# Average Moon Day and Night Temperatures

NASA’s Moon Fact Sheet doesn’t give the diurnal temperature range for the entire moon, just the equator:

``Diurnal temperature range (equator): 95 K to 390 K``

Strange. They have collected the data. Why didn’t they do the calculations? So I could do it?

I went through every 15 degree increment longitude data available here.

Day is the center hot spot +/- 90 degrees. Night is everything outside of that.

Here’s my result:

``````Lon    Day    Night
000: 303.914 099.629
015: 304.115 099.809
030: 304.250 099.569
045: 304.342 099.402
060: 303.527 099.818
075: 303.196 099.688
090: 302.704 099.543
105: 302.347 099.650
120: 301.705 099.676
135: 301.474 099.267
150: 301.550 099.314
165: 300.939 099.281
180: 300.458 099.378
195: 301.062 099.347
210: 301.293 099.516
225: 302.147 099.307
240: 303.114 099.249
255: 302.813 099.433
270: 302.921 099.221
285: 303.267 099.054
300: 303.318 099.161
315: 303.682 099.245
330: 303.588 099.397
345: 304.116 099.122

Avg: 302.743 099.420

Whole Moon:  201.082``````

As you can see the time-averaged whole moon goes from a nightly low of 99.42 K to a daily high of 302.743 K, with a 24 moon-hour average of 201.082 K.

I assume that both day and night is a 12 moon-hour period. This may not philosophically be so, but my whole purpose was to figure out the difference between light and dark equal-area hemispheres, not compare unequal light to dark areas.

Enjoy 🙂 -Zoe

```# Zoe Phin, 2021/02/15
# File: moont.sh

for l in {000..345..15}; do
echo http://luna1.diviner.ucla.edu/~jpierre/diviner/level4_raster_data/diviner_tbol_snapshot_\${l}E.xyz
done | wget -ci -
}

avg() {
awk '{ f=1/581.9; e=2*f-f^2; r=atan2(0,-1)/180
T[\$2]+=\$3; N[\$2]+=1; A[\$2]+=r/1438.355*(1-e)*cos(r*\$2)/(1-e*sin(r*\$2)^2)^2
} END {for (L in T) printf "%+06.2f %7.3f %.15f\n", L, T[L]/N[L], A[L]}' | awk '{
T+=\$2*\$3 } END { printf "%07.3f ", T }'
}

