WBH High and Low Temperature

[Dodo98]

@Geir
I've encountered a problem. When calculating the high and low temperatures of different moons, I get a low temperature of 0 K, which is physically impossible in principle. This result stems from a low luminosity of 0 for these moons. And this low luminosity comes from a luminosity modifier of 1. The problem seems to be the luminosity modifier, which forces the result to be 0. It seems to me that this modifier should never be 1 if you want to get valid results, but the rules say that it just can't be greater than 1 or less than 0, so 1 seems to be legal according to the rules. Have you ever encountered this problem?

 

[Geir]

@Geir
I've encountered a problem. When calculating the high and low temperatures of different moons, I get a low temperature of 0 K, which is physically impossible in principle. This result stems from a low luminosity of 0 for these moons. And this low luminosity comes from a luminosity modifier of 1. The problem seems to be the luminosity modifier, which forces the result to be 0. It seems to me that this modifier should never be 1 if you want to get valid results, but the rules say that it just can't be greater than 1 or less than 0, so 1 seems to be legal according to the rules. Have you ever encountered this problem?

Well, at least it doesn't give a negative value.

What the model doesn't account for is all cooling time, so it is likely to never get that low, but, hey, it's a simplified model.
You should probably consider 3K, (slightly above the background of about 2.72K) to be the lower limit. Most likely it would be closer to 10K, since that's about the temperature of cold molecular gas.

So, this lowest temperature should only apply to a moon when it's in its world's shadow and only on the far side (from the planet) of the moon, so it would be a reasonably short period of time. Even rock takes some time to cool off, but like I said, not that advanced.

Also, I don't think it's explicitly stated, but any seismic stress temperature should be applied to both hot and cold temperature, so if the moon has any tidal heating at all, it should kick in a Kelvin or two.

 

[Dodo98]

Well, at least it doesn't give a negative value.

What the model doesn't account for is all cooling time, so it is likely to never get that low, but, hey, it's a simplified model.
You should probably consider 3K, (slightly above the background of about 2.72K) to be the lower limit. Most likely it would be closer to 10K, since that's about the temperature of cold molecular gas.

So, this lowest temperature should only apply to a moon when it's in its world's shadow and only on the far side (from the planet) of the moon, so it would be a reasonably short period of time. Even rock takes some time to cool off, but like I said, not that advanced.

Also, I don't think it's explicitly stated, but any seismic stress temperature should be applied to both hot and cold temperature, so if the moon has any tidal heating at all, it should kick in a Kelvin or two.

Yes, I had also thought of a minimum value of around 10K. However, such values should actually only be achieved with moons that are very, very far away from the star. But I'll give you an example of one of the moons where I got 0K, which shouldn't actually be possible because it is very close to a fairly bright star. Maybe I miscalculated or made a mistake somewhere.

The star has a luminosity of 0.742 and the moon is orbiting a terrestrial planet every 10.23 days. The planet has an Orbit# of 1.91 or 0.673 AU.

The moon has a very heavy axial tilt of 87.72 degrees which I calculated to be a Basic Axial Tilt Factor of 0.999 (I used the formula) and after halving that value due to the short Orbital Period it has an Axial Tilt Factor of 0.5.

I calculated it's Rotational Factor to be 0.126 (The solar day is 39.69 hours).

It's Geographic Factor is 0.5 (Hydrographics Code 0).

Axial Tilt Factor + Rotational Factor + Geographic Factor = 1.126 (capped to a maximum of 1).

Next we have a 1 for Atmospheric Factor (no atmosphere) which leads to the Luminosity Modifier of 1.

The High Luminosity is 1.484 and the Low Luminosity is 0.

All these values combined with the Near AU of 0.596, Far AU of 0.75, Albedo 0.12 and Greenhouse Factor of 0 lead to a high value of 386 K and the mentioned Low Temperature of 0K. The mean temperature is calculated as 306K. And it's this enormous jump between Mean and Low Temperature that leads me to believe that I have an error somewhere.

