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There's a lot of information about figuring in the average albedo of a planet when calculating the base temperature for a planet. What if you want to get more granular, down to each (let's say) 10 degree latitude band?

We know different things have a different albedo. Estimates vary, but for the sake of this discussion, let's say our various albedo break down like this:

  • Land: .25
  • Water: .06
  • Land Ice: .9
  • Water Ice: .65

So if we have a 10-degree band of latitude consisting of...

  • .25 land
  • .15 land ice
  • .10 water ice
  • .50 water

...when plugging albedo into our temperature equation, what number does one use?

Do we take an average of the different albedo ratings, figuring in the percentage of surface for each albedo?

Or does the strongest albedo cancel out the rest? Or the weakest?

(Caveats to keep this question from being a super-complex climate modeling rabbit-hole: don't worry about all the other variables that you and I both know go into doing a good-enough-for-worldbuilding climate model. Assume 0 elevation; assume we're looking at the average temp across the year, don't worry about coasts vs interiors, don't worry about wind or ocean currents, etc. etc.)

Links to sources appreciated -- thanks very much!

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    $\begingroup$ You that the weighted average. Albedos combine linearly. (That's all I know. I have no idea why you would think that the albedo of a band of latitiude is relevant for the average temperature, or why you would think that the average tempeature is in any way useful. I live in Bucharest, where the average tempeature is about 11 °C; but this irrelevant when in July we regularly go above 40 °C, and in cold January nights we go below –15°C.) $\endgroup$ – AlexP Oct 7 at 0:48
  • $\begingroup$ Thanks, @AlexP. Getting a lot of input in various Worldbuilding groups that concur with your answer. As for why I'm doing this -- this is just one part of a larger process I'm doing to determine rough temperatures across the course of the year. The wide variation of temperature is figured in... elsewhere. Just needed this piece of the puzzle. $\endgroup$ – Matt Selznick Oct 7 at 2:29
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The weighted average. In this case, $0.25*0.25 + 0.06*0.15 + 0.9*0.1 + 0.65*0.5 = 0.49$.

To see why, we can look at what albedo is: the percentage of light coming in that gets bounced back out again. Wikipedia defines it as "diffuse reflection of solar radiation out of the total solar radiation", but at the end of the day, it's really just knowing how much sunlight is absorbed vs reflected. Let's imagine we're calculating the albedo of a chessboard and say the black squares have an albedo of zero, and the white squares have an albedo of one. For each photon that comes in, it has a 50% chance of hitting a black square and being absorbed, and a 50% chance of hitting a white square and being reflected. So, 50% of the incoming light is reflected and your albedo is 50%. Now, let's say we paint half of the white squares black, so on our chessboard we now have 75% black squares and 25% white squares. Because a photon now has a 25% chance of hitting a white square and being reflected, the albedo is 25%.

This same logic holds up when we look at your latitude band. A photon has a 25% chance of hitting somewhere where it has a 25% chance of being reflected (land), a 15% chance of hitting somewhere where it has a 6% chance of being reflected (water), and so on. The math is just combining all these probabilities together.

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  • $\begingroup$ Thanks, @Kofthefens! Weighted average it is! $\endgroup$ – Matt Selznick Oct 7 at 20:20

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