calc() {
for l in {000..345..15}; do
echo -n "\$l: "
cat *\${l}E.xyz | awk -vl=\$l '
(l=="000" && (\$1>-90 && \$1<90 ))			{ print }
(l=="015" && (\$1>-75 && \$1<105))			{ print }
(l=="030" && (\$1>-60 && \$1<120))			{ print }
(l=="045" && (\$1>-45 && \$1<135))			{ print }
(l=="060" && (\$1>-30 && \$1<150))			{ print }
(l=="075" && (\$1>-15 && \$1<165))			{ print }
(l=="090" && (\$1>0   && \$1<180))  			{ print }
(l=="105" && (\$1>15 && \$1<180 || \$1<-165))	{ print }
(l=="120" && (\$1>30 && \$1<180 || \$1<-150))	{ print }
(l=="135" && (\$1>45 && \$1<180 || \$1<-135))	{ print }
(l=="150" && (\$1>60 && \$1<180 || \$1<-120))	{ print }
(l=="165" && (\$1>75 && \$1<180 || \$1<-105))	{ print }
(l=="180" && (\$1>90 && \$1<180 || \$1<-90 ))	{ print }
(l=="195" && (\$1>105 || \$1<-75))			{ print }
(l=="210" && (\$1>120 || \$1<-60))			{ print }
(l=="225" && (\$1>135 || \$1<-45))			{ print }
(l=="240" && (\$1>150 || \$1<-30))			{ print }
(l=="255" && (\$1>165 || \$1<-15))			{ print }
(l=="270" && (\$1<0 ))						{ print }
(l=="285" && (\$1<15 && \$1>-165))			{ print }
(l=="300" && (\$1<30 && \$1>-150))			{ print }
(l=="315" && (\$1<45 && \$1>-135))			{ print }
(l=="330" && (\$1<60 && \$1>-120))			{ print }
(l=="345" && (\$1<75 && \$1>-105))			{ print }
' | avg
cat *\${l}E.xyz | awk -vl=\$l '
(l=="000" && !(\$1>-90 && \$1<90 ))			{ print }
(l=="015" && !(\$1>-75 && \$1<105))			{ print }
(l=="030" && !(\$1>-60 && \$1<120))			{ print }
(l=="045" && !(\$1>-45 && \$1<135))			{ print }
(l=="060" && !(\$1>-30 && \$1<150))			{ print }
(l=="075" && !(\$1>-15 && \$1<165))			{ print }
(l=="090" && !(\$1>0   && \$1<180))  			{ print }
(l=="105" && !(\$1>15 && \$1<180 || \$1<-165))	{ print }
(l=="120" && !(\$1>30 && \$1<180 || \$1<-150))	{ print }
(l=="135" && !(\$1>45 && \$1<180 || \$1<-135))	{ print }
(l=="150" && !(\$1>60 && \$1<180 || \$1<-120))	{ print }
(l=="165" && !(\$1>75 && \$1<180 || \$1<-105))	{ print }
(l=="180" && !(\$1>90 && \$1<180 || \$1<-90 ))	{ print }
(l=="195" && !(\$1>105 || \$1<-75))			{ print }
(l=="210" && !(\$1>120 || \$1<-60))			{ print }
(l=="225" && !(\$1>135 || \$1<-45))			{ print }
(l=="240" && !(\$1>150 || \$1<-30))			{ print }
(l=="255" && !(\$1>165 || \$1<-15))			{ print }
(l=="270" && !(\$1<0 ))						{ print }
(l=="285" && !(\$1<15 && \$1>-165))			{ print }
(l=="300" && !(\$1<30 && \$1>-150))			{ print }
(l=="315" && !(\$1<45 && \$1>-135))			{ print }
(l=="330" && !(\$1<60 && \$1>-120))			{ print }
(l=="345" && !(\$1<75 && \$1>-105))			{ print }
' | avg
echo
done | awk '
BEGIN { print "Lon    Day    Night" }
{ D+=\$2; N+=\$3; print }
END   { printf "\nAvg: %07.3f %07.3f\n\nWhole Moon:  %07.3f", D/24, N/24, D/48+N/48}'
}

### Blog Extra ###

require() { sudo apt-get install -y imagemagick; }

dlimgs() {
for l in {000..345..15}; do
wget -O L\$l.png http://luna1.diviner.ucla.edu/~jpierre/diviner/level4_raster_data/diviner_tbol_snapshot_\${l}E.png
done
}

scale() { n=0
for l in {000..345..15}; do
convert -quality 30% -scale 12% L\$l.png SL\$(printf "%03d" \$n).jpg
let n++
done
}