 

[Geir]

The moon has a very heavy axial tilt of 87.72 degrees which I calculated to be a Basic Axial Tilt Factor of 0.999 (I used the formula) and after halving that value due to the short Orbital Period it has an Axial Tilt Factor of 0.5.

That axial tilt on the moon, may be exposing the problem, but I don't think it's you math, but another limitation of the system.

If we break it apart, any world on its side and without an atmosphere is going to get you to 1.0 and 0 as luminosity modifiers, regardless of rotation. So simple addition (let's ignore the fact that this is a moon and not a planet, something that very likely complicates the picture a bit, but orbital inclination and tilt of the planet and its moon systems are also ignored) gets you to the limit of 1.0 fairly quick. The addition of factors probably ought to be modified into something more complicated to make the approach to a luminosity modifier of 1 more... asymptotic.

No going to test or advocate for this method, but if instead you took the square of each factor, then added the squares together and took the square root (sort of like a 3D distance computation, but not), then you'd get ~0.72 as in answer with your numbers, but that's just a shot in the dark. Might be better to just apply a 10K minimum in all cases, or take any low calculated value below say 20K and run it through the temperature addition equation with an added 10 K.

Again, probably too complicated, but easier than trying to determine how fast rocks cool or how heat from the hot side conducts to the cold side.

 

[Dodo98]

That axial tilt on the moon, may be exposing the problem, but I don't think it's you math, but another limitation of the system.

If we break it apart, any world on its side and without an atmosphere is going to get you to 1.0 and 0 as luminosity modifiers, regardless of rotation. So simple addition (let's ignore the fact that this is a moon and not a planet, something that very likely complicates the picture a bit, but orbital inclination and tilt of the planet and its moon systems are also ignored) gets you to the limit of 1.0 fairly quick. The addition of factors probably ought to be modified into something more complicated to make the approach to a luminosity modifier of 1 more... asymptotic.

No going to test or advocate for this method, but if instead you took the square of each factor, then added the squares together and took the square root (sort of like a 3D distance computation, but not), then you'd get ~0.72 as in answer with your numbers, but that's just a shot in the dark. Might be better to just apply a 10K minimum in all cases, or take any low calculated value below say 20K and run it through the temperature addition equation with an added 10 K.

Again, probably too complicated, but easier than trying to determine how fast rocks cool or how heat from the hot side conducts to the cold side.

Thank you, I will try the method of taking the square of each factor. And if that does not produce satisfactory results I still have the other option.

 

[Dodo98]

@Geir The option to take the square of each factor yields reasonable results. In some edge cases the Low Luminosity was still 0 because the Variance Factors added up to 1. So what I'm doing now is to use the square factor option + limiting the variance factors to 0.001 and 0.999. This ensures there will be no more 0K results.

I'm now on p. 125, doing the gas giant residual heat calculations. I noticed your example for Zed Prime's gas giant gives a result of 235K for the cloud top temperature. I calculated it as well to test my excel formula, but I'm getting a result of 260. I calculated: 279*POWER(1.419*(1-0.4)*(1+0)/1.06^2,1/4)

 

[Geir]

@Geir The option to take the square of each factor yields reasonable results. In some edge cases the Low Luminosity was still 0 because the Variance Factors added up to 1. So what I'm doing now is to use the square factor option + limiting the variance factors to 0.001 and 0.999. This ensures there will be no more 0K results.

Great!

I'm now on p. 125, doing the gas giant residual heat calculations. I noticed your example for Zed Prime's gas giant gives a result of 235K for the cloud top temperature. I calculated it as well to test my excel formula, but I'm getting a result of 260. I calculated: 279*POWER(1.419*(1-0.4)*(1+0)/1.06^2,1/4)

My bad. I read across the table crooked and used Aab V's distance of 1.30 AU by mistake.

 

[Dodo98]

Great!

My bad. I read across the table crooked and used Aab V's distance of 1.30 AU by mistake.

