Apollo and the Van Allen Belts
an estimate of the radiation dose received

Robert A. Braeunig
© September-2014


During the period 1968-1972, the United States sent nine manned missions, named Apollo, to the Moon. These missions remain the only flights in history to send human beings beyond low Earth orbit. To reach the Moon, these spacecraft had to pass through regions of intense ionizing radiation called the Van Allen radiation belts (VARB). The VARB are named after American space scientist Dr. James A. Van Allen (1914-2006), who designed the experiment that first detected them in 1958.

Because radiation is a scary word to those who really don't know much about it, the VARB are often targeted by conspiracy theorists who claim the Apollo lunar missions were nothing but a colossal hoax. They say the VARB are an impenetrable barrier deadly to any spacefarer who attempts to traverse them. The truth is much different and more complex. Whether or not the VARB are impassable is a problem with a calculable solution. In this article I will endeavor to solve the radiation problem by producing an estimate of the radiation dose received by the Apollo astronauts. Only then can we see the truth in uncertain terms.

This page is a companion to my previous article, Apollo 11's Translunar Trajectory (and how they avoided the heart of the radiation belts), in which I computed the trajectories that the Apollo 11 mission flew to and from the Moon. This gives us knowledge of the exact time and location of Apollo 11 within the VARB, allowing determination of the radiation levels to which the spacecraft was subjected during its passage through this region. Although Apollo 11 will be our test subject, it should be understood that all the Apollo missions flew similar trajectories. We can therefore consider Apollo 11 representative of all lunar flights.

Conspiracy Theories

Radiation is a favorite topic of many moon landing hoax theorists. No matter how badly they may lose the debate on other issues, they believe that space radiation is the one thing that trumps all others. If it were impossible for spacefarers to survive the Van Allen radiation belts then Apollo was hoaxed regardless of what all other evidence suggests. The argument typically goes like this:

Radiation is bad + There is radiation in space = Space is bad

That's generally all you'll get because that's all they know. It's no secret that space radiation exists, and we all know radiation can cause death, so, to the conspiracy theorist, the only conclusion is that space is a maelstrom of deadly radiation that will fry any outbound astronaut. Unfortunately this is a very simplistic and naive way of looking at the problem. Radiations vary dramatically in strength and intensity, with some being dangerous and some being harmless background radiation. Whether or not a human will experience illness or death is related to the radiation dose received. It is a problem with a very real answer that can be determined quantitatively.

I have seen only one hoax proponent attempt to prove the radiation claim mathematically, but his attempt was marred by many egregious errors and omissions, leading to wildly inaccurate numbers (see review here). It is fair to say that no conspiracy theorist possesses the expertise to correctly solve the problem, for if he did, he would know better than to subscribe to the hoax theory. Absent the ability to perform a proper quantitative solution, the conspiracists are forced to look for circumstantial evidence. They can generally do little more than point out what they believe to be suspicious activity or resort to quote mining.

Quote mining is the deceitful tactic of taking quotes out of context in order to make them seemingly agree with the quote miner's viewpoint. Below is one such example. In this case I've provided the accompanying text that the conspiracists are likely to omit, or at least draw your attention away from. The bolded text is the part the conspiracists highlight as being relevant to their argument.

"The successful operation of the solar batteries and the transmitter of Vanguard I (Satellite 1958 Beta) for over two years (as of the present date of writing) and the successful operation of similar equipment in Sputnik III (Satellite 1958 Delta) over a similar period provide the most direct evidence for the survival of electronic equipment in space vehicles. The integrated radiation exposures in these two cases are still much below the level at which serious deterioration may be expected.

"But, though mechanical and electronic equipment can operate within the high radiation areas, a living organism cannot survive this level of radiation damage. Hence, all manned space flight attempts must steer clear of these two belts of radiation until adequate means of safeguarding the astronauts has been developed."

Dr. James Van Allen, Space World, December 1961.

The bolded part is portrayed by the conspiracists as Dr. Van Allen agreeing with them that any venture into the VARB is lethal. However the context of that statement appears in the first paragraph, which the conspiracists deceptively conceal. Here we see that Dr. Van Allen is specifically discussing two missions that where each over two years in duration. He writes "a living organism cannot survive this level of radiation damage", where the 'level of radiation damage' is a reference to the integrated radiation exposures from the cited 2-year long missions. He also writes about "operat(ing) within the high radiation areas", which is in the context of extended missions, not rapid transits. Even then he doesn't say it is impossible, only that "adequate means of safeguarding the astronauts" is required. Nothing in his statement precludes the possibility of rapid transits through the region while on the way to or from the Moon.

Below is another quote highlighted by the conspiracists:

"So far, the most interesting and least expected result of man's exploration of the immediate vicinity of the earth is the discovery that our planet is ringed by a region–to be exact, two regions–of high-energy radiation extending many thousands of miles into space. This discovery is of course troubling to astronauts; somehow the human body will have to be shielded from this radiation, even on a rapid transit through the region."

Dr. James Van Allen, Scientific American, March 1959.

Here Dr. Van Allen specifically addresses rapid transits through the region, stating only that the human body will have be to be shielded. Also note that the dates of these quotes are many years prior to the first lunar flight in 1968, giving designers adequate time to further study the problem and devise solutions. In fact, Dr. Van Allen helped to design the Apollo lunar trajectories, which were engineered specifically to lessen radiation exposure. Despite the conspiracists' insistence that Dr. Van Allen agrees with them, he has rejected the claim that radiation exposure during the Apollo missions would have been fatal to the astronauts, calling it "nonsense".

This quote mining doesn't end with Dr. Van Allen. In recent years talk has ramped up about extended stay lunar missions and missions to Mars. Many of these conversations have addressed the severity of the radiation problem and how it must be solved before these missions can be undertaken. Of course these are just the kind of statements that a conspiracy theorist will embrace and take out of context to mean that Apollo was impossible. There is a big difference between a 10-day mission to the Moon and a 2.5-year mission to Mars – they are simply not comparable in terms of the radiation problem. The Apollo problem was solved decades ago, the Mars problem still needs some work.

When conspiracy theorists aren't mining for quotes, they go about pointing out actions that they suggest indicate evidence of wrongdoing. For instance, they love to call attention to the fact that the United States is the only nation to have sent astronauts beyond low Earth orbit, and last did so in 1972. The implication being that if it were possible to travel beyond the VARB, then surely somebody else would have done it by now. The simple truth is that no one has had adequate motivation to commit the massive funds and resources necessary to go back to the Moon or beyond. However there are indications this attitude may be changing, for instance Chinese officials have indicated they are interested in potential future manned lunar missions.

