Cooling down Venus will probably be by far the most efficient method to start terraforming the planet because then you wouldn't have to deal with the high temperatures and pressures at its surface.

I've chosen the final temperature of -50⁰ C since that is just above the triple point temperature of CO2 where this gas could change into a liquid ocean and because at the same time the lowest possible atmospheric pressure could be reached at the surface of Maxwell Montes, with about 10 km elevation the highest area on Venus. (light-brown on the image below)

Maxwell montes
Credits and author: Zamonin. Screenshot of a part of the File:VenusLanderTopo.jpg, licensed under the Creative Commons Attribution-Share Alike 3.0 Unported license.

Taking into account the total liquid mass of CO2 in Venus' atmosphere at -50⁰ C, the total surface area of the planet and the liquid carbon density of 1155 kg/m³, a 835 m deep liquid CO2 ocean could span the planet if its surface had the same elevation everywhere. So the surface of that ocean would then be at +835 m elevation, but since 80 % of the topography is within 1 km of the median radius the actual elevation would be somewhat lower.

With this peace software tool it can be calculated that at -50⁰ C CO2 becomes liquid at about 7 bar, while the scale height for Venus' atmosphere at -50⁰ C can be found out to be 4751 m.

So from the ocean's surface at 7 bar air pressure to the Maxwell Montes plateau there's a difference of at least 1.93 scale heights, meaning the pressure on the plateau would be 6.89 less or 1.02 bar (1.01 atm.) !

By comparison, at -40⁰ C CO2 would become liquid at 11 bar, and the scaleheight at that temperature 4964 m, while the CO2 ocean surface would then be at a max. +821 m elevation.

Then from the ocean's surface to the Maxwell Montes plateau a difference of 1.85 scale heights turns out to be, meaning the pressure on the plateau would become 1.73 bar (1.71 atm.),
With such low temperatures the CO2 ocean could be covered by about a 1 cm thick layer of water ice from all the water in the present atmosphere since during the cooling period probably most of the precipitated water would have flown to the lowest regions.

But can it be calculated how long it would take for Venus to cool down to a -50⁰ C surface temperature when it would receive no sunlight anymore ?

  • 2
    $\begingroup$ If answers don't surface here Earth Science StackExchange looks like they would accept it. "You can also ask about these topics [particularly climatology] in reference to other planets that fall under the umbrella of planetary science." $\endgroup$
    – Erin Anne
    Commented Oct 31, 2022 at 19:45
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    $\begingroup$ You have an awful lot of heat in the planet itself that's going to keep warming your ocean. $\endgroup$ Commented Oct 31, 2022 at 20:45
  • $\begingroup$ @LorenPechtel Yes, CO2 is already a supercritical fluid near the surface now, only below 31⁰ C it becomes a "real" liquid so by then a lot of heat should be gone. $\endgroup$
    – Cornelis
    Commented Oct 31, 2022 at 23:45
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    $\begingroup$ The Kelvin cooling time for Earth was about 30,000 years to get it from roughly 1000 degrees C to where it is today. Venus is very similar to Earth except for a much thicker atmosphere and closer proximity to the sun. I would count on needing at least that long, assuming you could block the sun from hitting the planet. But there are many, many variables. $\endgroup$
    – Dan Hanson
    Commented Nov 5, 2022 at 1:21
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    $\begingroup$ For anyone wanting to answer this, The unexpected temperature profile of Venus’s atmosphere might be useful. It's interesting to see that down to an altitude of about 60 km, the atmospheric temperature profile of Earth is similar to that of Venus. Below 60 km is where the temperature profile of the two diverges significantly - see figure 2. $\endgroup$
    – Fred
    Commented Nov 6, 2022 at 18:05

3 Answers 3


I found a YouTube video that talks about terraforming Venus. It just happens to include information about the temperature of Venus dropping when all sunlight is blocked.

If you skip to a certain part (4 minutes 30 seconds) you will see that it will show the temperature based on year.

It will take around 164 years for the temperature to get to -50 Celsius. The video has sources in the description.

  • $\begingroup$ Fine, "only" the calculation is still missing. Unfortunately, the time to build and place the mirrors will be many orders of magnitude greater than the 164 years of cooling ! $\endgroup$
    – Cornelis
    Commented Apr 8, 2023 at 9:17

At the target temperature of -50⁰ C, a black body radiates 140 W/m². The real emissions is more complicated, but that should be the order of magnitude we are working with.

$$P = \sigma T^4$$

It's enough to conclude that for instance the geothermal power from the deep of Venus (~30mW/m²) is too small to be relevant to the calculation.

