remember to have nickel and iron for the core. These can help produce that moon's magnetosphere - a crucial component to retaining an atmosphere. Mars' lack of a magnetosphere has contributed to it slowly losing its atmosphere.

posted over 9 years ago

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