Another hoax story that has cropped up from time to time is the assertion that the strength and intensity of the VARB is far greater than published. The conspiracists claim that NASA has hidden the real data and replaced it with false data. If we could only get our hands on the real data, they claim, it would show that Apollo was impossible. This is a ludicrous proposition. First, NASA is not the sole proprietor of VARB knowledge – other spacefaring nations have also studied the VARB – so for the assertion to be true it requires international cooperation. And second, space is a multi-billion dollar industry that depends on the accuracy of the VARB models. If the models were false, then billions of dollars of space hardware would be failing due to under designed radiation protection. Instead we find that satellites perform in a manner consistence with the published data.

No other spacefaring nation has expresses doubts about the ability to traverse translunar space, and they all acknowledge that Apollo succeeded in landing astronauts on the Moon. Furthermore, the United States is not the only nation to have sent living creatures to the Moon. In September 1968 the Soviet Union's Zond 5 became the first spacecraft to swing around the Moon and return to land on Earth. The mission was planned as a precursor to a manned lunar spacecraft. It carried a biological payload of two Russian tortoises, wine flies, meal worms, plants, seeds, bacteria, and other living matter. The biological payload was intact, proving that it was possible to survive a lunar flight and safely return to Earth. As a result of Zond 5, and Zond 7 in 1969, the Soviet Union concluded that, "seven-day flights along the trajectories of Zond-5 and 7 probes are safe from the radiation point of view."

It seems the only people who doubt the ability of manned spacecraft to safely traverse the VARB are conspiracy theorists.

Radiation Plan for Apollo

No one doubted that once humans ventured beyond Earth's atmosphere they would have to be provided with air and protected from the surrounding vacuum. But no one doubted, either, that this could be done by straightforward extensions of existing technology. Yet that last small cloak of air is of vital importance to life on Earth. It shields us almost completely from dangerous radiation, and from the incessant downpour of meteors. There were those who believed before 1957 that neither human nor vehicles could survive for long outside the protective blanket of the atmosphere.

But even the first satellites showed that the danger from meteorites had been grossly overestimated by the pessimists. And the danger from the known cosmic radiations had been greatly exaggerated; with the possible exception of solar flares, the hazard they presented was almost negligible to short duration space flight.

One of the most important discoveries made in space was reported by the very first American satellite, Explorer 1, in 1958. It was revealed that several huge layers of radiation particles are trapped by Earth's magnetic field, called the Van Allen radiations belts. This discovery came as a considerable shock as here was a previously unknown but very real peril, which made certain regions of space uninhabitable without a prohibitive weight of shielding. However, it was soon realized that the VARB represented an obstacle to be bypassed, not a complete roadblock. By choosing suitable orbits it was possible to avoid the most intense levels of radiation; and to outward-bound spacecraft the belts were no serious menace because the vehicles passed through them swiftly.

Observations from the ground and from spacecraft demonstrated that the space radiation hazard was one of the lesser engineering problems to be overcome in Apollo spacecraft design and mission planning. Flux maps of the Van Allen belts were developed, solar flare particle events were subjected to intensive statistical analyses, and techniques were developed to calculate radiation doses behind complex spacecraft structures. Van Allen belt radiation doses were kept small by use of low-altitude Earth orbits and rapid transits to the Moon along trajectories with inclinations of about 30 degrees. Only the very large (and consequently very rare) solar flare particle events constituted a hazard for moderately shielded spacecraft. Also, secondary radiation was not significant for such spacecraft.

Most of Apollo's radiation protection activity was directed towards providing protection against major solar flare particle events that might occur while astronauts were in the lunar module or on the lunar surface. The events, which start at the sun, were detected by ground-based instrumentation and were measured at the spacecraft by dosimeters and particle spectrometers. A prognosis of the radiation dose was prepared and continually updated by radiation environment specialists using a console in the Mission Control Center (MCC). Dose estimates were then provided for the use of the medical officer, who advised the Flight Director of the radiation effects to be expected.

The Solar Particle Alert Network (SPAN) was established to support the Apollo program. The network has seven stations around the world, situated so as to provide 24-hour coverage of the Sun at both optical (weather permitting) and microwave frequencies. There were three stations equipped with both optical and radio telescopes, namely, Houston, Texas, Carnarvon, Australia, and the Canary Islands. There were four additional sites equipped with optical telescopes only. SPAN's function was to monitor solar activity during the missions and provide warnings of particle events. Time of occurrence, area, and location of the flare were determined by SPAN observers and were teletyped to the MCC where the data were incorporated into the estimate of the particle event size. The radio burst profile was also teletyped to the MCC. Data from SPAN were augmented by data from the solar and ionospheric monitoring systems operated by the Environmental Science Services Administration and the Air Weather Service.

If SPAN detected that a large solar flare was imminent, there was a few hours' advance notice of the particle flux. This was adequate time for the astronauts on the Moon to get back to the LM, take off, rendezvous with the CSM, and take cover as best they could. While in lunar orbit, the Moon would protect the astronauts for half of each orbit. At other times the spacecraft would be turned so the bulk of the service module was between the astronauts and the incoming particles. The astronauts had a handheld Geiger counter so they could find the safest spot in the command module cabin should they have to ride out a solar flare.

During Apollo's operational period of 1969 to 1972, there were only three solar events that had a biological significance, none of which occurred during a mission. Only one event, in August 1972, was large enough to have caused severe illness.

Radiation Basics

Radiation may be defined as energy in transit in the form of high-speed particles and electromagnetic waves. Electromagnetic radiation is very common in our everyday lives in the form visible light, radio and television waves, and microwaves. Radiation is divided into two categories – ionizing radiation and non-ionizing radiation.

• Ionizing radiation is radiation with sufficient energy to remove electrons from the orbits of atoms resulting in charged particles, and it is this type of radiation that is evaluated for purposes of radiation protection. Examples of ionizing radiation include gamma rays, electrons, protons, and neutrons. Ionizing radiation is different from ion formation that occurs in ordinary chemical reactions, such as the generation of table salt from sodium and chlorine. In such a reaction, only the outermost electron is removed to form a positively charged ion. With ionizing radiation, if the energy is sufficient, electrons other than those in the outermost orbits can be released; this process renders the atom very unstable, and these ions are very chemically reactive. It is the subsequent reactions of these ionized atoms that initiate the biological effects that are observed.

• Non-ionizing radiation is radiation without sufficient energy to remove electrons from their orbits. Examples are microwaves, radio waves, and visible light.

Space radiation consists primarily of ionizing radiation that exists in the form of high-energy, charged particles; principally electrons, protons, and some heavier atomic nuclei – commonly called particulate radiations. There are three naturally occurring sources of space radiation: trapped radiation, galactic cosmic radiation, and solar particle events.

These particles when moving at high speed (i.e. when of high energy) can pass through matter. The depth of penetration depends on their speed and charge – the higher the charge the smaller the penetration. The energy of these radiations is expressed in a special unit known as the electron-volt (eV) and the multiples 1,000 eV = 1 keV; 1,000 keV = 1 MeV. One electron-volt corresponds to 1.60×10-19 joules.