@Geir Doing the Tidal Heating right now and I think I've come across a rare edge case. I have a Size 4 moon in Orbit around a small gas giant. The moon orbits very close at only 2.1 PD or 80275km. Doing the calculation produces a crazy Tidal Heating Factor of 18991. The calculation was as follows: (15^2*4^5*0.005^2) / (3000*0.080275^5*0.68*0.0446)

Is the calculation correct? I checked it twice and the result makes sense to me, considering the size of the moon and how close it orbits it's planet. But such a big Tidal Heating Factor means that the temperature would be hotter than the surface temperature of most stars

So I think this moon is gone...

 

[Geir]

@Geir Doing the Tidal Heating right now and I think I've come across a rare edge case. I have a Size 4 moon in Orbit around a small gas giant. The moon orbits very close at only 2.1 PD or 80275km. Doing the calculation produces a crazy Tidal Heating Factor of 18991. The calculation was as follows: (15^2*4^5*0.005^2) / (3000*0.080275^5*0.68*0.0446)

Is the calculation correct? I checked it twice and the result makes sense to me, considering the size of the moon and how close it orbits it's planet. But such a big Tidal Heating Factor means that the temperature would be hotter than the surface temperature of most stars

So I think this moon is gone...

The math looks correct. Well, either moon is a vapor ring (though the individual bodies in the ring with smaller size should drastically reduce the world size, since you'd check for each grain) or the eccentricity needs to go to zero. Otherwise, even the most highly temperature resistant material starts to sublimate at over ~3000K (ish) and if you did an atmosphere retention calculation on that temperature (not covered in the book at all), in all but the most massive world, the metal vapor would escape and from that ring of eventually freezing metal ring, which sort of means you get to a derived Roche Limit in another way - look at p. 76 and plug in the densities, with the moon likely much denser than the planet and see if the result makes for a higher Roche limit.

So, your options are:

1. Completely circularize the orbit (so the numerator becomes zero).
2. Treat it as ring, either because it works out so from p. 76, or just because it just can't be solid over time.
3. Delete all moons above ~5000K because even tungsten will start to vaporize at that temp - or lower - in a vacuum.

 

[Galadrion]

@Geir Doing the Tidal Heating right now and I think I've come across a rare edge case. I have a Size 4 moon in Orbit around a small gas giant. The moon orbits very close at only 2.1 PD or 80275km. Doing the calculation produces a crazy Tidal Heating Factor of 18991. The calculation was as follows: (15^2*4^5*0.005^2) / (3000*0.080275^5*0.68*0.0446)

Is the calculation correct? I checked it twice and the result makes sense to me, considering the size of the moon and how close it orbits it's planet. But such a big Tidal Heating Factor means that the temperature would be hotter than the surface temperature of most stars

So I think this moon is gone...

Personally, if I got a tidal heating result that extreme, the first thing I would check (besides a calculation verification pass - which you've already done) is whether the moon was inside the Roche Limit. Tidal stresses, after all, are the major mechanical driver behind why the Roche Limit works as it does.

 

[Dodo98]

The math looks correct. Well, either moon is a vapor ring (though the individual bodies in the ring with smaller size should drastically reduce the world size, since you'd check for each grain) or the eccentricity needs to go to zero. Otherwise, even the most highly temperature resistant material starts to sublimate at over ~3000K (ish) and if you did an atmosphere retention calculation on that temperature (not covered in the book at all), in all but the most massive world, the metal vapor would escape and from that ring of eventually freezing metal ring, which sort of means you get to a derived Roche Limit in another way - look at p. 76 and plug in the densities, with the moon likely much denser than the planet and see if the result makes for a higher Roche limit.

So, your options are:

1. Completely circularize the orbit (so the numerator becomes zero).
2. Treat it as ring, either because it works out so from p. 76, or just because it just can't be solid over time.
3. Delete all moons above ~5000K because even tungsten will start to vaporize at that temp - or lower - in a vacuum.

Thanks for checking my numbers. Yes, I've also thought about the last two options. I think I'll leave the eccentricity alone, since I'm basing it on the randomness of dice rolls. And if that means the moon will be torn apart by tidal forces, then so be it.