<p>Tidal lock would mean only one side of Mars would be exposed to solar irradiation, so its temperature would be mostly defined by the equilibrium between the energy received from the Sun and the black-body radiation.</p>

<p>Because of the orbit eccentricity, solar irradiation reaching Martian surface varies between 493 and 717 W/m² (<a href="http://ccar.colorado.edu/asen5050/projects/projects_2001/benoit/solar_irradiance_on_mars.htm" rel="nofollow">source</a>). According to <a href="https://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law" rel="nofollow">Stefan-Boltzmann law</a>, the temperature required for the black-body radiation to match these levels will vary between 305 and 335 K (32°C an 62°C), respectively.</p>

<p>Such temperatures will be reached in the area directly facing the Sun (so the Sun is seen in zenith), and gradually decline towards the background radiation temperature of space (about 3K) when approaching the dark side (where the Sun is just barely above the horizon). Such temperature distribution is due to the angle of incidence, not to the atmospheric "filter" or lack thereof.</p>

CC BY-SA 3.0
2016-08-23T12:55:36.713Z
2016-08-23T13:13:41.563Z
2016-08-23T13:13:41.563Z
Dmitry Grigoryev
<p>Since there's very little atmosphere on Mars, it doesn't matter much whether the Sun is in the zenith or hanging just above the horizon: without the atmospheric "filter," the sunlight would have roughly the same effect on all sunlight areas.</p>

<p>What would the temperatures be like on the sunlit portion of the planet?</p>

CC BY-SA 3.0
2016-08-23T11:04:47.097Z
It does matter if the Sun is at it's Zenith or Horizon.    See pretty picture:   http://www.windows2universe.org/earth/climate/images/sun_angle_insolation_big.jpg   and some mathematics on same principal, using the Moon.  http://www.lunarpedia.org/index.php?title=Lunar_Temperature   The atmosphere actually does the opposite of what you suggest, it regulates the temperature away from the extremes, and it also acts as a kind of blanket, trapping more heat than it filters or reflects.
userLTK
CC BY-SA 3.0
2016-08-23T12:36:34.070Z
@userLTK I think this question is subjective, namely the "doesn't matter as much". You interpret the difference in energy imparted by the Sun versus angle as "mattering", but the questioner considers such a small difference as "roughly the same effect". And really, the decrease based on angle is insignificant when compared to the decrease from the atmosphere. For Earth, the Sun on the horizon is effectively shining through 11 atmospheres of air, causing a significantly larger difference in surface temperature than caused by the angle change. I believe that's the root of their question.
zephyr
<the-sun><mars><temperature>
If Mars were tidally locked to the Sun, what would the temperatures on its surface be like?
Ricky
# If Mars were tidally locked to the Sun, what would the temperatures on its surface be like?

(Asked by: username_0 on 2016-08-23)

Since there's very little atmosphere on Mars, it doesn't matter much whether the Sun is in the zenith or hanging just above the horizon: without the atmospheric "filter," the sunlight would have roughly the same effect on all sunlight areas.

What would the temperatures be like on the sunlit portion of the planet?

**Tags:** the-sun, mars, temperature

> **Comments:**
> **username_1**: It does matter if the Sun is at it's Zenith or Horizon.    See pretty picture:   http://www.windows2universe.org/earth/climate/images/sun_angle_insolation_big.jpg   and some mathematics on same principal, using the Moon.  http://www.lunarpedia.org/index.php?title=Lunar_Temperature   The atmosphere actually does the opposite of what you suggest, it regulates the temperature away from the extremes, and it also acts as a kind of blanket, trapping more heat than it filters or reflects.  
> **username_2**: @username_1 I think this question is subjective, namely the "doesn't matter as much". You interpret the difference in energy imparted by the Sun versus angle as "mattering", but the questioner considers such a small difference as "roughly the same effect". And really, the decrease based on angle is insignificant when compared to the decrease from the atmosphere. For Earth, the Sun on the horizon is effectively shining through 11 atmospheres of air, causing a significantly larger difference in surface temperature than caused by the angle change. I believe that's the root of their question.  

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## 1 answers

### Answer by username_3 on 2016-08-23. Score: 4 | Accepted answer

Tidal lock would mean only one side of Mars would be exposed to solar irradiation, so its temperature would be mostly defined by the equilibrium between the energy received from the Sun and the black-body radiation.

Because of the orbit eccentricity, solar irradiation reaching Martian surface varies between 493 and 717 W/m² ([source](http://ccar.colorado.edu/asen5050/projects/projects_2001/benoit/solar_irradiance_on_mars.htm)). According to [Stefan-Boltzmann law](https://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law), the temperature required for the black-body radiation to match these levels will vary between 305 and 335 K (32°C an 62°C), respectively.

Such temperatures will be reached in the area directly facing the Sun (so the Sun is seen in zenith), and gradually decline towards the background radiation temperature of space (about 3K) when approaching the dark side (where the Sun is just barely above the horizon). Such temperature distribution is due to the angle of incidence, not to the atmospheric "filter" or lack thereof.


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