ApolloHoax.net - Van Allen Belts

ApolloHoax.net
« Van Allen Belts [help] »

Welcome Guest. Please Login or Register.
Feb 9, 2010, 2:33pm

ApolloHoax.net :: The Apollo Hoax :: The Hoax Theory :: Van Allen Belts [help]

Author

Topic: Van Allen Belts [help] (Read 153 times)

hungry4info
Mercury
member is offline

Joined: Sept 2009
Posts: 1

Van Allen Belts [help]
« Thread Started on Sept 7, 2009, 6:52pm »

First, let me state that I am not a conspiracy theorist, and am fully aware that the lunar landings (all of them, save 13 of course) did occur.

I have recently gotten into a debate with one who buys into the VA belts argument. This case seems unique as he's presenting math to back him up. Though he doesn't reference his source for the numbers, so for all I know he probably thought them up.

Quote:

Can you cite me with some numbers like
(1) flux of protrons/electron in the inner belt
(2) flux of protrons/electron in the outer belt
plus publications , to prove your points? Not wiki. But some journal pubs.
Tell me about the ability of the Apollo CSM to shield from VAB radiations?
Which VAB data are you talking about? tell me the data base link and the accessibility please.
Waiting

Outer belt region (2.5 -6)rE (5 MeV average) electrons (h1=5 rE) (h2=1.5 rE)
Use D1= Sqrt[(2 X 6rE - h1)h1]= 5.916 rE ~ 37862 Km
D2= Sqrt[(2 X 2.5rE - h2)h2]= 2.598 rE ~ 6627 Km
D2-D1=21235 Km
Minimum Transit time (outer belt) = 21235 Km/7.2 Km/sec ~ 2950 Sec

The side CM walls are 6 mm Aluminium+1.27 honey comb steel
hence negligible energy loss. The base walls are thicker (considerable energy loss)and
cover approx 2 Pi/3 Steradians.
Side walls cover approx 4Pi/3 Steradians.
Hence 1/3 rd of protons should have 40 %(approx)
energy lost while 2/3 (through side walls) freely pass (10 % loss).

--------------------------------------------

NOW proton flux (average 50 MeV) a minimum value is 10^5 protons/sec/cm^2.
for 1.6-1.9 rE particle flux approx 10^6 protons/sec/cm^2.
6 mm Aluminium conical side walls (CM)
will cause a small energy loss the protons.
Also remember the region in 1.6-1.9 rE has
proton flux above 10^6 p/cm^s/sec.
take minimum human body surface area = 2 X 10^4 cm^2.
---------------------

---------------------

approx 650 sec through region1
approx 350 sec through region2
region 1= 1.3 -1.6 rE & 1.9 -2.2 rE
(10^5 protons /sec / cm^2 )
region 2= 1.6 -1.9 rE (10^6 protons /sec / cm^2 )
Total protons = 8.3 X 10^12

Total energy absorbed by astronaut
= 8.3 (2/3 X50 % ) X 10^12 X 5 MeV X 1.6 X 10^-19 ~115 J.
5.3 J/70 Kg =0.076 J/Kg=76 rad X 2(RBE)= 150 rem (for protons).
When they return they will have 1/3 X 150 = 50 rem
(radial cut through inner VA belt)
total radiation due to inner VA belt is
200 rem (minimum).
---------------------

The inner VAB has
---------------------
(1.3-1.9 rE ..flux 10^6 e/sec/cm^2.)
for 3 -10 Mev electrons.average 5 MeV
Stopping range fro electrons.
3 MeV (Al)=1.5 cm
20 MeV (Al)=3 cm
50 MeV range (Al)= 6 cm
100 MeV range (Al)= 12 cm;
So most of the 3 MeV and above electrons cross through with energy loss(assume 50 %)
through the 6mm Al+1.27 cm stainless steel side walls.

------------------------
Inner belt radiations = 200 (protons) rem
------------------------
Outer VAB (electrons) using the same arguments

2X10^4 cm^2 X 2950 sec X 10^6 e/sec/cm^2 ~6 X 10^13 e
Energy absorbed
=6 X 10^13 (2/3 (loss factor)) X 5 MeV X 1.6X10^-19
=27 J ;
27 J/70 kg = 0.38 J/Kg= 38 rad =38 rem.

When they return they will have 1/3 X 38 = 12.6 rem
(radial cut through outer VA belt)
Total= 38+12.6= 50 rem.
------------------------
Thus INNER +OUTER =250 ~ 250 rem.

I suspect that he didn't preserve his units or something, MeV are itty bitty relative to rem, IIRC.

So far all I have is

( from http://engineering.dartmouth.edu/~Simon_....Simonsen_91.pdf )

Low-energy integral fluxes:
Electrons > 1.6 MeV
Protons > 30 MeV

High-energy fluxes:
Electrons > 40 KeV
Protons = 0.1 to 5 MeV

« Last Edit: Sept 7, 2009, 6:53pm by hungry4info »

Link to Post - Back to Top IP: