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Discussion > Numerical calculation of Planetary Black Body Temperature

Roger, the top 2cm of soil is a rather good insulator but not perfect. How else can one account for the changing soil and regolith temperature profiles when going from day to night. Heat is transported in from the surface down to a depth of 40 or 50cm during the day and it is slowly released at night. We see exactly the same thing on Earth except the variation is seasonal given our more rapid spin and the more extreme seasons on Earth

Jan 8, 2013 at 4:05 PM | Unregistered CommenterPaul Dennis

Paul, I take your point, but a quick skim of the V paper showed rapidly increasing conductance down to an isothermal (or boundary of zero energy transfer) at a few tens of cm. Therefore, during insolation, the surface layer of powder would show by far the geatest temperature drop. As Diviner was measuring the surface temperature, I am speculating that this will show little difference from the pefect SB temperature. As I said, not sure. But I think we now have enough data to do a calculation.

Jan 8, 2013 at 4:16 PM | Unregistered CommenterRoger Longstaff

The diviner hotside temps match my own calculated S-B temps. Note that there is next to no lag at peak noontime. That means there is not much heat flux. On the cold side we lose about 25K in fifteen earth days. That is not much, and bears no relation to what happens in Oxfordshire. It gets cold a lot faster than that, at least in the 288K area. And that is despite all that back-radiation. My opinion is that there is no way to map moon to earth. It's like they were different planets. :-)

Further, we ought to be looking at this as a heat problem. Take a unit mass of surface, work out the heat flux from all three actors (sun, S-B out, conduction). You are not going to do that without a dimension of time. Are we agreed that the static model is the coldest?

Jan 8, 2013 at 5:32 PM | Registered Commenterrhoda

Rhoda,

I don't think there is any need to do any further modelling. The locus of constant subsurface temperature is a good guide to the average grey body temperature since it defines the boundary across which there is no net transport of thermal energy. I think a calculation taking into account the variability of albedo (which is very large on the moon) will give an average temperature close to the measured temperatures.

As for the static planet. Well such a situation with heating on one side only will come out at the lowest average temperature as you surmise. It is also the situation that is furthest from equilibrium with a constant flux of solar energy as heat into the interior of the sphere. One could argue that if left long enough and there were no internal source of heat the sphere would eventually heat up to the S-B temperature and be of uniform temperature throughout.

Jan 8, 2013 at 5:44 PM | Unregistered CommenterPaul Dennis

Yin, as I remarked to Martin regarding his cube world example, should you not be averaging the radiated energy and then calculating the temperature from that, rather than averaging temperatures?

This is discussed at SoD

Jan 8, 2013 at 5:45 PM | Unregistered CommenterBitBucket

BitBucket, yes, for a rotating planet, but I was never modelling a rotating planet before, under the misapprehension that the IPCC model the N&Z criticized was using a static unrotating black body.

This means rewriting my code a little, and I haven't had time yet.

Jan 8, 2013 at 5:52 PM | Unregistered CommenterTheBigYinJames

BB, yes you should be averaging the incoming radiant energy and that is what I did when I did my calculation. In fact I think my every submission on this long thread has argued this when I've talked about heat storage etc. and the need to identify the constant temperature surface. For Earth the S-B temperature is 252K, for the moon it is 266K (as per IPCC and other estimates).

Note SoD does not talk about averaging the radiated energy. This is because the large distances mean there is little lateral heat flux and the bulk of the heating occurs immediately below the surface thus only local equilibrium is reached with a temperature gradient across the surface.

Now I really have spent too long on this!

Jan 8, 2013 at 6:04 PM | Unregistered CommenterPaul Dennis

Well.. I've rewritten the code to model a rotating planet, I'll post the code later for completeness, but basically I took a string of points on a line north pole to south pole, and for each point on that line, I iterated through 24 hours of minutes, half of them going through the usual sine wave of nothing to a peak at midday then back down to nothing, the other half no insolation. I added up all the insolation and averaged it out over those 24 hours, from which I got an average temperature for that point over 24 hours. Then I took the average of all those points. The results (if I've done it properly) doesn't look good for N&Z:

Average Grey Body Temperatures
------------------------------
Earth : 238.3 K
Moon : 252.2 K

The Moon is back up at it's 250K and the earth is only showing a GHE of around 50K.

