I've been making a solar system for my story based off the Keplar-47 solar system as my story has 2 suns like this solar system and since its easier to write a system based off a real one then to completely make one up from scratch. Here's a very artistic interpretation (the blue dot) of where I'm thinking the main planet, Portus, should be placed in this solar system (replacing Kepler-47c).

[![enter image description here][1]][1]
[1]: https://i.stack.imgur.com/Po5cL.jpg

Would it be possible for me to base a solar system off this system?


Would it be possible for my planet to be placed here and still be a habitable earth-like planet?

  • $\begingroup$ Is the question, essentially, "Can Kepler-47 system support life as we know it"? $\endgroup$
    – Alexander
    Nov 9, 2020 at 23:55
  • 3
    $\begingroup$ I'm sure God isn't going to sue you for copyright infringement if you use Keplar-47. $\endgroup$
    – Ángel
    Nov 10, 2020 at 0:00
  • $\begingroup$ @Ángel Yeah that was pretty stupid for me to worry about that $\endgroup$
    – WebbS_616
    Nov 10, 2020 at 23:49

1 Answer 1



In fact, using reality as much as you can to create the "infrastructure" of your world is a great way to get a realistic feeling to a fictional world. Better still, NASA happens to be a bit ahead of you.

The outer planet, Kepler-47c, orbits its host pair every 303 days, placing it in the so-called "habitable zone," the region in a planetary system where liquid water might exist on the surface of a planet. While not a world hospitable for life, Kepler-47c is thought to be a gaseous giant slightly larger than Neptune, where an atmosphere of thick bright water-vapor clouds might exist. (Source)

While the exoplanet Kepler-47c is not, itself, habitable (we think...), the statement points out that a planet already exists in kepler-47c's habitable or "goldilocks" zone (you know... the distance is "just right!").

The habitable zone (see chart below, click to enlarge, courtesy wikipedia) is a region of space based on the output of the star where liquid water can form. As you can see from the chart, Earth is smack dab in the middle of the zone for a star like our Sun, while Mars is at the outer edge of the zone.

Binary stars are a bit more complex because you need to determine the aggregate energy output of the star and need to worry about just how distant the two stars are from one another. In this case, thanks to Kepler-47c, it's easy. Since it's located at 0.989 AU compared to Earth's 1.0 AU, we can jump to the reasonable (if imprecise) conclusion that the aggregate energy output of Kepler-47 is equivalent to our own star and the two stars are, we believe, close enough together to hold the possibility of life.

enter image description here


In my opinion, your proposed solution exceeds the requirements of suspension-of-disbelief and is, in fact, plausible.

I am jumping to the conclusion that the location of your planet, as shown, is approximately 1AU from the binary stars. You don't have a scale on your picture.

  • $\begingroup$ Thanks so much for this, but yeah my planet would be approx 1AU from the stars. I sort of have another question though, would anything change if the stars were a brown dwarf doing a P-type orbit around a low mass yellow main-sequence star? Or would this be too complicated? $\endgroup$
    – WebbS_616
    Nov 10, 2020 at 23:40
  • 1
    $\begingroup$ @WebbS_616 the mass of the yellow main-sequence star would be irrelevant other than it would effect how many planets can be in orbit and where those orbits are. The addition of a brown dwarf would add only a bit of solar energy to the aggregate, allowing the planet to be a bit further from the pair, and probably makes up for the low mass of the yellow star, so in aggregate, there may be no difference at all. The orbit type would create a somewhat non-elliptical orbit for the planets, but at 1 AU it's likely insignificant. Short answer: No. $\endgroup$
    – JBH
    Nov 10, 2020 at 23:48

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