Mass attenuation problem, protecting the lunar base against radiation?

In summary, the astronauts covered the lunar base with a 4m thick layer of regolith to protect against radiation, but it was still 50% higher than on Earth. To achieve the same level of radiation as on Earth, they need to add approximately 5.333m of regolith. However, there is not enough information given to accurately solve for the necessary thickness of the regolith layer.
  • #1
carnivalcougar
40
0

Homework Statement


To protect the lunar base against space and sun radiation, astronauts covered it with 4m thick layer of regolith (moon soil). However, they found that they level of radiation was still 50% higher than on the Earth's surface, which they want to match. How many more meters of regolith do they need to add for proper protection?


Homework Equations



I = I˳e-µx


The Attempt at a Solution



I know that they want to reduce the radiation by 1/3, to 2/3 of the current value. However, no half layer value or absorption coefficient is given for regolith.
 
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  • #2
carnivalcougar said:

Homework Statement


To protect the lunar base against space and sun radiation, astronauts covered it with 4m thick layer of regolith (moon soil). However, they found that they level of radiation was still 50% higher than on the Earth's surface, which they want to match. How many more meters of regolith do they need to add for proper protection?


Homework Equations



I = I˳e-µx


The Attempt at a Solution



I know that they want to reduce the radiation by 1/3, to 2/3 of the current value. However, no half layer value or absorption coefficient is given for regolith.

If 4m of soil attenuates to 1.5 times the intensity at the earth, how many meters of soil is needed to attenuate to 1.0 times the intensity at the earth. The attenuation rate is constant, so...
 
  • #3
Would this just be 5.333m? if 4m attenuates to 1.5x Earth radiation, then 1/3 more regolith would attenuate to 1x Earth radiation. 4x(1/3) = 1.333 and 4+1.333 = 5.333
 
  • #4
carnivalcougar;47Ie?06819 said:
Would this just be 5.333m? if 4m attenuates to 1.5x Earth radiation, then 1/3 more regolith would attenuate to 1x Earth radiation. 4x(1/3) = 1.333 and 4+1.333 = 5.333

I'm not following your math, but it doesn't look exponential.

Draw a graph of I versus distance, with the plot having an exponential shape. The initial intensity with no attenuation is Io on the y-axis at distance d=0. The exponential I plot falls to a value of 1.5*Ie at a distance of d=4m. Can you then solve for how much more distance d it takes to have that exponential decay fall to 1.0*Ie?
 
  • #5
For this problem, I = I˳e-µx

I = 1 and I˳ = 1.5
x = 4m + some distance
μ = no idea

They do not give the half value layer λ which is related to μ by μ = ln2/λ

Therefore, I'm not seeing how to solve the problem.

If you start with I = 1.5 , I˳ = original intensity before the regolith (not sure what it would be, it's not given) then x = 4m and μ = unkown. There are still two unknowns in that problem.
 

1. What is the mass attenuation problem and why is it important for protecting a lunar base against radiation?

The mass attenuation problem refers to the decrease in the intensity of radiation as it passes through a material. This is important for protecting a lunar base against radiation because the thicker the material, the more radiation it can absorb, decreasing the amount that reaches the base and poses a threat to astronauts.

2. What types of radiation pose a threat to a lunar base and how does the mass attenuation problem affect them?

The types of radiation that pose a threat to a lunar base include solar wind, galactic cosmic rays, and secondary radiation from the lunar regolith. The mass attenuation problem affects these forms of radiation by reducing their intensity as they pass through protective materials, making them less harmful to humans.

3. How can we calculate the mass attenuation coefficient for different materials?

The mass attenuation coefficient can be calculated by dividing the linear attenuation coefficient by the density of the material. The linear attenuation coefficient can be measured through experiments with different materials and radiation sources, and the density can be determined through physical measurements or known values.

4. What are some effective materials for protecting a lunar base against radiation?

Some effective materials for protecting a lunar base against radiation include polyethylene, aluminum, and water. These materials have high mass attenuation coefficients and can effectively reduce the amount of radiation that reaches the base.

5. How can we design a lunar base to optimize protection against radiation?

To optimize protection against radiation, a lunar base can be designed with multiple layers of shielding materials, with thicker and denser materials closer to the exterior. The base can also be located in a naturally shielded area, such as a crater or underground, and have emergency shelters in case of solar storms. Additionally, incorporating self-shielding materials into the construction of the base can further reduce radiation exposure.

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