animate() { convert -delay 30 -loop 0 SL*.jpg animoon.gif; }
```

https://phzoe.com

## 11 thoughts on “Average Moon Day and Night Temperatures”

1. Peter D Grimshaw says:

So …….
Completely squews the notion of Earth’s av temp being caused by sunlight only?
Or at least throws a big cat amongst the pigeons?

Or does the moon’s A/V ratio affect the average temp?
Really interesting to fathom the consequences of this.

Are all planets over a certain size wannabe suns with internal furnaces, like Jupiter?

Liked by 1 person

1. Zoe Phin says:

I left the analysis for the future. Rotational velocity and specific heat play a role.

I’m not aware of any scientists that dispute that our 4 “gas giants” are mostly internally powered.

Like

2. gbaikie says:

The planet Mercury is cool place.
Very intense sunlight, but at given point in time, most of surface is very cold.
And region near poles is always cold.
A pole is 90 degree and if 10 degree away at 80 latitude, it’s cold at a level surface.
On Mercury a degree of latitude is 42.58 km. So, 10 degree is 425.8 km. And the area with radius of 425.8 km is 569,598 square km.
And area of California is 423,970 km. So got two regions on Mercury with area as big as California which is always cold.
Our Moon is similar, though it’s a lot smaller, and Moon has shorter day than Mercury.
Mercury day is 58 earth days plus 15h 30 minutes, our Moon is 29 earth days plus 12 hours and 44 minutes.
Here on Earth, Earth’s sun travels at about 1000 mph, Mercury sun travels at 4 or 5 mph- or you can walk faster than Mercury’s sun.
It’s cold because sun is at low angle, and really, really cold if in crater which doesn’t any sunlight reaching it.
But make something so it’s directly facing the sunlight it get lots heat/energy from the Sun. Or if have steep mountain side facing sun, it also gets hot.

But in terms having cold surface and access to tons of sunlight, all of Mercury surface has this if travel at 4 mph or even slower speeds closer to poles. So either near dawn or sunset. and that would much larger area than two Californians.
But other than region near poles, region above 45 degree latitude would seem better than lower than 45 degrees.

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2. Netko says:

As there are no atmosphere the temperatures on the moon should be the same every single moon day on given time. I used diviner data to calculate the temperatures for each Apollo mission for fun as exact hours of their EVA is well known. And yes, I enjoy reading this blog. Great work.

Liked by 1 person

3. Robert B says:

I did a quick calculation when someone used the argument that the Moon being 33°C cooler than the Earth means that there is a greenhouse effect.

I used two spheres where everything was the same but, for whatever reason, the temperature spread was similar to the spread of the Moon’s equator for one and Earth’s sea surface for the other. Keeping it simple, there were 8 parts to each surface of constant temperature. One was from 90 to 360 K, close to what NASA give for the equator. The other was from 273 to 305 K. The numbers were tweaked to give the same sum of T^4. Surprisingly, this just required changing the 273 to 275. The means came out close to the mean SST of Earth and the mean at the equator of the moon with about 33 K difference. A simple exercise to show that you would expect a large difference purely because the Moon is more like a sphere of individual black bodies while the Earth’s seas store and spread the heat so its more like a true black body.

Later, I did a much more complex calculation (since lost) although simplified, of what the difference in mean would be for a sphere of infinite seperate black bodies compared to a black body of constant temperature. I got the latter to be 2.7 times warmer for an equal sum of T^4.

For some reason, I haven’t noticed anyone do a similar calculation.

Just throwing out the challenge.

Like

1. gbaikie says:

” A simple exercise to show that you would expect a large difference purely because the Moon is more like a sphere of individual black bodies while the Earth’s seas store and spread the heat so its more like a true black body.”

The “greenhouse effect theory” cargo cult, does not even mention the ocean.
70% of Earth surface is ocean, and earth’s ocean controls global temperature.
But if Earth had as long of day as the Moon, it be colder.
The 24 hours day and ocean makes the tropical zone {the 23.5 degree latitude north and south is area which is 40% of the planet. With longer day of Moon, Earth would have a tropical “spot” rather than a tropical zone. And be area of about 78 million square km out Earth total area of about 510 million square km. Or be about 15% rather 40% of the Earth’s total surface area. But the remaining 85% would be warmer. Though need a lot of water {like Earth has- or all ends up as solid ice at the polar region].

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1. Robert B says:

They do mention the ocean, as in how many zetajoules goes into the sink. Never that so much heat actually equates to 1/100th of a degree warming. Never how changes in circulation of ocean heat can effect surface temperatures by a degree, the actual average as well as the reconstructed average by changing the distribution of temperatures around the globe.

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4. Bernd says:

Kramm, Dlugi and Mölders calculated the average temperature of the moon with two different methods. Their result (198 K) is very close to the Diviner result. They applied the same method to Earth. In case of an albedo of 0.30 the average surface temperature of Earth without atmosphere is 221 K, 67 K below the current average temperature of 288 K. http://www.scirp.org/Journal/PaperDownload.aspx?paperID=78836

The often mentioned value of 255 K for an Earth without atmosphere is wrongly calculated by assuming a globally uniform temperature.

Like

1. Zoe Phin says:

Kramm et al show absolutely no consideration for geo/luna-thermal, so their conclusion is suspect. The sun heats the soil up to a certain depth. The temperature at that depth matters very much. At lunar equator the solar undisturbed soil temperature is ~250 K. At earth equator it’s like 295 K.

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5. Bernd says:

Sorry I made a mistake. 221 K were calculated for Earth with an albedo of 0.10. The correct value for Earth with an albedo of 0.30 is 218.5 K, almost 70 K below the current average temperature.

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6. AC Osborn says:

The current temperature called the “Surface Temperature” is not the Average Temperature of the Earth.
The average temperature is somewhere between the Surface Temperature and the Core Temperature.
Why is this never taken in to consideration?
I am sure that Black Body calculations were designed for bodies in Thermal Equilibrium.

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