<p>For bodies without surface temperature you can guestimate the average temperature by considering the incoming energy and it being re-radiated into space via its complete surface (not only the surface which is exposed to the Sun (or host star) - that's what you link yourself.</p>
<p>You can, of course, do the same for only the illuminated part, and consider only that part to contribute to the re-radiation. However for a non-rating (or rather bound-rotating) planet you then will get on the night side a temperature equivalent to the background... 3K.</p>
<p>If you want to consider a realistic scenario, you have to calculate with the heat transfer equation and heat capacity. Not impossible (e.g. done <a href="https://arxiv.org/pdf/2211.12807.pdf" rel="nofollow noreferrer">here</a> by Johanna Bürger et al for 67/P based on observations and different assumed material properties), but can only be solved reasonably numerically.</p>
<p>When an atmosphere is involved, the calculation also becomes much more complicated as heat transfer and green house effects in it play also a major role.</p>

CC BY-SA 4.0
2023-04-19T18:47:23.950Z
I am wondering what do you mean by "temperature equivalent to the background... 3K"?  In my first guestimate of whole planet temperature range (Tday & Tnight), I had the min temperature (Tnight) computed by passing an albedo near 1.0.  Then I saw a paper that used the Tday = Teff + Teff / 3 so I switched to this for now.
VirtualPaul
CC BY-SA 4.0
2023-04-11T11:50:11.663Z
2023-04-11T11:50:11.663Z
planetmaker
<p>You Question includes an incorrect assumption.</p>
<p>There are no celestial objects that do not rotate faster or slower.</p>
<p>Celestial objects are formed when smaller objects collide at speed slow enough that they ca stick together.</p>
<p>Imagine two circular space objects each a centimeter in diameter.  Suppose that they collide head on with their centers exactly lined up.  Then they would not form a rotating object.</p>
<p>But suppose that the centers of the two objects are not lined up but are a millimeter apart.  Since the two objects were travelling in opposite directions they will stop moving in opposite directions but their momentum will make them rotate.  Each former particle will continue on in the same direction, but since they are now attached to each other they will rotate around each other.</p>
<p>Suppose that the center lines of the two objects are separated by a tenth of a millimeter. The two objects will not be perfectly lined up and so they will rotate after they are joined.</p>
<p>Suppose that the center lines of the two objects are separated by a hundredth of a millimeter. The two objects will not be perfectly lined up and so they will rotate after they are joined.</p>
<p>And so on and so on.</p>
<p>No two objects which meet in space and merge to form a larger object can ever be perfectly lined up.  Nor can they ever be travelling in exactly opposite directions when they meet.  Thus they must form a rotating object when they merge.</p>
<p>A planet forms as the result of gazillions (not to be overly precise) of collisions and mergers of tiny space objects.  The odds against all those gazillions of mergers producing a perfectly symmetrical angular momentum in the resulting planet are very, very, very, very, very, large - astronomical.</p>
<p>And so no planet can ever not rotate.</p>
<p>Celestial objects which are tidally locked have a rotation rate that has been slowed down by tidal interactions so that the rotation rate is now equal to their orbital period around a more massive object.  That results in one side of the tidally locked object always facing the more massive object and the other side always facing away from the more massive object.</p>
<p>If the object that the tidally locked object orbits around is a star that emits light, the side of the tidally locked object that faces the star will have eternal light and heat and the side of the tidally locked object that faces away from the star will have eternal dark and cold (except to the degree that the planetary hydrosphere and atmosphere may distribute heat from one half of the world to the other half).</p>
<p>But the tidally locked world will rotate.  Its rotation period will happen to be the same as its orbital period around the star, thus resulting in one side of the tidally locked object always facing the star and the other side always facing away from the star.  That is definitely different from the impossible case of a celestial object which has no rotation.</p>
<p>For example, the Moon is tidally locked to the Earth, and so it's sidereal rotation period is 27.321661 Earth days, the same as its sidereal orbital period around the Earth of 27.321661 Earth days.</p>
<p>So a tidally locked planet would rotate, but its rotation period would be the same as its orbital period around its star.</p>
<p>A rogue planet in interstellar space would have a dark side.  In fact every side would be a dark side, since it wouldn't have a star to light one side and would be only very dimly lie by starlight.  And if a rogue planet was in intergalactic space it wouldn't even have starlight to light it, just a little bit of light from other galaxies.  But rogue planets must rotate, like very other type of celestial object, from tiny motes of dust to super massive back holes and entire galaxies.</p>