Jan 8, 2013 at 6:32 PM | Unregistered CommenterTheBigYinJames

Welcome back TBYJ!

Jan 8, 2013 at 6:42 PM | Unregistered CommenterPaul Dennis

Paul Dennis, said, "Note SoD does not talk about averaging the radiated energy". Really? Here's a quote:

"The right way to calculate a planet’s average radiation is to calculate it for each and every location and average the results. The wrong way is to calculate the average temperature and then convert that to a radiation. In the case of the earth’s surface, it’s not such a noticeable problem.

In the case of the moon, because of the wide variation in temperature, the incorrect method produces a large error."

Are you sure that doesn't talk about averaging radiated energy?

Also, Yin, that says nothing about a rotating planet (although the text preceding it assumes this).

Jan 8, 2013 at 6:46 PM | Unregistered CommenterBitBucket

Not sure Roger will be pleased, however. He was championing my low static non-rotating values earlier :)

Jan 8, 2013 at 6:47 PM | Unregistered CommenterTheBigYinJames

I think any sensible model of the moon has to have it rotating, and I always intended to do this once I had the static model reflecting what other sources said. I was taking my lead from the N&Z paper, which appeared to be comparing a static non-rotating IPCC model with their own non-rotating model.

Now it looks like the non-rotating model will ALWAYS give you a much lower average temperature.

Is the explanation for N&Z thinking the IPCC model was non-rotating was because of a fluke that when you work out the very course incorrect value for average temp from average insolation on a flat disc model, then it gives a very similar magnitude to the case where you do a more sophisticated spherical integration with insolation values averages over a diurnal cycle? Is it mistaken identity on their part? Was there something in the IPCC description that made them think that's what they had done?

I don't know. Perhaps we should ask them. Once I've had my code checked, of course. Ned sounds like he would tear strips off me :/

Jan 8, 2013 at 6:55 PM | Unregistered CommenterTheBigYinJames

"I added up all the insolation and averaged it out over those 24 hours, from which I got an average temperature for that point over 24 hours." Hang on - surely that is back to the same error the IPCC made? What is surely needed is to calculate temperature in each grid and then average the temperatures? As Martin said on the other thread - to simulate thousands of thermometers equispaced across the whole surface of the planet.

I think if you try that you will get a completely different answer. Just calculate the SB temperature for each grid if in sunlight, and take it as 100K if not (to account for conduction on the darkside. I think this can be ignored to the first order on the hotside as the Diviner temperature at Zenith is the same (I think) as the theoretical SB temperature).

But I may have to do it myself, which I am dreading because my incompetence at programming is legendary.

Jan 8, 2013 at 7:10 PM | Unregistered CommenterRoger Longstaff

BB,

I think if you read through in detail what SoD says (it's a long time since I've read it) he talks about the incoming and outgoing radiation for each square metre and specifically excludes lateral heat transport. This means that he implicitly allows for the temperature to vary over the surface. It's not a big issue. You average the radiation over the surface and calculate a temperature you get 250K, you allow radiation to vary and average temperature you get 252K.

If you allow the radiation to be averaged over the surface you are then assuming a constant surface temperature everywhere. This of course is what would happen at true equilibrium.

Jan 8, 2013 at 7:20 PM | Unregistered CommenterPaul Dennis

I don't understand what all this talk is about non rotating disc models. The average solar radiation received by Earth, the moon and any other sphere is simply give by:

Pi*r^2*(1 - alpha)*S

From the point of view of the sun all it sees is a flat disc!

From the point of view of outgoing radiation that is from the whole surface of a sphere.

Surely this isn't where N&Z went wrong in simply not seeing that the average radiation received by a sphere is determined by the cross sectional area that intercepts the suns rays!

Jan 8, 2013 at 7:26 PM | Unregistered CommenterPaul Dennis

I don't think you can do that Roger, because you're back to having half of your thermometers permanently at 2K again, and this is not realistic for a rotating planet which never reaches that low thermal equilibrium point.