Since matter is made up of positively charged atomic nuclei and negatively charged electrons, it is clear that the electrically charged particles must interact with the atoms in the molecules that surround their path. It is very rare for heavy particles, such as protons or alpha particles, to hit an atomic nucleus or for electrons to collide with the orbital electrons. When such a collision has occurred the particles involved change direction. Except when scattered in this way the subatomic particles move through matter in straight lines. However, even when they do not come into physical contact with atoms they are continually slowed down by the local electric charges of the atoms, rather as air resistance will slow down a bullet. Because of the electrical interactions with the atoms through which they pass they lose energy until they can penetrate no further. At the very end of their tracks the path of the particles is no longer linear, since scattering occurs more readily when they move relatively slowly. The energy that the ionizing particles lose is taken up by the surrounding atoms, and some of the molecules of which they are part become chemically changed. When a person is exposed to the radiation, it is this radiochemical process that is responsible for the biological effects produced. Only a part of the energy lost in this way is used for producing such chemical changes. Much of the energy is dissipated as heat, though this does not bring about any of the observed biological changes, since a radiation dose sufficient to kill a mammal would raise its temperature by less than 1/100 of a degree.

Particulate radiations rarely exceed a few MeV, and therefore cannot penetrate very deeply into the body. Exposure to such radiations is therefore only likely to produce skin burns that are usually not serious except when the doses are very high. X-rays, gamma rays, and neutrons are much more penetrating. On exposure to these radiations the affected parts are not confined to the skin, but every part of the body will be irradiated. Under these conditions much smaller doses of radiation are dangerous.

The dose is the amount of energy that has been left in the irradiated material and is directly related to the amount of chemical alteration produced. It is typically given in units of rads or "radiation absorbed dose" for a particular material, defined as 1 rad = 0.01 J/kg. The SI unit of absorbed dose is the gray (Gy), where 1 Gy = 100 rad.

Different types of radiation have different biological effectiveness mainly because they transfer their energy to the tissue in different ways. The equivalent dose is calculated by multiplying the absorbed dose by a radiation weighting factor (WR) appropriate to the type and energy of radiation. The unit of equivalent dose is the rem, derived from the phrase "Roentgen equivalent man". The rem is now defined as the dosage in rads that will cause the same amount of biological injury as one rad of x-rays or gamma rays. For x-rays, gamma rays, and electrons, WR = 1, and for protons WR = 2. The SI unit of equivalent dose is the sievert (Sv), where 1 Sv = 100 rem.

In addition to the energy and composition of a particular particle, it is also necessary to describe how many of them there are. This is usually done in terms of flux, and when speaking in terms of a time interval, fluence. Flux is defined as the flow rate of particles per unit area, which has the dimensions [particles]×[time]-1×[area]-2. Fluence is the flux integrated over time, which is defined as the total number of particles that intersect a unit area in a specific time interval of interest, and has the units [particles]×[area]-2.

AE-8/AP-8 Radiation Belt Models

The Van Allen radiation belts are a torus (doughnut shape) of energetic charged particles circling Earth around its magnetic equator and held in place by Earth's magnetic field. The main belts extend from an altitude of about 1,000 to 60,000 kilometers above the surface in which region radiation levels vary. Most of the particles that form the belts are thought to come from solar wind and other particles by cosmic rays. The belts are located in the inner region of the Earth's magnetosphere. The belts contain energetic electrons that form the outer belt and a combination of protons and electrons that form the inner belt. The radiation belts additionally contain lesser amounts of other nuclei, such as alpha particles.

Figure 1 - Van Allen radiation belts

The description of the radiation environment requires a knowledge of the particle flux as a function of energy, species, location in space, and time. The AE-8 and AP-8 models consist of maps that contain omnidirectional, integral electron (AE maps) and proton (AP maps) fluxes in the energy range 0.04 MeV to 7 MeV for electrons and 0.1 MeV to 400 MeV for protons in the Earth's radiation belt. The fluxes are stored as functions of energy, L-value, and B/B0. The maps are based on data from more than twenty satellites from the early 1960s to the mid-1970s. AE-8 and AP-8 are the latest editions in a series of updates starting with AE-1 and AP-1 in 1966.

Figure 2 – Graphical representation of R, λ and B, L coordinates

Earth's magnetic field is the magnetic field that extends from the Earth's interior to where it meets the solar wind. Roughly speaking it is the field of a magnetic dipole currently tilted at an angle of about 10 degrees with respect to Earth's rotational axis, as if there were a bar magnet placed at that angle at the center of the Earth. A dipole line of force is defined by the equation,

(1) R = L × cos2 λ

where R is the radial distance from the dipole (i.e. from the center of Earth), λ is the magnetic latitude, and L is the maximum value of R, which is found at λ = 0o, the equator. The units of R and L are Earth radii.

B is the magnitude of the magnetic field, with B0 being the value of B at λ = 0o. The value of B0 is given by the equation,

(2) B0 = M / L3

where M is the dipole magnetic moment, which, for Earth, equals 0.311653 gauss. The magnetic field strength B is a function of B0 and λ, as follows:

(3) B = B0 × (4 – 3 cos2 λ)1/2 / cos6 λ

After computing the values of L and B/B0, the web page AE-8/AP-8 Radiation Belt Models, from the Community Coordination Modeling Center (CCMC), enables the computation of omnidirectional integral or differential electron and proton fluxes for specific energies. Electron and proton fluxes are available for solar maximum and solar minimum conditions.

From my previous work – Apollo 11's Translunar Trajectory* – the values of R and λ are known, from which it is a simple procedure to obtain L and B/B0 using equations (1) and (3). Since Apollo 11 occurred near the time of solar maximum, we'll use the AE8MAX and AP8MAX models to obtain the integral particle fluxes. Gathering the flux data involves the simple but tedious process of entering all the coordinates and particle energies into the web interface and transferring the output to a spreadsheet for further analysis. There is no need to show all the data in this article. An abridged version is shown below, which includes electron fluxes for the first 30 minutes of flight following translunar injection (TLI). This is enough data to allow demonstration of the procedures used in this analysis. The actual analysis used matrices that contained a total of over 8,000 flux values.

(* Note that the same computation methods were used to also derive the transearth trajectory.)

For any practical purpose, no one computes the B-L coordinates by hand or uses the CCMC web interface as I have. The practical way is to use computer programs from the National Space Science Data Center (NSSDC) that automate converting from geodetic coordinates or orbital elements to B-L coordinates, and further automate stuffing long sequences of B-L coordinates into the AE-8/AP-8 models. I elected to use the tedious manual method to demonstrate the procedure and make my work more transparent.