CC BY-SA 4.0
2023-04-19T18:32:07.767Z
Thanks for the clarification.  I knew about tidally locked planet, but I was not sure if there were other cases where a planet or moon would have a dark side.  So, I tried to include those cases in my question.
VirtualPaul
CC BY-SA 4.0
2023-04-11T17:35:46.620Z
2023-04-11T17:35:46.620Z
M. A. Golding
<p>I am trying to calculate the different temperatures (day, night, permanent day/night for tidally locked planets) for different surfaces of exoplanets like land, water, gas, ice.</p>
<p>I am using an exoplanet generator that I created so I have access to the altitudes and the type of surface based on different parameters I can choose.  At the moment, I only have atmosphere based on some parameters.</p>
<p>I have seen these which helped me:
<a href="https://astronomy.stackexchange.com/questions/10113/how-to-calculate-the-expected-surface-temperature-of-a-planet">How to calculate the expected surface temperature of a planet</a>
<a href="https://en.wikipedia.org/wiki/Planetary_equilibrium_temperature" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Planetary_equilibrium_temperature</a></p>
<p>Right now I am computing one temperature but that does not give me the day or night or dark side of a tidally locked planet.  I also don't know what is the water vs land surface temperature.  The lapse rate I am using is for Earth so I would need to compute this also for an exoplanet.</p>
<p>I am guessing this won't be exact science since there are a lot of unknowns in my scenario but any help toward getting more realistic estimates would be appreciated.</p>
<p>For the terran exoplanet, I compute the orbit that would make it in the habitable zone.  This allows me to use meaningful temperature colors when drawing it.  This is what I got so far:
<a href="https://i.sstatic.net/CRGzu.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/CRGzu.png" alt="This is what I got so far" /></a></p>

CC BY-SA 4.0
2023-04-10T19:08:16.273Z
On planets with atmospheres the effect of the atmosphere can be huge.  Consider Venus... Even on Earth.  What is the "night time temp on Earth" (well if you are in Ottawa in January, it is rather different from Singapore)
James K
CC BY-SA 4.0
2023-04-11T11:54:00.607Z
Thanks for your comment.  For sure I am not looking for a single temperature for the whole planet.  Although an average range would be a start.  Right now I am using the latitude to compute local temperature based on a planet temperature range.  I have added more information and a picture to clarify.
VirtualPaul
<exoplanet><temperature><surface>
How to calculate the day/permanently dark/night temperatures of different surfaces for exoplanets?
VirtualPaul
# How to calculate the day/permanently dark/night temperatures of different surfaces for exoplanets?

(Asked by: username_0 on 2023-04-10)

I am trying to calculate the different temperatures (day, night, permanent day/night for tidally locked planets) for different surfaces of exoplanets like land, water, gas, ice.

I am using an exoplanet generator that I created so I have access to the altitudes and the type of surface based on different parameters I can choose. At the moment, I only have atmosphere based on some parameters.

I have seen these which helped me:
[How to calculate the expected surface temperature of a planet](https://astronomy.stackexchange.com/questions/10113/how-to-calculate-the-expected-surface-temperature-of-a-planet)
<https://en.wikipedia.org/wiki/Planetary_equilibrium_temperature>

Right now I am computing one temperature but that does not give me the day or night or dark side of a tidally locked planet. I also don't know what is the water vs land surface temperature. The lapse rate I am using is for Earth so I would need to compute this also for an exoplanet.

I am guessing this won't be exact science since there are a lot of unknowns in my scenario but any help toward getting more realistic estimates would be appreciated.

For the terran exoplanet, I compute the orbit that would make it in the habitable zone. This allows me to use meaningful temperature colors when drawing it. This is what I got so far:
[![This is what I got so far](https://i.sstatic.net/CRGzu.png)](https://i.sstatic.net/CRGzu.png)

**Tags:** exoplanet, temperature, surface

> **Comments:**
> **username_1**: On planets with atmospheres the effect of the atmosphere can be huge.  Consider Venus... Even on Earth.  What is the "night time temp on Earth" (well if you are in Ottawa in January, it is rather different from Singapore)  
> **username_0**: Thanks for your comment.  For sure I am not looking for a single temperature for the whole planet.  Although an average range would be a start.  Right now I am using the latitude to compute local temperature based on a planet temperature range.  I have added more information and a picture to clarify.  