So for each square metre of Venus, we have to cool down 1040 tons of atmosphere, and some quantity of surface rock.

Basalt has a thermal conductivity of 2 W m^-1 K^-1, so at the final temperature, an equilibrium heat flow of 140 W/m² from the ground is at a thermal gradient of 7.2 K/m, meaning the original surface temperature is only 70 metres down. Conclusion: The mass of rock we need to cool is much lower than the mass of atmosphere we need to cool

$$Time \approx \frac{\Delta T \cdot R_{gas} \cdot n_{gas}}{P}$$

As a very pessimistic estimate, cooling down ~1000 tons of $CO_2$ takes 25 years at 140W/m².
That's of course assuming the radiation power at the end of the process, which is obviously way too low, as radiative heat loss scales with temperature to the fourth power, meaning the start of the process goes much faster.

In conclusion: A handful of years

  • $\begingroup$ That looks quite simple, could you add a reference for the formula ? And I assume the 1000 tons of CO2 is the weight of the gas above 1 square meter of surface ? $\endgroup$
    – Cornelis
    Commented Apr 11, 2023 at 8:11
  • $\begingroup$ The thermal conductivity of the rock should play a role as well, or? It is, after all hotter in the interior, and once the warm top has cooled, we need to estimate whether the interior of the planet can heat the top rock back up. $\endgroup$ Commented Apr 11, 2023 at 12:25
  • $\begingroup$ @AtmosphericPrisonEscape Why should the cooled surface be heated up again by the interior of the planet ? If the surface has cooled faster than the subsurface at a hundred m. below, then the conductivity of the rock in between will just lower the temperature there. And the radiative cooling of the surface stays much higher than the radiative power from the interior. $\endgroup$
    – Cornelis
    Commented Apr 16, 2023 at 14:32
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    $\begingroup$ @Cornelis: Radiation is irrelevant in the rock. Conduction and convection will diffuse the heat out, the question is only whether that's fast enough. $\endgroup$ Commented Apr 16, 2023 at 16:53
  • $\begingroup$ @AtmosphericPrisonEscape O.k., so let's say before the shading of Venus the surface temp. was 400⁰ C and the interiior was 5000⁰ C. And then with the shading the surface temp. becomes -50⁰ C. Isn't it then clear that the heat will flow faster because of the greater temp. difference between the surface and the interior ? $\endgroup$
    – Cornelis
    Commented Apr 18, 2023 at 9:20

Since the only guess we have is 30,000 years, perhaps we can get a little closer by observing seasonal temperature change on Earth. Some idea could be gained from data taken from the poles$^1$, which have no sunlight for many months in winter.

Although Venus has a CO2 atmosphere, data may be available to determine the amount of heat radiated from the planet as detected by an orbiter on the night side.

If the surface materials are known, a heat decay curve could be built based on the energy radiated vs heat holding capacity. This could actually be modeled in a laboratory by heating x amount of stone to 450C and determining its cooling rate. Atmospheric insulation would have to be accounted for.

If the entire planet were shrouded and 0 solar energy was able to get through, perhaps less than a century, maybe a decade or two, as a starting estimation for cooling time. Certainly worth investigating if one needed a spare planet in the future.

$^1$ Earth North pole average high: summer 0C, winter -40C

  • $\begingroup$ Did you make any (rough) calculation for this starting estimation or a worthly investigation ? $\endgroup$
    – Cornelis
    Commented Nov 6, 2022 at 17:26
  • $\begingroup$ @Cornelis some groundwork has been done. Need planet heat radiation data. I'm thinking it will be on the order of 10 years, rather than 30,000. $\endgroup$ Commented Nov 6, 2022 at 17:33
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    $\begingroup$ Using hyperphysics.phy-astr.gsu.edu/hbase/thermo/cootime.html I get 1.13295e12 seconds = 35,900 years. That's ignoring the atmosphere, perfect emissivity = 1, starting temp 737 K, final temp 223 K. $\endgroup$
    – PM 2Ring
    Commented Nov 7, 2022 at 12:13
  • $\begingroup$ @PM2Ring perhaps the modeling needs to consider that Venus surface is already solid and only the surface and the atmosphere needs to cool. We also have the CO2 phase change, but sorry, turning off solar energy input completely ... well ... didn't that happen in a Star Trek movie? (or during a solar eclipse?). $\endgroup$ Commented Nov 7, 2022 at 18:10

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