Robert A. Braeunig
© November2009
Most attempts to illustrate the trajectories of the Apollo flights to the Moon are intended to show only the primary elements of the mission, with little attention to geometric accuracy and scale. One such example is shown below. Unlike these common nottoscale drawings, this page takes a new approach. After determining the orbital elements of Apollo 11's translunar trajectory, I've calculated its position versus time and accurately plotted, in correct proportion and orientation, its flight path to the Moon. Furthermore, Apollo 11's trajectory through the region of the Van Allen Radiation Belts has been mapped to show how the trajectory was designed to bypass the most intense areas of the this potentially dangerous obstacle.
Orbital Elements
If you don't like math, I recommend you skip ahead because this next part is going to be math intensive. I've included the calculations for those who might be interested in checking my work and/or learning the computational methods.
All data used in the calculations comes from the following:
Apollo by the Numbers  Translunar Injection
And, unless noted otherwise, all mathematical equations and procedures come from another of my web pages:
Rocket & Space Technology  Orbital Mechanics
From the Apollo by the Numbers web page, we have the following data for Apollo 11 at the time of Translunar Injection (TLI):
GET: 002:50:13.03
KSC Date: 16Jul1969
GMT Date: 16Jul1969
KSC Time: 12:22:13 PM
Time Zone: EDT
GMT Time: 16:22:13
Altitude (ft): 1,097,229
Altitude (n mi): 180.581
EarthFixed Velocity (ft/sec): 34,195.6
SpaceFixed Velocity (ft/sec): 35,545.6
Geocentric Latitude (deg N): 9.9204
Geodetic Latitude (deg N): 9.983
Longitude (deg E): 164.8373
Flight Path Angle (deg): 7.367
Heading Angle (deg E of N): 60.073
Inclination (deg): 31.383
Descending Node (deg): 121.847
Eccentricity: 0.97696
C3 (ft^{2}/sec^{2}): 14,979,133
Although this is a good start, we need to do more to obtain the six elements needed to fully describe the orbit. Referring to my Orbital Mechanics web page, let's use the method described in the section titled "Launch of a Space Vehicle". The flight path angle, f, and velocity, v are both given, which are 7.367^{o} and 35,545.6 ft/s respectively. Let's first convert velocity to SI units (mks):
v = 35,545.6 × 0.3048 = 10,834.3 m/s
From the altitude we can calculate the length of the radius vector, knowing that altitude is the perpendicular distance from the vehicle center of gravity to the Fischer Ellipsoid of 1960. The radius of the ellipsoid is given by the formula,
R = a × b / SQRT[ (b × cos)^{2} + (a × sin)^{2} ]
Where a is the equatorial radius of Earth, b is the polar radius of Earth, and is the geocentric latitude. The accepted values of a and b in 1960 were 6,378,166 m and 6,356,784.3 meters, respectively. The ellipsoid radius at the given latitude of 9.9204^{o} is,
R = 6378166 × 6356784.3 / SQRT[ (6356784.3 × cos(9.9204))^{2} + (6378166 × sin(9.9204))^{2} ] = 6,377,528 m
Therefore, the magnitude of Apollo 11's radius vector at TLI was (converting nautical miles to meters),
r = 6,377,528 + (180.581 × 1,852) = 6,711,964 m
From equations (4.29), (4.30) and (4.31) we calculate the perigee and apogee radii, the eccentricity, and the true anomaly:
C = 2 × 3.986005E+14 / (6,711,964 × 10,834.3^{2}) = 1.011851
(R_{p}/r) = (C + SQRT[ C^{2}  4 × (1  C) × cos^{2}(7.367) ] ) / (2 × (1  C)) = 0.9833646
R_{p} = 0.9833646 × 6,711,964 = 6,600,308 m
(R_{a}/r) = (C  SQRT[ C^{2}  4 × (1  C) × cos^{2}(7.367) ] ) / (2 × (1  C)) = 84.39725
R_{a} = 84.39725 × 6,711,964 = 566,471,300 m
e = SQRT[ (6,711,964 × 10,834.3^{2} / 3.986005E+14  1)^{2} × cos^{2}(7.367) + sin^{2}(7.367) ] = 0.976965
n = arctan[ (6,711,964 × 10,834.3^{2} / 3.986005E+14) × cos(7.367) × sin(7.367) / ( (6,711,964 × 10,834.3^{2} / 3.986005E+14) × cos^{2}(7.367)  1 ) ] = 14.9087^{o}
The semimajor axis is simply the average of the perigee and apogee radii:
a = ( 6,600,308 + 566,471,300 ) / 2 = 286,535,800 m
Next, we need a method to transform coordinates from one plane to another, which we accomplish using the following (from Astronomical Formulae for Calculators, 4th edition, by Jean Meeus):
tan a = ( sin l × cos i  tan b × sin i ) / cos l
sin d = sin b × cos i + cos b × sin i × sin l
where a = equatorial longitude, d = equatorial latitude, l = orbital longitude, b = orbital latitude, and i = inclination of the orbital plane. b is, of course, always equal to zero because the spacecraft never leaves the orbital plane. Setting b = 0, the equations simplify to the following:
tan a = tan l × cos i
sin d = sin l × sin i
These transformations use the location of the ascending node as the origin for measuring longitude. Once the transformation is complete, the resulting answer must be converted to either geographic longitude or celestial longitude, as required.
