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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 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.

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 of a mass similar to that of Titan, at $1.3452 \times 10^{23}$ kilograms.

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.

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.

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.

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.

• 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.

• 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.