By working out the temps all around the planet you've effectively frozen time and stopped the rotation, and I proved it by doing what you said, working out the temps for all time zones and averaging the temps, and it comes out to within 2 degrees of my non-rotating model.

By arbitrarily setting the temperature of the dark side to 100K, you're effectively doing the same smearing - only calibrating it by a measurement instead of working it out from the heat capacity of moon regolith, (which I'd love to do)

Jan 8, 2013 at 7:33 PM | Unregistered CommenterTheBigYinJames

Paul said, "If you allow the radiation to be averaged over the surface you are then assuming a constant surface temperature everywhere. ". I don't want to be argumentative (really) but that is not how I understand it. The problem is that the temperature does vary and it is the extent of those variations that makes the difference between averaging temperatures (wrong way) and averaging radiation (correct way). For Earth, the variations in radiation are relatively small and the two methods give similar results. For the Moon, the variations are larger and the choice of method determines whether you get the correct answer or not.

When you use the radiation-averaging method, you get an overall average radiation level per sqm that is different from the individual radiation values from which it is computed (because the temperature varies). That level can then be converted into a temperature that represents the average temp for the body. But, as I just said, it isn't the temperature of every point on the body.

Note that it seems to me that the rate of rotation should make no difference at all; Martin's cube example makes it appear to matter only because he averaged the temperatures, not the radiation. But as I know nothing about radiation physics, my interpretation is not worth diddly-squat.

Jan 8, 2013 at 7:45 PM | Unregistered CommenterBitBucket

"By working out the temps all around the planet you've effectively frozen time and stopped the rotation".

No, I was thinking of doing this for every time step of the integration, for a rotating planet (but the more I think about it, a non-rotating planet would give the same result for a body with no atmosphere).

Are we not just right back to Holder's inequality again?

Jan 8, 2013 at 7:52 PM | Unregistered CommenterRoger Longstaff

Roger, yes that's what I did.

I took a point.... At hour 0, I work out a temperature of the point, at hour 1 I work out the temperature of the point, at hour 2.... Half of those hours the point is in darkness and will be 2K because this algorithm has no memory. It is only able to work out the temperature of a point by reference to incident insolation, not by the temperature it was a moment ago. This brings down the average temp a lot.

If you wanted to model more realistically, you would create an array of points which were persistent.
At every second you would work out what temperature drop or increase that point would experience with reference to incident insolation, current temperature, and heat capacity of the point. Your points would heat up and cool down during a diurnal cycle. Averaging all those points across all of their time states would give you an average temperature.

But.... your starting condition is hard... you need to start with some warm rock in the system. Or you can run 1000 diurnal cycles and hope the initial 'all cold points' conditions work themselves out of the system and it settles into a cycle.

It would be an interesting thing to model, I may give it a go.

Jan 8, 2013 at 8:11 PM | Unregistered CommenterTheBigYinJames

James,

I would do it with crude approximations for conduction - none in insolation (and an instantaneous, equilibrium SB derived temperature) and an automatic default to 100K in darkness. It would be far better to do it with the proper conduction and starting condidions, and that could be the next step, but given what we know of the surface of the regolith from Diviner measurements I think that we should get a correct result to within a few degrees.

Do you (or anybody else) agree with this, or have I got things completely muddled?

Jan 8, 2013 at 8:36 PM | Unregistered CommenterRoger Longstaff

BB,

thanks for redirecting me back to SoD you've given me something to think about. My immediate reaction is to say that you're right here but there is something that confuses me and I'm not sure why my method of averaging temperatures yields the correct result for BOTH the moon and Earth. I see what SoD is saying and agree with him and you. A subtlety that is easy to trip one up.

However, what confuses me is that the results are so similar in my model to the standard calculation. For Earth I get 252K using my method and for the moon I get 266K. Using the standard method as SoD describes and others have used I get 258K for Earth and 269K for the moon.

So the take home message is we need to average the incoming radiant energy then calculate the temperature. The temperatures either way are about 258K for Earth and 269K for the moon.

