3
$\begingroup$

Suppose a planet like Earth or Venus is totally isolated from sunlight. How quick would it cool down?

On Earth it seems to me that the air temperature drops significantly even during a night. Does it mean that if we screen Venus of Sun's radiation with some untransparent cloud, it would cool down rather quickly, and not for millions of years as some may expect?

Improve this question
$\endgroup$
1
  • $\begingroup$ Please clarify in your question title, that you are asking about a planet's atmosphere, not the bulk planet. The bulk planet which consists of solid material takes way longer to cool by conduction and convection than the gaseous atmosphere which cools radiatively and convectively. $\endgroup$
    – AtmosphericPrisonEscape
    Nov 25, 2016 at 15:31

2 Answers 2

Reset to default
3
$\begingroup$

Not a full answer as that is hard to give. But:

This is a diagram by Kevin Trenberth picturing energy flows in the Earth climate system:

Energy flows
(Source: Skeptical Science)

In the hypothetical extreme case of blocking all incoming solar radiation, the Earth would — initially — have a net heat loss of roughly 340 $W m^{-2}$.

93% of net radiation change on Earth enters the oceans. In the past 20 years, this has led to an increase in ocean heat content (OHC) of roughly $10^{23}$ J. With a 340 W m^{-2} heat loss, undoing that would take $10^{23} J / ((340 W/m^2) * .7 * 510 * 10^6 km^2$ which is roughly 10 days. Although this is a severe oversimplification, it does give us a first order of magnitude.

It might be possible to run a climate model and switch off the Sun, just for fun. You don't need a heavy climate model to do so. PlaSim is good enough to get a handle on it.

My guesstimate is that it would take a couple of months before mid-latitude oceans start to freeze.

enter image description here
(Source: EPA)

Improve this answer
$\endgroup$
5
  • 1
    $\begingroup$ An estimate by a user of another site showed that if we shield Venus from Sun with a non-transparent cloud, it will cool down to 300K in about 6 years or slightly more. At the point of 31 C the CO2 will separate into liquid and gas phase, creating global ocean, further reducing pressure and greenhouse gas effect, and allowing making floating habitats (rather than flying ones). $\endgroup$
    – Anixx
    Nov 23, 2016 at 22:55
  • $\begingroup$ Six years, that is longer than what I would estimate. Venus has no oceans and the atmosphere doesn't really have the mass to contain a huge amount of energy. It is possible that my estimate is off by orders of magnitude. $\endgroup$
    – gerrit ♦
    Nov 23, 2016 at 23:05
  • $\begingroup$ But does either estimate include heat transferred from surface rocks to the atmosphere? $\endgroup$
    – jamesqf
    Nov 24, 2016 at 18:43
  • $\begingroup$ @jamesqf Mine doesn't. As I said, it's very basic. The land does not hold nearly as much land as the oceans, though. That's why continental climates have much wider temperature fluctuations than oceanic climates. $\endgroup$
    – gerrit ♦
    Nov 24, 2016 at 22:54
  • $\begingroup$ @gerrit: On Earth, presumably because (except for geothermal features) the land is roughly at equilibrium with the atmosphere. Which would be true of Venus, too, except the equilibrium temperature is "really, really hot", and it's probably hot for a long way down. So in the cooldown period, the land would be something like a planetwide geothermal feature. $\endgroup$
    – jamesqf
    Nov 25, 2016 at 5:33
1
$\begingroup$

The estimate by "gerrit" looks reasonable to me. I'd like to add an estimation of resulting surface temperature based on geothermal flux.

Current surface balance is achieved at about 240W/m2 given current solar irradiation minus albedo. If the Earth will be shielded from Sun, the only source to counter-balance the radiation loss would be the geothermal flux, which is about 0.1W/m2

How low the surface temperature will drop? The OLR as a function of temperature as sigma*T^4. The ratio between Sun/no-Sun in energy flux is therefore 240/0.1=2400, and this ratio is proportional to (T1/T2)^4. So the temperature should drop by a factor of (2400)^1/4, or about 7 times. Since the equilibrium emission temperature is 255K, the surface temperature will drop to 36-40K in a matter of days, as gerrit estimated, and will stay there for millions of years.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.