Edit: I added a grid and updated the map, the red dot on the western equator being the location where the moon's surface comes closest to the parental planet.

enter image description here Im working on a project based on a system with several habitable moons orbiting a Gas Giant. Since i decided to use a star which has it's habitable zone quite far away, it would normally be one with a particularly high Mass, Luminosity and Surface Temperature. However, they seem to have a UV-Radiation Output that tends to dissipate a potential world's Atmosphere.

For this question i only want to focus on the third moon seen from the gas giant in the simulation in below's post:

Fixing my multi-lunar/planetary System

Gas Giant's Parameters

  • Masses: 11 Jupiter Masses
  • Semi Major Axis: 7.75 AU
  • Orbit: Nearly Circular
  • Axial Tilt: 23,5 Degree
  • Axial Rotation: 1 Earth Year

Moon's parameters

  • Mass: 0.6 Earth Masses
  • Diameter: 11000 km
  • Density: Same as Earth
  • Semi Major Axis to the Central Planet: 900000 km
  • Orbit: Nearly Circular
  • Rotational/Orbital Period: 39 Hours
  • Day Length: 39 Hours (compared to Earth's 24 hours)
  • Type of Rotation: Tidally Locked to the Central Gas Giant.
  • Axial Tilt: 23.5 Degree
  • Axial Rotation: 1 Earth Year
  • Atmosphere Composition: 78% Nitrogen, 21% Oxygen, small amount of water vapour and greenhouse gasses, pretty much like earth.
  • Ocean Coverage: 65%
  • Surface Pressure: Same as Earth

Imaginary Star's Parameters:

  • Mass of the Star: 3 Solar Masses
  • Diameter: 10 times the sun's.
  • Luminosity: 25 times the sun's
  • Spectral Class: G
  • Surface Temperature of the Star: 5900 Kelvin
  • Radiation emited by the star: 10% Ultra-Violet, 40% Infrared, 50% visible light
  • Radiation arriving at the moons surface: 5% Ultra-Violet, 53% Infrared, 42% visible light.
  • Star's Lifespan: Same as the sun.
  • Star's Age: Same as the sun.

My Question is: Is it possible to give any idea about ocean currents, climate zones and their location on this moon? Could they still be at least earthlike or would they be completely different? Coriolis Force is likely going to be smaller, any other differences?

  • $\begingroup$ Keep in mind that there's a difference between "earth-like" (which usually means habitable within the range of temps and weather humans are used to) and "completely different" (which the climate will be due to differences in geography, atmosphere, gravity, etc). Therefore, are you only asking if the moon's parameters would permit, insofar as we understand science today, an inhabitable climate? $\endgroup$
    – JBH
    Commented Apr 23 at 14:38
  • $\begingroup$ Nope, nope, nope, nope. Nope$^{9^{9!}}$ Currents, climate distribution and all the shabang strongly depends on the local and global topography that you haven't defined. Just to give you an earthly example, Napoli in Italy is a tad northern than New York, yet the big apple is not celebrated for its mild climate year long like its Italian sister. And the existence of mild climate in Europe is caused by the Gulf stream, which is not granted but strongly depends on global factors. $\endgroup$
    – L.Dutch
    Commented Apr 23 at 14:39
  • $\begingroup$ I have deleted my answer. Add all the needed information. $\endgroup$
    – L.Dutch
    Commented Apr 23 at 15:22
  • $\begingroup$ Thanks a lot! Okay so maybe we need a diagram for my map which indicates what color equals what elevation and maybe a system of ocean currents? What other info is missing? $\endgroup$ Commented Apr 23 at 15:33
  • $\begingroup$ Well maybe this region in space is really rich in hydrogen and thus the star possesses a bigger amount of it, therefore the star has a longer live span before it burned out. $\endgroup$ Commented Apr 23 at 18:17

2 Answers 2


I think L.Dutch's initial reaction is probably right. You'll have areas that could be rationalized as Earth-like, but not the entire moon.

  1. (I'm assuming your moon orbits your planet along the planetary ecliptic plane. See #5 for details.) Your moon is tidally locked to the planet. That means the sun will only ever bathe half of the moon — the same half every time — and do so every 20 hours. The daylight side will experience closer to Earth-like conditions. The planet-side of the moon will be a lot colder. I doubt that enough reflected light from the planet exists to compensate for this.

