[witherllooll (OP)]

I want to know the gamma and cosmic radiation flux/watts per square meter on the moon's surface. Specifically in the south pole of the moon, but data for any location is appreciated. For example, on the Earth's surface, the solar flux is approximately 1000 watts/square meter. I want to find this value, but for gamma rays and cosmic rays on the moon. So far, I've found that in the recent Chang'e 4 mission (advances.sciencemag.org/content/6/39/eaaz1334), there is a dose rate of 13.29 micrograys per hour as converted to water and to the lunar surface. However, I can't find a way to convert this value to watts/meter^2. The main reason why is because I am working on a project where I am hypothesizing what would happen if a solar power tower was adapted to work with gamma/cosmic rays and was placed on the moon's surface. I need to know the watts/meter^2 to calculate how much gamma/cosmic ray energy would be reflected to the central tank of water. All help is appreciated, thanks!

[Kryptid]

According to this source: https://www.int.washington.edu/PHYS554/winter_2004/chapter8_04.pdf, the energy density of cosmic rays is on the order of 1 electron-volt per cubic centimeter (1.60218 x 10^-19 joules per cubic centimeter). For simplicity's sake, I'll assume that cosmic rays travel at the speed of light (the more energetic ones do travel very close to it). So I can model a series of cubic centimeter-sized cubes traveling through a square meter of space at the speed of light. Light travels at 2.998 x 10¹⁰ centimeters per second.

A square meter contains 10,000 square centimeters, so 10,000 x 2.998 x 10¹⁰ = 2.998 x 10¹⁴ of these cubes travel through the square meter every second. Multiply that by the energy of 1.60218 x 10^-19 joules per cube and you get 4.8 x 10^-5 watts per square meter. That's very, very small.

[evan_au]

For energy collection on the Moon, it is better to use solar cells or an infra-red thermal collector while the Sun is shining.
- Cosmic rays and gamma rays have too little intensity

For astronomical purposes, it is desirable to reflect these particles...

how much gamma/cosmic ray energy would be reflected to the central tank of water

Gamma rays and cosmic rays are very energetic, and tend to pass through anything you would use as a mirror.

Even X-Rays are very difficult to reflect - astronomical X-Ray telescopes work by the X-rays striking the mirror at a very small angle (grazing angle)
See: https://en.wikipedia.org/wiki/X-ray_telescope#Focusing_mirrors

For gamma-ray telescopes, they typically use an absorbing plate (eg lead), with a pseudo-random holes cut into it.
- By correlating the intensity of observed gamma rays with the pattern of holes, they can work out the source of the gamma rays.
- The Fermi Gamma Ray telescope used detectors similar to those used in particle accelerators.
See: https://en.wikipedia.org/wiki/Fermi_Gamma-ray_Space_Telescope#Large_Area_Telescope

As charged particles, cosmic rays will be bent by the magnetic field of the Sun and the magnetic field of the galaxy. So it is thought to be very difficult to trace a cosmic ray back to its source (at present we have only sampled the magnetic field on the edges of the Sun's influence).

[witherllooll (OP)]

For energy collection on the Moon, it is better to use solar cells or an infra-red thermal collector while the Sun is shining.
- Cosmic rays and gamma rays have too little intensity

While this is true, my team's main reason for using cosmic and gamma rays is because we've found processes and manufacturing methods specific to our idea that can really decrease the cost of an gamma/cosmic ray energy capture system.

For astronomical purposes, it is desirable to reflect these particles...

how much gamma/cosmic ray energy would be reflected to the central tank of water

Gamma rays and cosmic rays are very energetic, and tend to pass through anything you would use as a mirror.

Even X-Rays are very difficult to reflect - astronomical X-Ray telescopes work by the X-rays striking the mirror at a very small angle (grazing angle)
See: en.wikipedia.org/wiki/X-ray_telescope#Focusing_mirrors

We're planning to use smoothened lunar surface material/lunar regolith because, according to our research, it is able to reflect cosmic and gamma rays as shown in the diagram, but with some loss of energy. We're not sure exactly how much energy the inelastic collision takes away, but as per this journal agupubs.onlinelibrary.wiley.com/doi/10.1029/2005JE002656, we're pretty sure that it is not a very significant amount.
ps. Thanks for your response!


Screenshot 2021-01-05 101843.png (77.24 kB . 708x406 - viewed 448 times)

[Bored chemist]

it is able to reflect cosmic and gamma rays as shown in the diagram

The diagram does not show reflection of gamma rays.

[Kryptid]

Since cosmic rays contain fast-moving, electrically-charged particles, they should be affected by (and therefore affect) magnets and thus you might be able to use those as a means to tap some of that energy. But, as I've pointed out, the extreme low power level makes that a dubious undertaking at best.

[evan_au]

We're planning to use smoothened lunar surface material/lunar regolith

When you say "smoothened", it needs to be smooth within a wavelength in order to reflect electromagnetic waves effectively.
- Rolling out Lunar Regolith with a heavy weight will still leave dust grains that are large compared to the wavelength of visible light, let alone X-Rays or gamma rays. So this does not meet the criterion of "smooth".

the central tank of water.

Humans are effectively a tank of water. Humans have some impurities in their water (like DNA) which are adversely affected by cosmic rays and gamma rays.
- So putting a meter or two of lunar regolith over the top of a human habitation on the Moon is a wise precaution, as it will reduce the level of ionising radiation to a safer level.
- It doesn't need to be really smooth, just as long as it is a meter or two thick
- This layer of dirt will also help moderate the extreme temperature swings on the Lunar surface, between the 2-week day and the 2-week night

[alancalverd]

Current estimates of human tissue dose are about 60 microsievert/hour at the moon's surface.

Assuming most of this is beta and gamma radiation, that is pretty close to 60 microgray/hour of prospective absorbed dose, where 1 gray = 1 joule per kilogram.

Cosmic gamma radiation is very energetic so you will need something like 5 cm of lead to absorb 90% of the incoming flux. So a kilogram will occupy an area of around 20 cm² and will deliver about one sixtieth of a microwatt, assuming 100% conversion efficiency, or a bit less than 10 microwatt for a 55 tonne (1 square meter) collector.

I think you would do better to collect the visible and infrared light from the sun, which is several times the intensity at the earth's surface (and more reliable due to the absence of clouds).