Table 1
Apollo 11 – Translunar Phase
Elapsed
Time*
(min)
Coordinates AE8MAX Electron Flux (e-/cm2-s)
R
(radii)
λ
(deg.)
L
(radii)
B/B0 ≥0.1 MeV ≥0.5 MeV ≥1 MeV ≥2 MeV ≥3 MeV ≥4 MeV ≥5 MeV ≥6 MeV ≥7 MeV
0 1.05352 8.8090 1.07882 1.11092 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00
1 1.06893 12.3280 1.11999 1.22637 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00
2 1.08878 15.6666 1.17442 1.38551 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00
3 1.11275 18.7966 1.24166 1.59107 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00
4 1.14049 21.6964 1.32103 1.84536 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00
5 1.17162 24.3541 1.41168 2.14962 3.931E+03 9.206E+01 1.560E+01 2.296E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00
6 1.20578 26.7659 1.51254 2.50334 2.342E+06 4.738E+04 6.804E+03 1.308E+03 8.369E+01 0.000E+00 0.000E+00 0.000E+00 0.000E+00
7 1.24260 28.9349 1.62236 2.90373 6.452E+06 1.159E+05 9.227E+03 1.393E+03 7.225E+01 0.000E+00 0.000E+00 0.000E+00 0.000E+00
8 1.28176 30.8695 1.73976 3.34539 1.106E+07 1.847E+05 6.917E+03 7.051E+02 3.312E+01 0.000E+00 0.000E+00 0.000E+00 0.000E+00
9 1.32294 32.5817 1.86325 3.82043 1.212E+07 1.951E+05 3.245E+03 1.911E+02 8.691E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00
10 1.36585 34.0864 1.99132 4.31882 1.333E+07 1.765E+05 1.831E+03 7.636E+01 2.647E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00
11 1.41026 35.3995 2.12248 4.82912 1.167E+07 1.094E+05 1.042E+03 3.566E+01 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00
12 1.45593 36.5378 2.25532 5.33932 7.511E+06 7.681E+04 7.928E+02 2.270E+01 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00
13 1.50267 37.5179 2.38857 5.83768 2.368E+06 5.436E+04 7.486E+02 1.732E+01 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00
14 1.55030 38.3558 2.52112 6.31360 2.248E+06 6.031E+04 2.131E+03 2.586E+02 9.491E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00
15 1.59868 39.0668 2.65202 6.75809 2.476E+06 7.977E+04 7.268E+03 1.391E+03 1.349E+02 4.868E+00 0.000E+00 0.000E+00 0.000E+00
16 1.64767 39.6651 2.78052 7.16422 2.085E+06 1.107E+05 2.266E+04 4.065E+03 6.064E+02 6.910E+01 4.423E+00 0.000E+00 0.000E+00
17 1.69715 40.1636 2.90603 7.52720 1.455E+06 1.848E+05 5.967E+04 1.194E+04 2.320E+03 3.894E+02 4.417E+01 2.726E+00 0.000E+00
18 1.74703 40.5743 3.02813 7.84429 2.354E+06 3.270E+05 1.181E+05 2.452E+04 4.842E+03 8.811E+02 1.675E+02 2.012E+01 0.000E+00
19 1.79723 40.9079 3.14654 8.11462 4.345E+06 6.509E+05 2.374E+05 5.352E+04 1.099E+04 2.136E+03 3.641E+02 2.803E+01 0.000E+00
20 1.84767 41.1739 3.26110 8.33885 6.204E+06 9.794E+05 3.610E+05 8.457E+04 1.798E+04 3.413E+03 4.706E+02 2.728E+01 0.000E+00
21 1.89829 41.3809 3.37175 8.51886 7.248E+06 1.186E+06 4.427E+05 1.044E+05 2.295E+04 4.016E+03 4.177E+02 2.046E+01 0.000E+00
22 1.94903 41.5364 3.47852 8.65742 8.218E+06 1.411E+06 5.170E+05 1.262E+05 2.848E+04 4.293E+03 3.578E+02 1.482E+01 0.000E+00
23 1.99984 41.6472 3.58148 8.75784 9.220E+06 1.618E+06 5.960E+05 1.466E+05 3.146E+04 4.326E+03 3.129E+02 1.078E+01 0.000E+00
24 2.05069 41.7190 3.68075 8.82380 9.884E+06 1.804E+06 6.667E+05 1.590E+05 3.239E+04 4.063E+03 2.685E+02 8.376E+00 0.000E+00
25 2.10153 41.7571 3.77649 8.85905 1.048E+07 1.995E+06 7.381E+05 1.700E+05 3.307E+04 3.786E+03 2.316E+02 6.718E+00 0.000E+00
26 2.15235 41.7661 3.86888 8.86733 1.116E+07 2.203E+06 8.011E+05 1.792E+05 3.202E+04 3.533E+03 2.031E+02 4.836E+00 0.000E+00
27 2.20311 41.7498 3.95811 8.85223 1.189E+07 2.427E+06 8.623E+05 1.879E+05 3.055E+04 3.310E+03 1.801E+02 3.409E+00 0.000E+00
28 2.25378 41.7117 4.04436 8.81709 1.256E+07 2.611E+06 9.260E+05 1.887E+05 2.929E+04 3.098E+03 1.590E+02 2.524E+00 0.000E+00
29 2.30436 41.6550 4.12783 8.76498 1.314E+07 2.751E+06 9.924E+05 1.821E+05 2.821E+04 2.895E+03 1.398E+02 1.839E+00 0.000E+00
30 2.35482 41.5822 4.20871 8.69871 1.373E+07 2.894E+06 1.057E+06 1.763E+05 2.715E+04 2.715E+03 1.235E+02 1.322E+00 0.000E+00

* Elapsed time is measured from translunar injection. In mission time, T0 = GET 002:50:13.03.

Computing the Unprotected Dose

Now that we have the particle fluxes we can proceed to compute the radiation dose. In this article I will demonstrate the method by performing a small number of sample computations. The actual computations are quite voluminous and are performed using an Excel spreadsheet.

Let's first take a look at the trajectories. Figures 2 and 3 below show the path of Apollo 11 in relation to the VARB. The blue line represents the spacecraft's path and the red dots indicate 10-minute intervals of time. The red and orange areas are the most powerful regions of the VARB, and the violet and blue areas are the least powerful. You can see that the trajectories were designed to bypass the VARB to the greatest extent possible. Please note that Figures 2 and 3 show the path of Apollo 11 as a simple plot of distance above or below the magnetic equator versus distance from Earth. The figures are not meant to depict the actual shape or appearance of the orbits (as Apollo 11 moved away or toward Earth it was also swinging around the planet).

Figure 3 - Apollo 11's outbound trajectory (translunar)

Figure 4 - Apollo 11's inbound trajectory (transearth)

Before considering the spacecraft shielding, let's first compute the dose that an exposed and unprotected astronaut would receive if he were to follow the Apollo 11 trajectories through the VARB.

Computing Particle Flux

Figure 5

The data in Table 1 can be broken down into a group of blocks (see Figure 5), each bounded by T1, T2, E1 and E2, where T is time and E is particle energy. At the intersections we have the integral fluxes F1(1), F2(1), F1(2) and F2(2). We must compute the energy fluence, Ψ, for each block, which we do as follows:

We first calculate the particle fluxes at times T1 and T2 for particles with energies between E1 and E2. The fluxes computed by the AE-8/AP-8 model, and as shown in Table 1, are for a given energy and greater. Therefore, if we want the flux, φ, for particles between E1 and E2, we subtract the integral flux at E1 from the integral flux at E2.

(4) φ = F1 – F2

Computing Average Particle Energy

Next we calculate the average energy for particles between E1 and E2 at times T1 and T2. Since the particle flux increases exponentially with decreasing energy, we cannot use a simple average. Figure 6 below is a typical flux versus energy diagram.

Figure 6

The average energy of the particles between E1 and E2 can be closely approximated using the following:

(5)* Eavg = E1 + B × (E2 – E1)

where,

B = 0.029 A2 – 0.205 A + 0.5

A = log(F1 / F2)

(* R. A. Braeunig. Derived by computing Eavg, via integration, for sample values of F1 and F2; producing a plot of B versus A; and fitting a curve to the resulting data.)

Computing Energy Flux and Fluence

We now calculate the energy flux, ψ (lower case), for times T1 and T2.

(6) ψ = φ × Eavg

Next we calculate the energy fluence, Ψ (upper case), for the block bounded by T1, T2, E1 and E2.

(7) Ψ = (T2 – T1) × (ψ1 + ψ2) / 2


Before going further, let's demonstrate what we've learned to this point by computing the energy fluence for electrons between the energies of 1 and 2 MeV and for the time interval of 20 to 21 minutes. From Table 1, the integral fluxes are,

F1(1) = 3.610E+05 e-/cm2-s
F2(1) = 8.457E+04 e-/cm2-s
F1(2) = 4.427E+05 e-/cm2-s
F2(2) = 1.044E+05 e-/cm2-s

The fluxes for 1-2 MeV electrons are,

φ1 = 3.610E+05 – 8.457E+04 = 2.764E+05 e-/cm2-s
φ2 = 4.427E+05 – 1.044E+05 = 3.383E+05 e-/cm2-s

The average electron energies are,

A1 = log(3.610E+05 / 8.457E+04) = 0.63029
B1 = 0.029 × 0.630292 – 0.205 × 0.63029 + 0.5 = 0.3823
Eavg1 = 1 + 0.3823 × (2 – 1) = 1.3823 MeV

A2 = log(4.427E+05 / 1.044E+04) = 0.62741
B2 = 0.029 × 0.630292 – 0.205 × 0.63029 + 0.5 = 0.3828
Eavg2 = 1 + 0.3823 × (2 – 1) = 1.3828 MeV

The energy fluxes are,

ψ1 = 2.764E+05 × 1.3823 = 3.821E+05 MeV/cm2-s
ψ2 = 3.383E+05 × 1.3828 = 4.678E+05 MeV/cm2-s

Finally, the energy fluence for our selected block is,

Ψ = (1260 – 1200 s) × (3.821E+05 + 4.678E+05) / 2 = 2.550E+07 MeV/cm2


Our next step is to compute the total energy fluence, which is found by summing Ψ for all the individual blocks, i.e. Ψtotal = Σ Ψn. Since this involves thousands of computations, I can only report the results. Table 2 below gives the electron and proton energy fluences for both the outbound and inbound transits of the Van Allen radiation belts.

Table 2
Mission Phase Elapsed
Time
(min)
Energy Fluence, Ψ
(MeV/cm2)
Electrons Protons
Outbound VARB Transit 214 2.358E+10 7.848E+09
Inbound VARB Transit 140 4.913E+09 1.472E+09

Computing the Unprotected Dose

The absorbed dose, D, is calculated from the equation,

(8) D = Ψ × A / m

where D is in units of grays (Gy), Ψ is in units if J/cm2 (where 1 MeV = 1.60217657×10-13 J), and A and m are the surface area (cm2) and mass (kg) of the body absorbing the dose.

Since the flux is omnidirectional, A must equal the body's total surface area. For an adult human the total surface area is about 1.8 m2. The average mass of the Apollo 11 crew varied from about 76 kg at launch to about 74 kg at recovery. To demonstrate the method, below is the dose calculation for electrons on the outbound trip.

D = (2.358E+10 × 1.60217657E-13) × (1.8 × 1002) / 76 = 0.8948 Gy

The absorbed doses are summarized in Table 3. Since the gray wasn't adopted as part of the SI system of units until 1975, Apollo literature uses the older unit of rads, where 1 rad = 0.01 Gy.

Table 3
Mission Phase Absorbed Dose, D (rads)
Electrons Protons
Outbound VARB Transit 89.48 29.78
Inbound VARB Transit 19.15 5.74

The final step is to convert the absorbed dose to equivalent dose, H, by multiplying by a quality factor appropriate to the type of radiation.

(9) H = D × WR

where WR = 1 for electrons, 2 for protons

The final doses are summarized in Table 4 below. Note that the total dose is about 180 rem. A person will experience radiation sickness with a dose of 100-200 rem, and death with a dose of 300+ rem. Clearly the calculated dose is significant enough to cause serious illness, but it is below that typically regarded as causing death. We see, therefore, that even a completely exposed and unprotected astronaut, i.e. one naked and outside the spacecraft, could survive Apollo 11's trip through the VARB from the radiation point of view.

Table 4
Mission Phase Equivalent Dose, H (rem)
Electrons Protons Total
Outbound VARB Transit 89.48 59.56 149.04
Inbound VARB Transit 19.15 11.48 30.63
Total 108.63 71.04 179.67

Unprotected Dose—Concluding Remarks

Figures 2 and 3 show Apollo 11 traversing the region of the VARB in about 90 and 60 minutes respectively, while Table 2 shows the elapsed times to complete the transits as 214 and 140 minutes. Figures 2 and 3 show only the most intense part of VARB, which account for about 90% of the total dose. The longer durations are that needed to reach the far outer edge of the belts, which extends out to about 9.5 radii. The fluxes in the extended region are low and contribute only about 10% of the dose.

We can see from Tables 2, 3 and 4 that the dose received from the outbound transit is about five times greater than the inbound transit. This is because the return trajectory had a higher orbital inclination, 39.925o, than the outbound trajectory, 31.383o. Taking into account the additional 11o (1969) tilt of the geomagnetic axis and we find that the maximum magnetic latitude for the inbound trip was about 51o versus about 42o for the outbound trip. This is significant because the return trajectory was further out toward the less intense and less expansive edges of the VARB, resulting in the much lower dose.

Had Apollo 11's orbit been in the plane of, rather than inclined to, the geomagnetic equator, then the unshielded dose would have exceeded 3200 rem for each transit. We see, therefore, that the selected trajectories reduced the potential worst exposure by 95% and 99% respectively for the outbound and inbound transits.