Note: it's actually more than just one half. The entire moon will see some daylight each orbit around the planet. Half of the planetside hemisphere will see daylight starting from a point coincident with the planet's orbital path. That half will very quickly lose light as the moon orbits the planet. The other half of the planetside hemisphere will see sunlight as the moon again crosses the planet's orbital path. A teenie bit more light as the moon moves into our out of shadow. But the effect of that sunlight is more akin to twilight here on Earth than it is full daylight. Further, the axial tilt of the moon doesn't change this analysis based on my previous assumption. So I continue to assert the fundamental problem of #1.

  1. Given #1, you won't have ice at the north and south poles (at least I'm pretty sure you won't). You'll have an ice cap at the equator on the planet side.

  2. Generally, hot air rises and cold air falls. You won't have the equivalent of Earth's prevailing westerlies because the moon isn't rotating fast enough to have them and the planet's gravity will have its own affect holding the atmosphere in place. You'll have wind, but I believe you won't have those "westerlies." You will have some atmospheric cells, but the moon's somewhat small so they'll be larger compared to the surface area of the moon than you'd expect to see on Earth. What's really different, though, is the heat expansion during the daylight pass pushing into the colder zones on the planet side of the moon. Maybe those will replace your equivalent of Earth's "westerlies," but they'd push from all directions around the sphere — strongest at the equator and weakest at the poles. That could produce some nasty storms.

  3. Your currents will depend strongly on what geography is facing the planet. Could you update your map to indicate the equatorial position that always faces the planet? Water works a bit like wind other than it must move around things (like continents...). So your equatorial hot water will want to move to polar cool water — except you'll have a strong tendency to move hot water to the planet side of the moon.

  4. Despite the moon's axial tilt, you're unlikely to have seasons due to the rapid orbit around the primary... unless the moon's orbit follows the axial tilt of the planet. It needn't. If it follows the axial tilt then a lot of what I just said is meaningless and the analysis of your moon's climate will be whomping difficult because sunlight will bathe different areas of the moon with each orbit of the moon and those areas will shift during the orbit of the planet. If this is the case, your guess is as good as ours and you might as well declare the climate to be anything you want.

As I mentioned to you in a comment to an earlier question, questions about climate are the hardest to answer. You'll rarely get detailed answers because there are so many variables and there's so much work. There are generalizations that can be made based on the weather of Earth — but the further you get away from Earth planetary norms, the less those generalizations apply. Even drawing in oceanic current lines are normally problematic... and if that moon orbits along the planet's equator rather than within the planet's ecliptic plane... my head's starting to hurt.

  • $\begingroup$ Thank you for your in depth answer. There are a few follow up questions i have. For example number 2: the ice caps. Why exactly wont there be any ice caps on the pole? And why would there be one at the equator? According to my simulations every area of the moon has a day/night cycle. Is it really non believable if i just said: The climate is similar to earths but the temperature differences are a bit higher due to the longer nights and days. Also due to the lower coriolis forces there will still be westerlies, but no polar easterlies. (en.wikipedia.org/wiki/Polar_easterlies) $\endgroup$ Commented Apr 23 at 19:20
  • $\begingroup$ The 2 continents at the west and the east coast of the big north eastern continent. These are the areas that i declare the parental planet to be seen from. $\endgroup$ Commented Apr 23 at 19:23
  • $\begingroup$ Also yes. the planetary ecliptic plane. I assume that means that there will be an eclipse every day once the moon, planet and the star will be placed in a linear order. $\endgroup$ Commented Apr 23 at 19:27
  • $\begingroup$ @Zadai.Fehbiab It's not enough to have a day/night cycle. You need enough daylight (energy) to melt the ice. As I mentioned in my first note, every point on the moon will see light, but the planet side hemisphere will receive only a fraction of the energy the sunside of the moon will. This is due to the tidally locked nature of the moon and it is why I don't believe there will be ice at the poles. The heat of the sunside of the moon will keep the poles clear and the coldest point on the moon is the planet side equator closest to the planet. $\endgroup$
    – JBH
    Commented Apr 23 at 22:11
  • $\begingroup$ Could you do me a favor? Modify your map to place a dot identifying the point on the moon that is closest to the planet. You're tidally locked, so this will always be the same point (if! If the moon is orbiting in the planet's ecliptic plane. If it's orbiting with the planet's axial tilt then that point will move... complicated). $\endgroup$
    – JBH
    Commented Apr 23 at 22:13

That's not a star, it's an impossible thing that breaks Physics

The star does not work for many problems, all boiling down to that almost anything in a star relates to mass. In fact, you can take the Mass of a star and its classification and find out almost all other needed factors from physics. With your values, you threw out all of physics.

  • By increasing the diameter by 10, you increase volume by 1000. By only increasing mass by 3, you reduced density to 0.003 of the sun. As a result, the item has a density of 0.00423 g/cm³. Taking that as pure hydrogen, we are talking about a cloud that is either during gravitational collapse into a star, or a star that just turned supernova... however, that thing is still denser than earth atmosphere by a factor of 100! There is no way that these factors work together as any stable thing - if anything, it would collapse into a star within a very short time.
  • Luminosity and Mass are directly linked with $L=M^{3.5}$. You said 3 times mass, so luminosity needs to be 46.76 times, or you violate physics.
  • Stars strive for a density close to 1.4 g/cm³, and thus Mass links to Diameter! - Stars are found on or close to the line on the Mass-Radius graph, and because density is mass per volume and volume is scaling with radius³, your 3 times solar mass demands about 2.5 solar diameters, not more.
  • G colored Stars are Main sequence, they are in a very narrow band of 0.9 to 1.1 solar masses. According to the Herzsprung-Russel-Diagram, 46.76 solar luminosity makes you expect a star with color A, so A-IV Subgiant or A-V Main Sequence, burning with about 7500 to 10000 K. Those stars are white to yellow-white, and for a Subgiant one might end at the F-IV, which would indicate a star that is on the last legs of its life cycle.

By throwing out those fundamental things, you throw out so many basics of thermodynamics and gravity, that we can't say anything we know applies to your universe.

Because there is no star, we can't model the planet.

As said, all assumptions are off. We can't assume anything anymore, physics in your world must be different. As such, we can't even assume that items are attracted to one another in the same fashion. All formulas for how the average temperature of the planet turns out? invalid for what you want to model.

A Realistic star would mean that is a rock.

Let me pick solar mass as the one thing you are right about. The rest is a 46 Luminosity, 2.5 solar radii, 8k000 Kelvin surface temperature. Now let me pin the gas giant at 8 AU for easier math.

Throwing Luminosity and distance into the wattage per square meter formula, you get a roundabout 10 W/m² at 8 AU with the 46 solar luminostities. That is roughly 1/150th of what earth gets. Using an Albedo of 0.3, not unlike earth, you get an expected average temperature of... 463.3 Kelvin, or too hot to have liquid water, so we need to recalculate with no albedo, coming to about 379.8 Kelvin - or about 106°C. No liquid water, and the sun side is scorched rock that is just a little smaller than earth.

How to fix it all?

Sorry but you need to start from scratch and should abide by physic stringently. You might use Artifexian's Worldsmith to help you fix a lot of your problems and estimate the basic temperatures.

  • $\begingroup$ Thank you for your take. I'm wondering what user JBH has to say about it. $\endgroup$ Commented Apr 24 at 10:20
  • $\begingroup$ If he shares the same view, I'm willing to downsize my star to something more sun like and turn one or 2 of the moons that i might add into planets. $\endgroup$ Commented Apr 24 at 10:29
  • $\begingroup$ @Zadai.Fehbiab you need to downsize massively $\endgroup$
    – Trish
    Commented Apr 24 at 10:39
  • $\begingroup$ Massively means from 3 solar masses to maybe 1.2? $\endgroup$ Commented Apr 24 at 10:44
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    $\begingroup$ @Zadai.Fehbiab One thing more deserves to be said. The only way to write science fiction that's any fun at all to read is for some aspect of it to be handwaved. You're building a moon that's important to your story, so you want it to be reasonably realistic. The sun isn't important to your story, so it's the handwaved part. If writers didn't do this there would be no aliens, no life other than on Earth in the universe, nor any habitable planets as independent evolution is likely to be as poisonous to humanity as trying to live underwater. Worldbuilding is how we manage that process. $\endgroup$
    – JBH
    Commented Apr 28 at 21:07

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