Kudos BB...thanks for the heads up and redirecting me back to SoD

Jan 8, 2013 at 8:41 PM | Unregistered CommenterPaul Dennis

Paul, BB,

"So the take home message is we need to average the incoming radiant energy then calculate the temperature." But surely what that gives you is the temperature of averaged insolation - isn't the whole point of this that Holder's inequality tells us that this is an incorrect way to calculate average temperature?

If I am wrong about this I will admit it (with a bit of grumbling) but now I really want to know!

Jan 8, 2013 at 9:02 PM | Unregistered CommenterRoger Longstaff

Roger,

I'll use the example at SoD. Consider three different surfaces, each 1 square metre at temperatures of 1K, 10K and 100K. They are each in equilibrium with the incident radiation and the average temperature is (1+10+100)/3 which is 37K. The average outgoing radiation for this temperature is 37^4 x 5.67x10^-8 or 0.106 W/m^2.

Now the actual average outgoing radiation is given by:

((1^4 x 5.67x10^-8) + (10^4 x 5.67x10^-8) + (100^4 x 5.67x10^-8))/3

which is 1.89 W/m^2

By averaging temperature one finds that the incoming and outgoing radiation is not balanced. One has to average the radiation over the sphere and then calculate a single temperature. If one does not do this, but averages temperature then the energy flux in and out does not balance.

I recommend that you visit the SoD link that BB put up earlier on this page of the thread.

Jan 8, 2013 at 9:29 PM | Unregistered CommenterPaul Dennis

I hesitate to comment, ariving late in the day and not having the time to go through what's been said.

I hope the following clarifies rather than adding confusion.

BB said "Note that it seems to me that the rate of rotation should make no difference at all; Martin's cube example makes it appear to matter only because he averaged the temperatures, not the radiation."
Jan 8, 2013 at 7:45 PM BitBucket

Case 1. Ordinary Arithmetic Average

If you are interested in the average temperature in the normal sense ie the integral of the temperature over the entire surface, divided by the area of the entire surface, then I believe that the cubic planet example shows that, in general, the result will depend on the speed of rotation. Note that this measure is what climate scientists use in talking about the average global temperature, as I understand things.

Case 2. Mean = "fourth root of the average of the fourth power"

Maybe you are interested in the "fourth root of the mean of the fourth power" of temperature.

I think that this is the temperature of a planet whose temperature is the same at all points (because it is made of thermally superconducting blackbody stuff) and which radiates the same total power as the planet you are studying. (It's vaguely analogous to the root mean square value of an electric current - the square root of the average of the square of the current.)

In that case you need to integrate the outgoing power over the surface to find the total outgoing power.

However (it's in equilibrium):

the total outgoing power = the total incoming power,

so you simply need to multiply the watts/m2 arriving from the Sun by the area of your planet's disk. This is obviously independent of the rotational speed of the planet.

Bottom line. It seems that whether or not rotation speed makes a difference depends on the kind of "average temperature" you are calculating:

- Normal arithmetic average does depend on rotation speed.

- Equivalent black body temperature does not.

I hope this makes sense. Please point out errors/misconceptions - I was making it up on the fly and it's late here. I'll be interested to see if it makes sense tomorrow morning.

Jan 8, 2013 at 9:33 PM | Registered CommenterMartin A

Paul, I think the problem comes because N&Z used a parameter in their thesis that required a true mean temperature for an airless planet, they did this by first calculating the temperature at every point by taking the fourth root of the absorbed radiation at that point and averaging the resulting temperature field across the planet surface.

It is to most people a meaningless number, but they then develop their theory from it. Whether or not the rest of their theory makes sense or not is not the question here, which is only the calculation of the average surface temperature.

Now, look at the NASA Diviner data (of 27 March 2012). It shows an avarage equatorial tempwerature of 213K. Now, the equatorial belt must be the hottest temperature field on the moon, so the average temperature of the whole planet must be less than this (from what I gather, if you use 2K for the unlit surface you get about 154K, but if you include conduction effects you get 197K).

I cannot think of any other way of explaining this. Read their paper again - what we are discussing is Holder's inequality!

Jan 8, 2013 at 9:52 PM | Unregistered CommenterRoger Longstaff