We're given the inclination and the geocentric latitude of the TLI launch point, which are 31.383^{o} and 9.9204^{o} N respectively. From this we calculate l and a as follows:l = arcsin[ sin(9.9204) / sin(31.383) ] = 19.3189^{o}
a = arctan[ tan(19.3189) × cos(31.383) ] = 16.6614^{o}
The argument of perigee, w, is equal to l of the launch point minus the true anomaly:
w = 19.3189  14.9087 = 4.4102^{o}
The geographic longitude of the ascending node equals the geographic longitude of the launch point, 164.8373^{o} E, minus the launch point a:
Geographic longitude of ascending node = 164.8373  16.6614 + 360 = 178.5013^{o} E
The celestial longitude of the ascending node, W, is equal to the local sidereal time at the geographic longitude at the time of TLI:
W = LST @ 178.5013^{o} E, 16Jul69, 16:22:13 GMT = 23^{h} 53' 31" = 358.380^{o}
Apollo by the Numbers gives the longitude of the descending node as 121.847^{o}, though exactly what this number represents is not explained. After a considerable amount of investigation and number crunching, I've concluded this is the degrees of longitude east of the launch site at GET=0. Since the sidereal time over Launch Pad 39A at GET=0 was 56.604^{o}, the celestial longitude of the descending node is 56.604 + 121.847 = 178.451. The ascending node is 180^{o} opposed at 358.451^{o}. This checks very closely with my calculated value of 358.380^{o}. The reason for the difference is unknown; however, I'm using the calculated valve to maintain internal consistency between all the numbers.
The only orbital element remaining is the time of perigee passage. Referring to the Orbital Mechanics page, we'll use the method described in the section titled "Position in an Elliptical Orbit". First, we calculate the time between the last perigee and the spacecraft's true anomaly at TLI:
E = arccos[ (0.976965 + cos(14.9087)) / (1 + 0.976965 × cos(14.9087)) ] = 0.0282451 radians
M = 0.0282451  0.976965 × sin(0.0282451) = 0.000654295 radians
n = SQRT[ 3.986005E+14 / 286,535,800^{3} ] = 0.00000411624 rad/s
t = 0.000654295 / 0.00000411624 = 159 s
The time of perigee passage is, therefore, the time of TLI minus 159 seconds:
T = 002:47:34 GET = 16:19:34 GMT 16Jul69
Recapping, Apollo 11's translunar orbital elements were as follows:
Locating Apollo 11
Now that we have the orbital elements, we can produce the calculations needed to locate Apollo 11 in threedimensions. In general, the procedure is: (1) select a time, (2) calculate the true anomaly (by first calculating the mean anomaly and eccentric anomaly), (3) calculate the radial distance, (4) calculate the longitude in the orbital plane, (5) transform to equatorial longitude, (6) convert to celestial longitude, and (7) calculate the equatorial latitude. And, for information, let's also (8) calculate the flight path angle and (9) calculate the spacefixed velocity. For example, suppose we want to perform the calculations for 004:00:00 GET; we go about it like this:
Time since perigee passage = 4 × 3600  (2 × 3600 + 47 × 60 + 34) = 4,346 s
Mean anomaly, M = 4,346 × 0.00000411624 = 0.0178892 radians
Eccentric anomaly, E = 0.0178892 + 0.976965 × sin E
by iteration, E = 0.382848 radians
True anomaly, n = arccos[ (cos(0.382848)  0.976965) / (1  0.976965 × cos(0.382848)) ] = 121.765^{o}
Radius, r = 286,535,800 × (1  0.976965^{2}) / (1 + 0.976965 × cos(121.765)) = 26,866,200 m
Flight path angle, f = arctan[ 0.976965 × sin(121.765) / (1 + 0.976965 × cos(121.765)) ] = 59.684^{o}
Velocity, v = SQRT[ 3.986005e+14 × 286,535,800 × (1  0.976965^{2}) ] / (26,866,200 × cos(59.684)) = 5,318 m/s
Orbit longitude, l (from ascending node) = 121.765 + 4.4102 = 126.175^{o}
Equatorial longitude, a (from ascending node) = arctan[ tan(126.175) × cos(31.383) ] + 180* = 130.581^{o}
Equatorial longitude, a (celestial) = 358.380 + 130.581  360 = 128.961^{o}
Equatorial latitude, d (celestial) = arcsin[ sin(126.175) × sin(31.383) ] = 24.858^{o}
(* The arctan function is problematic because calculators and computers cannot distinguish the quadrant of the angle. For instance, arctan(1) = 45^{o} and 225^{o}. Care must be taken when working with coordinates to make certain the angle is placed in the correct quadrant.)