While most particles would be stopped inside the human body, the most energetic protons (>100 MeV) will pass clean through the body. These high-energy particles will deposit only a fraction of their energy in the tissue through which they pass. This has been ignored above; the computations assume that 100% of the particle energy is deposited. Thus the real doses would be slightly lower than calculated.

Spacecraft Shielding

Computing the dose received by an unshielded astronaut is interesting in that it provides a worst case scenario, but otherwise it has little practical meaning. To know an astronaut's actual exposure we must consider the amount and type of shielding that lies between him and outer space.

The part of the Apollo spacecraft inhabited by the astronauts during their passage through the VARB was called the command module (CM). The command module hull, which completely surrounded the astronauts, provided the first layer of radiation shielding. Referring to Figure 7 we see that the hull was a multi-layered composite consisting of both metallic and non-metallic components.


From Virtual Apollo, Apogee Books, Scott P. Sullivan.

Figure 7

The command module is generally quoted as having a shielding rating of 7 to 8 g/cm2. Let's see if we can compute for ourselves the area density of the hull.

The heat shield consisted of an epoxy novalac resin injected into the cells of a fiberglass-phenolic honeycomb matrix. The product is commercially known by the name AVCOAT 5026-39-HCG. The resin is very light, having a density of only 32 lb/ft3 (0.51 g/cm3), while glass-phenolic typically has a density of about 1.8 g/cm3. Let's say we have a composite density of 0.7 g/cm3. The thickness of the heat shield varied from 0.75 to 2.75 inches, with the least thickness on the sides and the greatest thickness on the base. I estimate an average thickness of about 1.3 inches (3.3 cm).

Behind the heat shield was a structural shell consisting of stainless steel honeycomb sandwiched between two stainless steel face sheets. These face sheets each had a thickness of 0.075 inches (1.90 mm). Inside this outer hull was an aluminum pressure hull, consisting of an aluminum honeycomb sandwiched between aluminum face sheets. These face sheets varied in thickness between 0.035" and 0.065", so let's call it 0.050 inches (1.27 mm) per sheet. The densities of stainless steel and aluminum are 8 g/cm3 and 2.7 g/cm3 respectively.

Adding up what we have so far gives us an area density of,

ρA = (3.3 × 0.7) + (0.38 × 8) + (0.254 × 2.7) = 6.04 g/cm2

To this we must add the stainless steel honeycomb, aluminum honeycomb, fibrous insulation, and miscellaneous other materials. The mass of these additional items is not much, but we can expect to get up to a total area density of about 7 g/cm2.

To confirm this result we can use an alternative method to compute the area density. We know the mass of the command module heat shield was 848 kg, the mass of the spacecraft structure was 1,567 kg, and the total exterior surface area was 36 m2. Computing the area density is a simple matter of dividing the mass by the area. We therefore have,

ρA = ((848 + 1567) × 1000) / (36 × 1002) = 6.71 g/cm2

This should include the honeycomb, so we need only to add for the insulation and miscellaneous materials. Again we get up to a total area density of about 7 g/cm2, confirming our previous estimate.

This is, or course, an average – the area density will be greater in some areas and less in others. A rough estimate gives a variation of about 5 to 10 g/cm2 between the thinnest and thickest parts of the hull. Since the radiation flux is omnidirectional, using the average area density should give a good integrated solution to the radiation problem.

Figure 8 – Apollo command module

It must also be noted that, because of the shape of the spacecraft, the path that most particles must take to reach that astronauts requires them to strike the hull at an incident angle, which effectively increases the shielding thickness. For instance, if a particle strikes the hull at an angle of incidence of 30o, the effective area density is increased by a factor of, 1/cos(30o) = 1.155. This can explain why the documented shielding rating (7 to 8 g/cm2) is higher than our computed value.

Furthermore, we must consider that the hull was only the primary radiation shield, i.e. the first line of defense. There was a considerable amount of secondary shielding located between the hull and the astronauts in the form of equipment, instrument displays, propellant tanks, etc. In fact, the entire mass of the command module was distributed around the astronauts forming a shell between them and outer space. The command module had a total mass of about 5,800 kg, thus it provided both primary and secondary shielding with an overall mean area density of,

ρA = (5800 × 1000) / (36 × 1002) = 16 g/cm2

Note also that attached to the aft end of the CM was the service module (SM), which provided many tons of additional shielding covering about 1/3 of the CM's exterior. We see, therefore, that the Apollo spacecraft afforded the astronauts a considerable amount of protection from space radiation. Based on the numbers above, our best estimates of the command module's shielding is summarized in Table 5.

Table 5

Shielding Material

Area Density
(g/cm2)
Spacecraft Hull: ----
  Heat shield 2.36
  Stainless steel 3.50
  Aluminum 0.85
  Other 0.3
  Total hull 7
Secondary shielding 9

Computing the Shielded Dose

It is now time to figure out how much of the surrounding space radiation can penetrate the spacecraft's shielding. We'll first analyze electrons, then protons, and then something we haven't talked about yet—secondary radiation, or bremsstrahlung.

Electrons

The maximum range, R, that an electron can penetrate in a material can be computed from an empirical formula (given by L. Katz and A. S. Penfold),

(10a) R = 0.412 × E1.265 – 0.0954 ln(E)

where 0.01 ≤ E ≤ 2.5 MeV

(10b) R = 0.530 × E – 0.106

where E > 2.5 MeV

where R is in g/cm2 and is material independent, and E is the maximum electron energy in MeV. The ability to stop electrons depends primarily on the number of electrons in the absorber. Hence, the range when expressed as a density thickness of the material gives a generic quantifier by which various absorbers can be compared. With the maximum range known, the actual shielding thickness required can be computed,

(11) t = R / ρ

where t is the material thickness and ρ is the material density.

From Table 1 we see that the integral flux for electrons greater than or equal to 7 MeV is recorded as zero. This means that electrons ≥7 MeV are present in insignificantly small numbers and can be ignored. Although Table 7 is only a partial list, this situation holds true for the entire flight of Apollo 11. Since we know all of our electrons are <7 MeV, this establishes an upper limit. Let's compute the maximum range of a 7 MeV electron using equation (10b),

Rmax = 0.530 × 7 – 0.106 = 3.604 g/cm2

This density thickness is considerably less than even the thinnest part of the command module hull. Therefore all electrons will be absorbed by the shielding with none penetrating through to the habitable interior. The radiation dose from electrons goes to zero.

Electrons will, however, produce secondary radiation, which we'll get to later.

Protons

Table 6 below gives the penetration range of high-energy protons in aluminum. The units are mg/cm2, so divide by 1000 to get g/cm2. We can see that, at the same energy, protons are far less penetrating than electrons. However, while our most energetic electrons are <7 MeV, the VARB are home to protons with energies ≥400 MeV. Fortunately the fluxes of these very high-energy particles are low.

Table 6

The range depends on the type of material. The value of R is higher than aluminum in stainless steel, and less than aluminum in the heat shield and insulation. When these materials are layered, they combine to produce approximately the same value of R as aluminum alone. We can therefore use aluminum as an analog for the multi-material composite hull.

We can see from Table 6 that protons with energies ≥100 MeV have sufficient energy to penetrate the hull, though many of these will likely be absorbed by the secondary shielding. The data also reveals that as a proton's energy doubles, its range increases by a factor of about 3.4. Hence, a 200 MeV proton will have a range of about 34 g/cm2, which is enough to pass completely through the spacecraft. It appears that 100 MeV is a logical transition, where we assume protons <100 MeV are absorbed by the shielding, and protons ≥100 MeV penetrate through to the crew cabin. The working assumption, therefore, is that we have 9.854 g/cm2 of shielding, i.e. exactly the value of R for a 100 MeV proton. Although this is a greater rating than the hull alone, it's a conservative assumption since we're ignoring most of the secondary shielding.

Although these high-energy protons will penetrate the shielding, they will lose some of their energy while doing so. For example, our assumption that a 100 MeV proton has just enough energy to penetrate the shielding means it will have effectively no energy remaining afterward, while a proton >100 MeV will retain some or most of its energy. We need a method to determine how much of a proton's energy remains after passing through the shielding.

Figure 9

Figure 9 shows the magnitude of a particle's energy loss relative to its range. The retarding force that causes a particle to loss energy is called stopping power. The force usually increases toward the end of the range and reaches a maximum, the Bragg peak, shortly before the energy drops to zero. The curve that describes the force as function of the material depth is called the Bragg curve. The stopping power of the material is numerically equal to the loss of energy E per unit path length, x.

(12) S = –dE / dx

The equation above defines the linear stopping power, which is in units MeV/cm or similar. Linear stopping power is often divided by the density of the material to obtain the mass stopping power, which is expressed in units MeV-cm2/g or similar.

The deposited energy can be obtained by integrating the stopping power over the entire path length of the particle while it moves in the material. From Figure 9 (and others like it) the following approximating equations have been derived:

(13a) dE = E • [ 0.35 (x/R)2 + 0.45 (x/R) ]

where x/R ≤ 0.8

(13b) dE = E • [ 5.1 (x/R)2 – 7.1 (x/R) + 3 ]

where x/R > 0.8

dE is the change in energy (i.e. the deposited energy), E is the initial energy of the proton, x is the depth of penetration in the material (which, in this case, is the shielding thickness), and R is the proton's range in the material.

For protons ≥100 MeV, the relative relationship between range and energy is approximately, R1/R2 ≈ (E1/E2)1.7. Since the shielding thickness is assumed to equal the range of a 100 MeV proton, we have R1 = x and E1 = 100, therefore,

(14) x / R = (100 / E)1.7

Let's now perform an example by computing the energy lost by a 200 MeV while penetrating the shielding. We have,

x/R = (100/200)1.7 = 0.3078

dE = 200 × [ 0.35 × 0.30782 + 0.45 × 0.3078 ] = 34.3 MeV

And the energy retained by the proton as it makes its way into the crew cabin is,

E = 200 – 34.3 = 165.7 MeV

Just as high-energy protons pass through the hull, they will also pass through the body. Therefore we must compute which protons will be stopped in the body, thus depositing all of their energy, and which will pass through the body, thereby depositing only a fraction of their energy. First let's compute the area density of the human body. As before, this is done by dividing the mass by the surface area. However, since a proton that passes through the body will penetrate the skin twice, we multiply by 2 (this is analogous to a particle having to penetrate the hull twice to pass through the spacecraft).

ρA = (75 × 1000) / (1.8 × 1002) × 2 = 8.33 g/cm2

The range of a 100 MeV proton in body tissue is about 8 g/cm2, so again it looks like 100 MeV is the magic number. We'll assume protons <100 MeV will deposit all their energy in the body, and protons ≥100 MeV will pass through the body. As before we use equation (12) and equation (13a) or (13b) to compute the deposited energy of the ≥100 MeV protons. Be advised that the energy we use for this computation is the energy of the proton after it has penetrated into the spacecraft, not its original energy.

For example, our previous calculation found that the energy of a 200 MeV proton is reduced to 165.7 MeV after penetrating into the crew cabin. Since this proton's energy is still above 100 MeV, it will also penetrate the astronaut's body. The amount of energy deposited in the body is,

x/R = (100/165.7)1.7 = 0.4238

dE = 165.7 × [ 0.35 × 0.42382 + 0.45 × 0.4238 ] = 42.0 MeV

Interestingly we see that the higher the velocity of a particle, the lower the stopping power, i.e. the less energy it deposits per length of track. Therefore, when passing through a given thickness of material, a lower velocity particle will deposit more energy than a higher velocity particle. Counter-intuitively, a 400 MeV proton produces less biological damage than a 100 MeV proton.

Applying the above computational method to the case of Apollo 11, we find that the energy fluence from protons ≥100 MeV is about 4.2×106 MeV/cm2 (total for both VARB transits). This is about 40% of these protons' original energy, with the balance being either deposited in the shielding or retained after passing through the body.

Using equations (8) and (9) to calculate the doses we have,

D = (4.2E+06 × 1.60217657E-13) × (1.8 × 1002) / 75 = 0.00016 Gy = 0.016 rad

H = 0.016 × 2 = 0.032 rem

Thus, we have a radiation dose from protons of 32 mrem, which is less than the dose one would receive from an abdominal x-ray.

Bremsstrahlung

Bremsstrahlung, which is German for braking radiation, is secondary photon radiation produced by the deceleration of charged particles passing through matter. Bremsstrahlung is primarily electron braking radiation. The fraction of the total electron energy that is given up as bremsstrahlung by an electron in the energy range of interest and stopped in a material having atomic number Z may be expressed approximately as,

(15)* f = k × Z × E

where k is a semi-empirical constant having a value of 7×10-4 MeV-1. The energy given up as bremsstrahlung is the product of the electron energy and the fraction converted, E × f.

An approximate energy spectrum of the bremsstrahlung is given by,

(16)* N(Eγ)d(Eγ) ≈ [ 2 × k × Z × (E – Eγ) / Eγ ] × d(Eγ)

where N(Eγ)d(Eγ) is the number of photons having energies between Eγ and Eγ + dEγ. This spectrum is shown graphically below.

(* Evans, R. D., "The Atomic Nucleus," McGraw-Hill Book Co., Inc., New York, 1955)

Figure 10 – Bremsstrahlung spectrum

From Figure 10 it is seen that the spectrum is quite soft; consequently, a material having a high photoelectric cross section would constitute a very effective shield. The most effective shield for electrons in the VARB would be composed of a layer of low-Z material to stop the primary particles followed by a layer of high-Z material to attenuate the bremsstrahlung.

