Back to Math And Terraforming

Temperature is very important for any celestial body, be it a terraformed planet, an unterraformed one, an asteroid or just a space colony.

## Basic Formula Edit

The formula I use for temperature is the following:

**Tv = LOG10(1+Ks*(1e+100))/((4.6e+22)*0.58674^(LOG10(1+Ks*(1e+100))))**

The temperature you will get is what I call *void temperature*, temperature of an object which is 25% gray, placed in outer space and exposed to solar light with no tilt angle.

where **Ks** is the Solar Constant. To see how you can determine the solar constant, go to Planetary Parameters. I made this formula after weeks of working with data gathered from books and sites. it works well for a planet around one or many stars. For distant objects, such as Sedna or in case of a Rough planet, the formula will give you smaller values. This so, because radiation from distant stars, from the interstellar space and universal background radiation also heats a celestial body up to a certain limit.

Also, the formula does not count internal heat of a planet, tidal heating or other processes, including atmosphere greenhouse effects.

To get the Solar Constant, see the related article. Then, apply the formula:

**Ks = KS/d^2**

where **Ks** is the solar constant for your planet, **KS** is the star's own solar constant and d is distance in millions km.

## Adapted Formula For Planets Edit

For each celestial body, the formula can be adapted:

### Objects with no atmosphere Edit

In case of asteroids and other celestial bodies that have no significant atmosphere, as they are directly illuminated by the Sun, their temperature comes close to the value you got by the formula. However, during night, they get very cold. On the Moon, during night, temperature drops abruptly to -180 degrees C and are kept above -200 because Earth has a tiny infrared radiation.

Basically, if you want to calculate the temperature of an object with no atmosphere (let's say an asteroid), you need to calculate it when exposed to different amounts of light. If you want to get deep into this, you can calculate temperature from the solar constant of each object sending light or heat towards your asteroid. Also, you can determine average temperature for each latitude. How? With this formula:

**Ksl = Ks*(90-a)/90**

Here, **Ksl** is the local solar constant, **Ks** is the solar constant and **a** is the angle (latitude). Then, insert Ksl into the temperature equation and you get the data.

Even more, since different celestial bodies reflect more or less light, you can use the albedo correction. Albedo is a number that shows the amount of light reflected by a celestial body. 0 means no light is reflected, 1 means all light is reflected. So, to get the correct value of light absorbed by the asteroid, you get the next formula:

**Ksa = Ks/4*(1-alb)**

Here, **Ksa** is the amount of absorbed light, **Ks** is the solar constant and **alb** is the albedo.

### Gas giants Edit

In case of gas giants, if the amount of heat reaching out from the interior is negligible, directly use the temperature formula. Since they are made by huge layers of gas, their atmosphere mixes fast and is able to absorb large amounts of heat.

### Objects with atmosphere Edit

They have a limited greenhouse effect and are able to send a limited amount of heat to the dark side. Even the tiny atmosphere that exists around Mercury is able to keep the dark side from reaching temperatures below -200 C.

The average temperature for an Earth-like planet is:

**Teff = (Tv*Kt)-273.15**

where **Teff** is the effective average temperature of a celestial body, **Tv** is the *void temperature* extracted with previous formula and **Kt** is a correction parameter. For Earth, it is 0.79.

On a rocky planet with atmosphere, the greenhouse effect plays an important role. Since we are terraformers, we suppose that the planets we talk about have been terraformed and contain greenhouse or anti-greenhouse gasses. Also, we assume that the average temperature is about +15 C, similar to our Earth. What will these gasses do?

**Greenhouse** gasses will be transparent for visible light, but will reflect infrared (heat). The planet receives less energy then Earth does, but also emits less energy in the infrared. Their atmospheres will trap heat for much longer time then Earth's. Nobody knows exactly how they will behave, but I have an empirical formula that might show you what to expect:

**Tl = (Te-T)*(Ks/1.98)+T**

All values should be in Celsius. **Tl** is local temperature, **Te** is the temperature that should be on an Earth - like planet, **T** is the average global temperature (+15 C) and **Ks** is the solar constant for the specified planet. This way, you can discover interesting properties. You can give for *Te* values like Earth's day or night temperature, average temperature for a specified latitude or winter/summer average temperature. Assuming your planet has the same day length, the same year length, same axial tilt and same diameter (and you only vary the solar constant), you will get interesting results. For example, assuming on Earth for a certain spot you got a 10 degrees variation between day and night, on the orbit of Saturn, you will only get a variation of 0.1 degrees. The lower the solar constant, the smaller will be temperature changes. This is the reason why on an Outer Planet, even assuming a very long polar night, temperatures will not change dramatically.

Using this formula, for a terraformed Titan, where days are 15 times longer then on Earth, you will get the following: Since on an average Earth day-night cycle you have a temperature variation of 10 degrees on Earth, on Titan, for the same timespan, you will get 0.1 C. For a 15 times longer day, we can suppose the temperature change will be 15X0.1=1.5 degrees C. Again, on Earth we get at the polar nights temperatures of -60 C (so, 75 C below average temperature). On Titan, if the polar night were having the same length as on Earth, we would get only 0.75 C temperature decrease. Given the length of a Titanian year (29 Earth years), we get 0.75X29=21.75 degrees. So, on the poles you will only see temperatures dropping to -6 C. This is in theory. However, since Titan is much smaller then Earth and gasses will circle faster around, I suppose polar temperatures will not drop below zero.

**Anti-Greenhouse** gasses or a similar technology would be needed for inner planets. They should reflect light (or at least infrared light) during day and allow infrared light to pass during night. Nobody knows for sure how would they work and I have no formula for this. The only thing I can speculate is that small holes in the anti-greenhouse gas layer will result, depending on distance to the host planet, in heat increase, radiations, possibly destroying all life directly exposed to light.

## Another formula Edit

There is another formula which I derived from Stefan - Boltzmann Law and adapted for our measurement units:

**Tv = (Ks^0.25)*307.7-273.15**

Here, **Tv** is *void temperature* (temperature of a body with color 25% gray, exposed directly to radiation, with no tilt) and **Ks** is the local solar constant. Result is in degrees Celsius.

If you want more and if you know the albedo, then you can use the following formula:

**T = (Ks^0.25)*((1-alb)*4*307.7-273.15**

The only difference is that **alb** is the bolometric albedo of the target object.

I prefer the previous formula, even if it is more complicated then this one, because results are closer to reality.

## Adapted formula for terraformed planets Edit

For more details, see Climate.