# 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

https://phzoe.com

## 22 thoughts on “Geothermal to the Moon!”

1. Jarle says:

Very interesting. Congrats on the new hypothesis!
Could we detect the inevitable 27 day variation in this number somehow?

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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. 😦

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1. Jarle says:

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)

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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.

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2. Jarle says:

If it is possible, the answer is statistics.

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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 …

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1. Jarle says:

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 🙂

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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.

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3. gbaikie says:

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?

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4. Jarle says:

Swift work 🙂 0.0023K difference. Moon only occupies 0.00077% of the sky.

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1. Jarle says:

Or, thinking about it. If the theory is that the earth can only radiate energy to the moon, then the moon is all there is, and as such comprises 100% of the sky

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1. Heat can only be transferred from matter to other matter, but … inverse square law applies. The Earth sends radiation to all other cooler matter in the universe, but … they are so far away that almost no HEAT is transferred. Orbiting satellites detect a lot of radiation. The moon gets a tiny amount. And further objects get virtually nothing – a negligible amount. 🙂

At L2 (~1.5 mil. km away), the Earth sends out as much as it receives from space. What is within that zone? Just the moon, and maybe passing asteroids.

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1. Jarle says:

Ok. Can you elaborate som more on the first sentence in the last paragraph?

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2. The Earth receives ~2.725 K from all the stars and galaxies. This is equivalent to 0.0000031 W/m^2. At ~1.5 million km away from Earth, Earth could send that same amount to an object sitting there.

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5. Jarle says:

Is it a coincidence that this is L2?

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1. Turns out I’m completely wrong. It’s more like 66 mil km away. I thought L2 was chosen for COBE, Planck, WMAP to measure CMB because of radiation quality, but apparently not. The radiation math works for ~66,000,000 km away, not 1,500,000 km. Ignore what I previously said. L2 is purely a gravity quality point. 🙂

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1. gbaikie says:

There is two L-2s
Earth/Sun L-2 and Earth/Moon L-2
Earth/Moon L-1 is the closest L-point. Earth/Moon L-3 and L-4/5 is next, then fourth closest is Earth/Moon L-2.
Only Earth/Sun L-1 and L-2, have a number of spacecraft/satellite at them, at the moment.
And Earth/Moon L-5 {or L-4} would good spot for space colonies. Or it’s Earth/Moon L-5 which meant by “L5 Society”:
https://en.wikipedia.org/wiki/L5_Society

It seems unlikely we will we get settlements in L-4/5 within the next 50 years, But if get Mars settlements {possible within 50 years] then after this we could get settlements in L-4/5.

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6. Guenter Zietkowiak says:

You have an own “model of the global energy flow”.On the chart at the upper rigth corner there is a formular. The left side is W/mm and the rigth side is W. You have to divide the rigth side by an area. And it will be clever to ad on the left side a source term such as energy by human production or energy by geothermie. Then the global energy balance is perfect with the exception there is no additional energy. And that is not true. With best wishes
Guenter Zietkowiak
retired engineer

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7. Peter D Grimshaw says:

Wow, so this is your string theory of heat transfer?
The string talked about in the other post.
The moon is the nearest thing the earth can love to send heat to in the vacuum of space?

I wonder if there are any other examples of this phenomenon ?
The same would be true for all objects in space, such as the sun and it’s sunlight?

What a well of creative thinking !

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1. The sun is said to have a mostly linear-by-radius thermal gradient, but if emission occurs by T^4 at the surface, such a thing would be impossible, right?

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1. Peter D Grimshaw says:

Thanks Zoe just getting to grips with all this, it’s like really good food and brain-exercise!
And the reason climate science (as it is generally taught) is so complicated, is because it doesn’t actually make sense. Hmm.

One reason I’ve started getting busy on this was because Facebook told me a post was “against community standards”. The post was a website from Peter L Ward PhD the volcanologist who set up the early warning system for earth rumbles. I was discussing Climate with a friend who worked on the prototype for the Tokamak Fusion reactor in the Uk. And Facebook wants to close down discussion of climate because we might misinterpret it.

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1. Facebook isn’t banning Flat Earthers. Tells you everything you need to know.

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