Geothermal to the Moon!

The Earth and the Moon have been neighbors for a very long time. We should expect long term steady state heat transfer from Earth’s internal energy (without Sun) to the Moon to be equal to Earth’s geothermal heat flux. To my knowledge, no one has ever made this claim. Sound outrageous? Let’s see…

According to [Davies 2010], the geothermal flux is 46.7 TW:

We conclude by discussing our preferred estimate of 47 TW, (rounded from 46.7 TW given that our error estimate is ± 2 TW)

— [Davies 2010]

When we divide this figure by the surface area of the Earth, we get:

(46700000000000 ± 2000000000000)/510065728777855 = 0.09155683 ± 0.0039 W/m²

This is not an emergent radiative flux, but a conductive heat flux.

I will now argue that this number is not accidental, but is based on heat transfer from the Earth to the Moon.

In my article Measuring Geothermal …, I extracted 335.643 W/m² as an emergent flux equivalent from geothermal.

In my article Deducing Geothermal, I deduced 332.46 W/m² as an emergent flux equivalent from geothermal.

Let’s average those two values to get: 334.05 W/m²

Now we apply the inverse square law for distant radiation, using data from NASA:

334.05 W/m² × (6371 km/ 378000 km)² = 0.09489495 W/m²

This value falls within the error range of measured geothermal flux. Do you think this is a coincidence? I think not. It makes perfect sense.

Enjoy 🙂 -Zoe

10 thoughts on “Geothermal to the Moon!

    1. Thank you very much 🙂
      Ha, good one! That geothermal heat flux number was determined by data collected over generations. It’s the farthest thing from real-time imaginable. It’s a long term average. I don’t even know the average depth used or temperature at that depth, but I would really like to. 😦


      1. What I mean is, if earth does not lose heat to all of space, but ony to the concentrated little yellow dot on the sky, then maybe one could expect to see a 27 day variation in the satellite temperature data (UAH/RSS). I suppose nights with a full moon should be less cool than nights with a new moon. (No heat would be lost to the fully litt moon having a temperature of 127C)

        Liked by 1 person

        1. Oh, I agree with you there completely. There should be 2 main possibilities:
          1) Colder moon side facing hotter earth side
          2) Hotter moon side facing colder earth side

          I’m leaving out the equal temperature case as it’s rare.

          In main though, as the Earth is bigger, the NET heat flow is from Earth to Moon.

          The problem at looking at UAH data is how to distinguish that 0.09155 W/m^2 from natural variability? It’s not like we know ahead of time what the temperature should’ve been on those nights.

          Maybe it is possible? I don’t see how yet.


    1. Hmm. I suppose if you isolate the two cases I mentioned and use a long term average over time … you should see the small difference. Is this close enough to what you were thinking?

      Note: There’s a difference in the deep geo heat flux and the shallow geo heat flux. The shallow heat flux varies quite wildly. More on this soon …


      1. You mean comparing nighttime data facing hot moon, to nighttime data facing cold moon, and likewise comparing daytime data facing hot moon to daytime data facing cold moon? Easy for you to make the algorithm for processing the data 🙂


        1. I need minute by minute temperature data going back a long time. I don’t trust those night-time averages compiled by temperature stations that really only sample at 12 and 6 AM. Same goes for daytime.

          As for UAH, they have global compiled daily data from 1978 to 2017. I made a program to measure global difference between a full moon and a new moon.

          The result is: 0.0023 K difference.

          Not sure this needs its own published article. It’s not exactly what either of us wanted to know.


  1. If you detonated a US 1.2 megaton bomb on surface of the Moon- you wouldn’t see it from Earth.
    If you detonated a US 1.2 megaton bomb 1 km above lunar surface- you have a better chance to see it.
    The moon has a surface area of 38 million square kilometers the half of moon facing earth: 19 million square km.
    If had 100 1.2 megaton bombs exploding in middle of 1 square km in 10 km square at 1 km above surface lunar
    surface, I guess one can see that from Earth surface. But energy from the blast of these 100 1.2 megaton bomb
    exploding 1 km above lunar surface in the 100 km square km reaching earth’s surface would be fairly insignificant.

    Now, if exploded 19 million 1 metaton 1 km above the lunar surface in square km of the moon facing Earth.
    Seeing it would not be vaguely a problem. The Moon would be quite bright for a second or so. And a significant amount energy would reach the Earth’s surface.
    How much?

    Nuclear bomb in space are not like nuclear bombs in Earth atmosphere.
    A large amount energy of nuclear bomb is in neutrons which heat Earth’s atmosphere to very temperature.- these neutrons would heat up the lunar surface as the Moon has little atmosphere.
    Maybe asking question which give no answer, and one could design bombs in different ways-
    One can focus the energy in various ways. Or didn’t focus the blast, very little portion of total energy would hit the lunar surface. If 1/2 km above lunar surface, a higher amount hit the Moon- less area is heated
    If buried the bomb one get 100%.
    Burying bomb say 4 meter under lunar surface would get near 100% of energy absorbed by Moon, but not sure you even see it from Earth {“it” being the 19 million 1 megaton explosions].
    A way which work best it seems to me, is if covered Moon with 19 million domes of air, then exploded.
    Or cover moon with steel greenhouse and a try to vaporized as much of it {in terms of area} with the nukes.
    So design bombs and steel greenhouse to be vaporized by the bombs.

    Anyhow question is how much energy gets to Earth’s surface?


Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

%d bloggers like this: