# If Mars had a denser atmosphere, how warm would it be?

Looking at Venus, Earth and Mars, the closer you are to the Sun the warmer you are. At the same time the warmer planets also have thicker atmospheres.

Planet Minimum Maximum
Venus 870 °F (465 °C) 870 °F (465 °C)
Earth -129 °F (- 89 °C) 136 °F (58 °C)
Mars -195 °F (- 125 °C) 70 °F (20 °C)

Source for temperatures in the table: earthguide.ucsd.edu.

Mars does not have an significant magnetic field which is problematic on several levels for terraforming. There are several issue with keeping an atmosphere on Mars and some potential solutions. For this question please assume the issues with maintaining an atmosphere are addressed.

Given sufficient atmosphere, would Mars have temperature ranges similar to Earth? If so, would the atmosphere required to keep it warm, be problematic for any human to breathe?

• Mars is at about 1.4 astronomical units from the sun. I expect that the density of sunlight therefor is about half what we get on Earth. I guess that we should be able to caculate from there but no idea how to actually do that. Nov 4, 2015 at 12:19
• And then there's cold Titan with 40% thicker atmosphere (surface pressure) than Earth, and a temperature average of -94°C, 10 AU from the Sun. In certain aspects remarkably Earth like since the other two atmospheric rocky worlds Venus and Mars differ in surface pressure with factor 15,000. Nov 4, 2015 at 12:30
• There is also the lack of Oceans which have a huge impact on global temperatures. I suspect that even with atmosphere you would need large bodies of water to moderate temperature fluctuations Nov 4, 2015 at 13:27
• It is true that 5x the greenhouse effect ≠ 5 x the greenhouse gases. More like 2^5 = 32x. Mars AFAIK has no significant volcanoes but if it did, the CO_2 level would steadily increase until the planet was warm enough for liquid water. Nov 16 at 17:17

## 1 Answer

Other factors being equal, planetary temperature (measured on an absolute scale e.g. degrees Kelvin) falls with the square root of distance from the sun, so a merely Earth-like atmosphere isn't enough to warm Mars.

According to this planetary temperature calculator, an Earth-like planet (similar albedo and greenhouse-effect atmosphere) at Mars's orbital distance would have an average surface temperature of -40ºC as compared to Earth's 15ºC.

We can check this; Mars' semi-major axis is 1.52 AU, Earth's average temperature is about 288K (15ºC). So we expect:

$$\frac{288K}{\sqrt{1.52}} = 233K = -40ºC$$

The atmosphere would moderate the extreme high and low temperatures on Mars substantially, though.

Mars's albedo (reflectivity) is slightly lower than Earth, which would increase temperatures a few degrees above that, but not enough to make it a comfortable place. If you increased the greenhouse gases in the atmosphere substantially (~5 times the greenhouse effect seen on Earth) the temperature would be similar to Earth's. There's probably a mixture of gases which would provide that effect while remaining human-breathable, but I couldn't tell you what it was.

Producing and maintaining that atmosphere, of course, is far beyond our capabilities today.

• space.stackexchange.com/a/26879/12102
– uhoh
Nov 17, 2018 at 8:03
• If you increased the greenhouse gases in the atmosphere substantially (~5 times the greenhouse effect seen on Earth) the temperature would be similar to Earth, are you taking into account the logarithmic nature of the greenhouse effect? 5 times the GHG concentration does not yield 5 times the temperature difference. Nov 30, 2018 at 11:05
• @Russel Borogrove Perfluorocarbon gases would work, being much stronger greenhouse gases than CO2 while also being non-toxic. They have been proposed for terraforming Mars (astrobiology.nasa.gov/news/thawing-mars). Dec 1, 2020 at 23:48
• For those interested, the inverse square root relation is derived on this Wikipedia page ( en.wikipedia.org/wiki/Planetary_equilibrium_temperature ). Briefly, the absorbed power scales as $\mathsf{distance}^{-2}$ and the temperature as $T^4 \sim \mathsf{power}$ (according to the Stefan-Boltzmann law), so you end up with $T \sim \mathsf{distance}^{-2 \times 1/4} \sim \mathsf{distance}^{-1/2}$. Dec 22, 2021 at 15:10