If we installed an opaque screen in space in the vicinity of Earth blocking all sun rays, how long will it take to cool the planet back to pre-industrial times?

Earth sunblock

I know global warming is not just about temp, but how heat is unable to dissipate because of greenhouse gases. Yet, I suppose if you turn the incoming heat off, this must have an impact.

So, rephrasing: after suppressing all sunlight, how ground temperature on Earth would vary?

To state the obvious: of course such a screen would not be easy to make nor deploy in space, but this is not the point.

  • $\begingroup$ Re, "after suppressing all sunlight..." Have you tried looking up the surface temperatures of some of the outer planets? That might be a good starting point. $\endgroup$ Aug 19, 2020 at 20:37
  • $\begingroup$ Re, "heat is unable to dissipate because of greenhouse gases." The greenhouse effect is more subtle than that. It does not prevent the Earth from radiating energy to space, but it does slow the process down by just a little bit. The problem is, just that "little" bit is enough to raise the global average temperature by a couple of degrees C, and those couple of degrees are enough to de-stabilize all of the systems that our lives depend on. $\endgroup$ Aug 19, 2020 at 20:46
  • $\begingroup$ You probably know from experience how fast the earth cools at night. It will most likely be similar to that but on the entire planet. $\endgroup$ Aug 19, 2020 at 21:24
  • $\begingroup$ It's not necessary to build this large screen. Simply get a few large volcanoes to explode and the world would cool soon enough. When Mount Pinatubo exploded a few decades ago, the earth cooled by 1 degree F for a few years. And scientists are seriously looking at simply using high flying jets to spray chemicals into the upper atmosphere to do the same thing. But such geoengineering it replete with problems that we haven't even though about. $\endgroup$ Aug 19, 2020 at 21:33
  • 1
    $\begingroup$ @AccidentalTaylorExpansion: no, because sun heat is stored underground. This is why seasons are shifted relatively to sunlight exposure (otherwise the coldest/hottest day of the year would be respectively dec 21 and june 21 in northern hemisphere, and that is not the case). So the true answer is more elaborate than your assumption. $\endgroup$
    – Winston
    Aug 20, 2020 at 12:30

1 Answer 1


The average heat input from the sun according to this reference is $164 J/(s.m^2) = 39 cal/(s.m^2) = 3400 kcal/(day.m^2)$.

Water has much bigger thermal capacity than rocks, and most of the planet surface is water. The average depth of the oceans are about 3600 m.

The mass of the column of water corresponding to $1 m^2$ of surface is: $3,6 *10^6 kg$.

$$\Delta T = \frac{3400}{3,6 *10^6} = 9,4*10^{-4} K$$

According to that calculations, each 3 years the oceans would lose 1K. But the temperature of the surface would probably fall faster, mainly when ice started to cover most of it.

  • $\begingroup$ Thank you for the effort. I don't understand your calculation though. How 3 thousands divided by 3 millions would return around 100 thousand? Are you sure you did not forget a minus sign in the exponent? And how do you go from 100 thousands K to a decrease of 1 K every 3 years? Please elaborate (what is the calculation about exactly?) $\endgroup$
    – Winston
    Aug 20, 2020 at 12:27
  • $\begingroup$ Oops. I forgot the minus sign. Now you can see it let about 1000 days to 1 degree. $\endgroup$ Aug 20, 2020 at 14:37
  • $\begingroup$ Yes, thanks for fixing that. So basically your calculation tells how water would cool down if the sun was the only provider of heat. But radioactive material in the crust, mantle, and a hot core, provide internal heat to the planet. So the planet's surface will not reach 0°K even after 1000 years without sunlight. And because of high pressure at the bottom of oceans and lower density of ice, water cannot freeze there, which probably add another factor. $\endgroup$
    – Winston
    Aug 20, 2020 at 14:58
  • $\begingroup$ I think one important info is missing. Assuming that the earth is almost in equilibrium, i.e. the mean temperature of the earth is not increasing or decreasing vastly: This means that the energy coming to earth from the sun is about the same, as the energy leaving the earth. That is why Claudio Saspinski takes this rate of 3400kcal/(day.m2) to be the rate of energy that the earth will lose. However, cooling down 3.6km thick ocean is different from mean temp on the ground... $\endgroup$
    – Martin
    Feb 5 at 14:42

Your Answer

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

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