---

## 2 answers

### Answer by username_2 on 2023-04-11. Score: 1

For bodies without surface temperature you can guestimate the average temperature by considering the incoming energy and it being re-radiated into space via its complete surface (not only the surface which is exposed to the Sun (or host star) - that's what you link yourself.

You can, of course, do the same for only the illuminated part, and consider only that part to contribute to the re-radiation. However for a non-rating (or rather bound-rotating) planet you then will get on the night side a temperature equivalent to the background... 3K.

If you want to consider a realistic scenario, you have to calculate with the heat transfer equation and heat capacity. Not impossible (e.g. done [here](https://arxiv.org/pdf/2211.12807.pdf) by Johanna Bürger et al for 67/P based on observations and different assumed material properties), but can only be solved reasonably numerically.

When an atmosphere is involved, the calculation also becomes much more complicated as heat transfer and green house effects in it play also a major role.

> **Comments:**
> **username_0**: I am wondering what do you mean by "temperature equivalent to the background... 3K"?  In my first guestimate of whole planet temperature range (Tday & Tnight), I had the min temperature (Tnight) computed by passing an albedo near 1.0.  Then I saw a paper that used the Tday = Teff + Teff / 3 so I switched to this for now.  

---

### Answer by username_3 on 2023-04-11. Score: 1

You Question includes an incorrect assumption.

There are no celestial objects that do not rotate faster or slower.

Celestial objects are formed when smaller objects collide at speed slow enough that they ca stick together.

Imagine two circular space objects each a centimeter in diameter. Suppose that they collide head on with their centers exactly lined up. Then they would not form a rotating object.

But suppose that the centers of the two objects are not lined up but are a millimeter apart. Since the two objects were travelling in opposite directions they will stop moving in opposite directions but their momentum will make them rotate. Each former particle will continue on in the same direction, but since they are now attached to each other they will rotate around each other.

Suppose that the center lines of the two objects are separated by a tenth of a millimeter. The two objects will not be perfectly lined up and so they will rotate after they are joined.

Suppose that the center lines of the two objects are separated by a hundredth of a millimeter. The two objects will not be perfectly lined up and so they will rotate after they are joined.

And so on and so on.

No two objects which meet in space and merge to form a larger object can ever be perfectly lined up. Nor can they ever be travelling in exactly opposite directions when they meet. Thus they must form a rotating object when they merge.

A planet forms as the result of gazillions (not to be overly precise) of collisions and mergers of tiny space objects. The odds against all those gazillions of mergers producing a perfectly symmetrical angular momentum in the resulting planet are very, very, very, very, very, large - astronomical.

And so no planet can ever not rotate.

Celestial objects which are tidally locked have a rotation rate that has been slowed down by tidal interactions so that the rotation rate is now equal to their orbital period around a more massive object. That results in one side of the tidally locked object always facing the more massive object and the other side always facing away from the more massive object.

If the object that the tidally locked object orbits around is a star that emits light, the side of the tidally locked object that faces the star will have eternal light and heat and the side of the tidally locked object that faces away from the star will have eternal dark and cold (except to the degree that the planetary hydrosphere and atmosphere may distribute heat from one half of the world to the other half).

But the tidally locked world will rotate. Its rotation period will happen to be the same as its orbital period around the star, thus resulting in one side of the tidally locked object always facing the star and the other side always facing away from the star. That is definitely different from the impossible case of a celestial object which has no rotation.

For example, the Moon is tidally locked to the Earth, and so it's sidereal rotation period is 27.321661 Earth days, the same as its sidereal orbital period around the Earth of 27.321661 Earth days.

So a tidally locked planet would rotate, but its rotation period would be the same as its orbital period around its star.

A rogue planet in interstellar space would have a dark side. In fact every side would be a dark side, since it wouldn't have a star to light one side and would be only very dimly lie by starlight. And if a rogue planet was in intergalactic space it wouldn't even have starlight to light it, just a little bit of light from other galaxies. But rogue planets must rotate, like very other type of celestial object, from tiny motes of dust to super massive back holes and entire galaxies.

> **Comments:**
> **username_0**: Thanks for the clarification.  I knew about tidally locked planet, but I was not sure if there were other cases where a planet or moon would have a dark side.  So, I tried to include those cases in my question.  

---

