Mass attenuation problem, protecting the lunar base against radiation?

Click For Summary

Homework Help Overview

The discussion revolves around a problem related to radiation protection for a lunar base, specifically focusing on the effectiveness of a 4-meter layer of regolith in attenuating radiation from space and the sun. Participants are exploring how much additional regolith is needed to match the radiation levels found on Earth's surface.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between the thickness of regolith and the resulting radiation intensity, referencing the exponential attenuation formula. Some express uncertainty regarding the absence of specific values such as the absorption coefficient or half-value layer for regolith.

Discussion Status

The conversation includes various attempts to calculate the required thickness of regolith, with some participants suggesting numerical estimates based on the given information. However, there is a recognition of the limitations due to missing parameters, and the discussion remains open with no clear consensus on the solution.

Contextual Notes

Key constraints include the lack of specific values for the absorption coefficient and half-value layer of regolith, which complicates the calculations. Participants are also questioning the assumptions made regarding the attenuation rate and the initial intensity of radiation.

carnivalcougar
Messages
40
Reaction score
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.
 
Physics news on Phys.org
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...
 
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
 
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?
 
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.
 

Similar threads

  • · Replies 21 ·
Replies
21
Views
6K
Replies
17
Views
10K
Replies
1
Views
2K
  • · Replies 48 ·
2
Replies
48
Views
14K
  • · Replies 9 ·
Replies
9
Views
5K
  • · Replies 9 ·
Replies
9
Views
4K
  • · Replies 52 ·
2
Replies
52
Views
8K
  • · Replies 9 ·
Replies
9
Views
10K
Replies
23
Views
13K
  • · Replies 15 ·
Replies
15
Views
6K