2
$\begingroup$

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?

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ It does matter if the Sun is at it's Zenith or Horizon. See pretty picture: windows2universe.org/earth/climate/images/… and some mathematics on same principal, using the Moon. 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. $\endgroup$
    – userLTK
    Aug 23, 2016 at 11:04
  • $\begingroup$ @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. $\endgroup$
    – zephyr
    Aug 23, 2016 at 12:36

1 Answer 1

Reset to default
4
$\begingroup$

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). According to Stefan-Boltzmann 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.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .