# 3D Model of radiation in earth orbit?

I am trying to estimate required shielding masses to maintain a radiation level at or below earths average radiation of 3 mSv per year at different earth orbits (LEO, MEO, GEO and 50000 km+).

Initially I reverse engineered the radiation estimation method used to compute the estimated received radiation for the apollo mission to determine the required shielding in the highest (50000 km+) earth orbit.

Unfortunately this document has been removed.

So my second approach is to use figures like:  and find a dose rate experienced in the most intense radiation (in mSv/hr), link it to the value with 3*10^8 [units?] and then use the other 10^x values as a scaling factor for the dose rate experienced at those orbit heights in mSv/hr.

And then dividing that with the 7 cm of water required to half the radiation as mentioned by KeithS in this StackExchange question.

However, this method

1. Does not take into account the different composition of radiation particles of the different orbits.
2. Does not take into account the different composition of the radiation particles in the shielding effectivity.
3. Does not take into account the effects of Bremzstralung.
4. Is highly inaccurate due to the inaccurate data (1 static picture)
5. Needs verification on the units of the used picture as datasource.

So a 3d model that converts the radiation measurements into either mSv or Gy to a volume (sphere) with 1 kg of material as a function of the additional shielding mass with density rho, and shielding of y grams/cm^2 would highly improve the accuracy of the estimate.

The data is available as appears from the figures, but I can not find such a model (I understand the actual radiation is time dependent but even an average or instance of the data would significantly increase estimate accuracy.)

Do you know any such models?

First iteration solution: Convert to required shielding level using:

Yielding

• ${(10 \cdot 9.33 \cdot 365 \cdot 24)}\cdot {(\frac{1}{2})}^{n_{worst}}=3$
• ${(1 \cdot 9.33 \cdot 365 \cdot 24)}\cdot {(\frac{1}{2})}^{n_{best}}=3$

Resulting in

-$n_{worst} = 18.06 -> t_{worst} = 18.06 \cdot 7 = 126.5 grams/cm2$ -$n_{best} = 14.8 -> t_{best} = 14.8 \cdot 7 = 103.6 grams/cm2$

Assuming gold as a shielding material with density of $\rho_{gold}$ = 19.3 g/cm^3, for the 1L sphere with radius $(\frac{4}{3} \cdot Pi \cdot r^3) =0.001 -> r_v = 0.062035 m =6.2035 cm$, the shielding mass (-the 1 liter of non shielded sphere) becomes:

-worst case mass = $(\frac{4}{3} \cdot Pi \cdot (r_v+\frac{t_{worst}}{\rho_{gold}}/)^3)-19.3=\frac{4}{3} \cdot Pi \cdot (0.062035+0.01 \cdot \frac{126.5}{19.3})^3)19300-19.3 =148.6 kg$

-best case mass = $(\frac{4}{3} \cdot Pi \cdot (r_v+\frac{t_{best}}{\rho_{gold}}/)^3)-19.3=\frac{4}{3} \cdot Pi \cdot (0.062035+0.01 \cdot \frac{103.6}{19.3})^3)19300-19.3 = 106 kg$

Doubts:

Validity of assumption 1, the conversion of Roentgen to mSv is estimated at: 10 to a 100 [Roentgen/hour] = 0.01 to 0.04 Gy/hour according to van Allen in 1958. Where 0.01 to 0.04 Gy/hour would convert to 0.01 mSv per 10 roentgen in stead of 0.01 mSv per 1 Roentgen as assumed.

Applicability of assumption 2: The required shielding will in reality be optimized for different orbits, since for example bremsstraling is most efficiently shielded different than the high energy protons, yielding to different values than the simple 7 grams/cm^2.

• according to this highly up voted and authoritative-sounding answer all you have to do is find a random survivalist's blog about building a bomb shelter and hiding in it during a nuclear war, and use their numbers (humor). In reality, I think that this is a really well written and researched question. I like the way you've pointed out that "different composition of the radiation particles" as well as their energy distribution all have to be treated with care in order to do a reasonable shielding calculation. – uhoh Apr 30 '18 at 16:34