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The Idealized greenhouse model article on Wikipedia gives the following solution for finding the global surface temperature on an idealized planet with an atmosphere:

$$T_s = \left[ \frac{S_0(1 - \alpha_p)}{4\sigma} \frac{1}{1 - \frac{\epsilon}{2}} \right]^{1/4} $$

  • $T_s$ is the surface temperature on the planet,
  • $S_0$ is the solar constant, in this case for Mars,
  • $\alpha_p$ is the planetary albedo,
  • $\sigma$ is the Stefan-Boltzmann constant,
  • $\epsilon$ is the emissivity or absorptivity of the atmosphere.

The number 4 in the denominator is accounting for the fact that the surface area of a sphere is 4 times the area of its intercept (its shadow), so the average incoming radiation is $S_0/4$.

But for a local area near the equator that would stretch $y$ km along its longitude, for a full day/night cycle the total surface area for the incoming radiation along its latitude would be $2\pi ry$ km², while the intercept (or shadow) of that total surface area would be $2ry$ km², $r$ being the radius of the circle at that latitude of the planet.

So couldn't one use the expression above for a local area near the equator when the number 4 in the denominator would be replaced by $\pi$, and also take into account the latitude of the local area and the axial tilt of Mars ?

It still would be an idealized model of course, not accounting for the winds for example.

And could this adjusted model be still more refined, or localized ?

And are there other models to predict the average local temperatures on Mars ?

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You obviously could do this, by fudging $\epsilon$ suitably. However this idealised greenhouse model is wrong for Earth and is probably wrong for Mars as well. It's wrong for Earth because the atmosphere is not optically thin for long-wave radiation (infrared), so you simply can't consider the atmosphere as a single layer like this. Instead you have to model radiative transfer of longwave radiation up through the atmosphere to a height where the atmosphere, upwards, is optically thin, at which point the longwave radiation can escape to space. It's this height, and the temperature of the atmosphere at this height which matters. And, of course, in order to know both the height and the temperature of the atmosphere at the height where it becomes optically thin you have to model the pressure & temperature profiles of the atmosphere as well as knowing its composition.

So if you actually want to model the surface temperature of Mars and how it depends on the atmosphere and its composition, you almost certainly have to bite the bullet and run a real climate model. People do this, for instance NASA as well as others.

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Of course the temperature will depend on the incoming solar radiation and that can be calculated with the formula in this answer:
$$I_{solar} \approx = 717 \ \text{W/m}^2 \left(\frac{1.38138027 \text{AU}} {r}\right)^2 \cos(lat - lat_{ss})$$
where $lat_{ss}$ is the sub-solar latitude.

To try out the suitability of the adjusted idealized greenhouse model in the question we could compare the average air temperature of -62⁰ C that was measured on september 25, 2020 by the Mars Insight Mission lander. (latitude 4.5⁰ N) with the temperature calculated with that model.

With the values of $r$ = 1.40077721 and $lat_{ss}$ = - 24.632289 ⁰ obtained from the JPL Horizons website on that day the $I_{solar}$ = 609 W/m².

The regional albedo is also an important factor to consider.
On Mars Global Surveyor Thermal Emission Spectrometer experiment: Investigation description and surface science results Figure 8 we can see the albedo on Mars mainly varies between 0.1 and 0.3.
And on the MARS Global Data Sets albedo map the Insight's landing site looks rather bright, so an albedo of 0.3 could be appropriate.
Using the adjusted idealized greenhouse model (with $\pi$ in the denominator instead of 4) with $\epsilon$ = 0, $\alpha_p$ = 0.3, and $S_0$ = $I_{solar}$ = 609 W/m² gives -52⁰ C for a Mars with no atmosphere !

It has to be noted that Insight's deck has a height of over 83 cm, so that could also explain the difference in calculated and measured temperature.
That apparent difference is confirmed in this article.

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