X- or gamma radiation cannot be completely absorbed, but only reduced in intensity, when passing through matter. If mono-energetic gamma radiation attenuation measurements are made under conditions of good geometry, a straight-line relationship between the logarithm of the intensity versus the thickness d of the shield is obtained,

(17) I = I0 × e-μd

where I is the gamma radiation intensity transmitted through an absorber of thickness d, I0 is the gamma radiation intensity at zero absorber thickness, d is the absorber thickness, and μ is the attenuation coefficient.

Since the product μd in the above relation must be dimensionless, if the absorber thickness is measured in cm, then the attenuation coefficient is called the linear attenuation coefficient and has dimension cm-1. If the thickness d is in g/cm2 then the attenuation coefficient is called the mass attenuation coefficient and has units of cm2/g.

Table 7A – Mass attenuation coefficients for ALUMINUM

Table 7B – Mass attenuation coefficients for IRON

(There is no universal consensus for a definition distinguishing between x-rays and gamma rays. One common practice is to distinguish between the two types of radiation based on their source: x-rays are emitted by electrons, while gamma rays are emitted by the atomic nucleus. A common alternative is to distinguish x- and gamma radiation on the basis of wavelength, with radiation shorter than some arbitrary wavelength, such as 10-11 m, defined as gamma radiation.)


We've already computed that negligible electrons will penetrate the hull, but let's now see how much bremsstrahlung they will produce.

The Apollo command module hull was ideally composited to function as a shield for electrons. The heat shield consisted of low-Z hydrogenous material (Z < 4), which overlaid a stainless steel structural hull (Z = 26). Inside of this was another layer of low-Z insulation and then an inner aluminum pressure hull (Z = 13).

To determine the fraction of energy given up as bremsstrahlung, we must know into what shielding layer the electrons will penetrate. From Table 5 we see that the area density of the heat shield is 2.36 g/cm2, thus we can use equation (10b) the compute the energy needed to reach that range,

E = (2.36 + 0.106) / 0.530 = 4.65 MeV

We can therefore say that bremsstrahlung from <5 MeV electrons is produced in the low-Z heat shield, and that from ≥5 MeV electrons is produced in the high-Z stainless steel. No electrons of significance will penetrate beyond the first stainless steel face sheet.

To reach the inside habitable area of the spacecraft, bremsstrahlung generated in the heat shield must penetrate both the stainless steel hull (from Table 5, ~3.5 g/cm2) and the aluminum hull (~0.85 g/cm2). Bremsstrahlung generated in the outer stainless steel sheet must pass through what remains of the stainless steel hull (~2 g/cm2) and the aluminum hull (~0.85 g/cm2). The bremsstrahlung must also make its way through the secondary shielding (~9 g/cm2).

Bremsstrahlung is computed in the same way that we computed the electron energy fluence, except we now include in the analysis equation (15) to convert the electron energy into the appropriate amount of bremsstrahlung. Performing these computations we find that there is a bremsstrahlung energy fluence of 4.14×107 MeV/cm2 generated in the heat shield, and 1.67×105 MeV/cm2 generated in the stainless steel.

Even if every bit of this radiation were absorbed, it would amount to a dose of only 0.16 rem, or 160 mrem. However, before it can become an absorbed dose it must penetrate the shielding.

The amount of penetration depends, in part, upon the energy of the individual photons. For an x-ray beam that contains a spectrum of photon energies, the penetration is different for each energy. The overall penetration generally corresponds to the penetration of a photon between the minimum and maximum energies of the spectrum called the effective energy. The effective energy of an x-ray spectra is the energy of a mono-energetic beam of photons that has the same attenuation as the spectrum of photons. Its exact value depends upon the shape of the spectrum.

To be conservative, it is better to estimate the effective energy too high rather than too low. Referring to the spectrum in Figure 10, it looks like we would be safe in selecting Eeff = 10 keV, which is generally considered the transition between soft and hard x-rays.

We next look up the attenuation coefficients for aluminum and stainless steel; for stainless steel we use iron. From tables 7A and 7B, we find that, for an energy of 10 keV (1.00E-02 MeV), the mass attenuation coefficient for aluminum is 26.2 cm2/g and for iron is 171 cm2/g.

Using equation (17) we compute the fraction of x-ray intensity that is transmitted through the absorbers,

For aluminum:   I / I0 = e-(26.2 × 0.85) = 2×10-10

For stainless steel:   I / I0 = e-(171 × 2.00) = 3×10-149

We can obviously stop at this point. These low energy x-rays will be attenuated to effectively zero long before they ever reach the crew cabin. In fact, a mere 1 millimeter of aluminum will attenuate 99.9% of 10 keV x-rays.

The penetration of x-rays goes up dramatically with energy. Suppose the bremsstrahlung spectrum includes a small number of 50 keV photons; let's see how many of these will make it through the shielding. From Tables 7A and 7B, we have μ = 0.368 cm2/g for aluminum and μ = 1.96 cm2/g for iron. We now have,

For aluminum:   I / I0 = e-(0.368 × 0.85) = 0.731

For stainless steel:   I / I0 = e-(1.96 × 2.00) = 0.0198   (only one face sheet)

The radiation intensity transmitted through both absorbers is the product of the individual fractions,

I / I0 = 0.731 × 0.0198 = 0.0145

So we see that a small but measurable amount of photons from the highest end of the bremsstrahlung spectrum could conceivably penetrate the shielding. However, the above computation assumes the x-rays are traveling in a well collimated beam along the shortest path through the shielding. In actuality the bremsstrahlung will scatter in many directions, with some photons being forced to travel long pathways though the shielding, and others being directed harmlessly out into space. The above computations overestimate the fraction of bremsstrahlung intensity that will actually reach the crew cabin. Even though some hard x-rays are likely to make it through the shielding, these are so small in number that the dose will be negligible.

Although bremsstrahlung is usually associated with electrons, the protons will also give up some of their energy as bremsstrahlung. Even though there exists some high-energy protons, on average they have energies comparable to the electrons. Because protons are less penetrating than electrons, nearly all will be stopped in the heat shield, with only those having energies >45 MeV able to penetrate deeper. Like electron bremsstrahlung, proton bremsstrahlung will be effectively attenuated to zero by the inner layers of the hull.

We can conclude that the radiation dose from bremsstrahlung is near zero.

Summary

Based on my analysis of electrons, protons, and bremsstrahlung, the predicted total dose received by the Apollo 11 astronauts as a consequence of their transits of the Van Allen radiation belts was only about 32 mrem, or 0.016 rads (all from protons ≥100 MeV). This shows that the Apollo trajectories though the VARB were not only survivable, but that the radiation doses received were inconsequential. Of course the VARB were not the only source of radiation to which the crews were exposed. To record the actual skin doses, the astronauts worn dosimeters. These dosimeter measurements for all the Apollo missions are summarized in Table 8 (Apollo 7 and 9 were Earth orbit missions).

Table 8