Using the procedure outlined above, the following table was produced. The polar coordinates of distance, longitude and latitude can be used to pinpoint Apollo 11's position in space at any time. It should be noted that all the calculations in this web page treat the orbit as a simple twobody program, i.e. Earth and the spacecraft. Apollo 11 is assumed to coast freely in space under the gravitational influence of only Earth. The gravity of the Moon, and all other perturbing forces, are ignored.
Apollo 11  Translunar Phase  
GET (h:m:s) 
Greenwich Mean Time (date, time) 
True Anomaly (degrees) 
Flight Path Angle (deg) 
Velocity (m/s) 
Distance from Earth (m) 
Equatorial Coordinates (Celestial)  
Longitude (^{o})  Latitude (^{o})  
002:50:13  7/16/69, 16:22:13  14.9087  7.367  10,834.3  6,711,964  15.0416  9.9204 
003:00:00  7/16/69, 16:32:00  58.5356  28.894  9,532.0  8,641,760  57.4909  27.6309 
003:15:00  7/16/69, 16:47:00  91.1895  44.913  7,646.2  13,318,712  94.9318  31.2164 
003:30:00  7/16/69, 17:02:00  106.5866  52.398  6,531.9  18,095,084  112.5874  29.0897 
003:45:00  7/16/69, 17:17:00  115.6544  56.766  5,819.2  22,613,282  122.5194  26.7882 
004:00:00  7/16/69, 17:32:00  121.7654  59.684  5,318.1  26,866,368  128.9617  24.8574 
004:15:00  7/16/69, 17:47:00  126.2411  61.804  4,941.4  30,889,122  133.5460  23.2717 
004:30:00  7/16/69, 18:02:00  129.7060  63.431  4,644.6  34,715,890  137.0177  21.9543 
004:45:00  7/16/69, 18:17:00  132.4955  64.731  4,402.6  38,374,976  139.7650  20.8413 
005:00:00  7/16/69, 18:32:00  134.8077  65.800  4,200.0  41,888,995  142.0111  19.8860 
005:15:00  7/16/69, 18:47:00  136.7677  66.700  4,027.0  45,276,043  143.8935  19.0543 
005:30:00  7/16/69, 19:02:00  138.4588  67.470  3,876.7  48,550,764  145.5025  18.3215 
005:45:00  7/16/69, 19:17:00  139.9392  68.139  3,744.5  51,725,157  146.8997  17.6690 
006:00:00  7/16/69, 19:32:00  141.2507  68.728  3,626.8  54,809,187  148.1289  17.0828 
006:30:00  7/16/69, 20:02:00  143.4830  69.719  3,425.5  60,738,374  150.2036  16.0682 
007:00:00  7/16/69, 20:32:00  145.3262  70.526  3,258.3  66,391,605  151.9006  15.2155 
007:30:00  7/16/69, 21:02:00  146.8855  71.198  3,116.2  71,808,685  153.3254  14.4842 
008:00:00  7/16/69, 21:32:00  148.2297  71.770  2,993.2  77,020,296  154.5461  13.8470 
008:30:00  7/16/69, 22:02:00  149.4063  72.264  2,885.3  82,050,701  155.6090  13.2843 
009:00:00  7/16/69, 22:32:00  150.4489  72.695  2,789.4  86,919,495  156.5468  12.7819 
009:30:00  7/16/69, 23:02:00  151.3824  73.076  2,703.3  91,642,791  157.3832  12.3294 
010:00:00  7/16/69, 23:32:00  152.2255  73.415  2,625.4  96,234,035  158.1363  11.9184 
010:30:00  7/17/69, 00:02:00  152.9929  73.719  2,554.4  100,704,600  158.8196  11.5427 
011:00:00  7/17/69, 00:32:00  153.6957  73.993  2,489.3  105,064,209  159.4439  11.1971 
011:30:00  7/17/69, 01:02:00  154.3432  74.242  2,429.2  109,321,264  160.0177  10.8777 
012:00:00  7/17/69, 01:32:00  154.9426  74.470  2,373.6  113,483,080  160.5478  10.5810 
013:00:00  7/17/69, 02:32:00  156.0206  74.870  2,273.3  121,545,956  161.4986  10.0452 
014:00:00  7/17/69, 03:32:00  156.9668  75.211  2,185.1  129,295,990  162.3305  9.5726 
015:00:00  7/17/69, 04:32:00  157.8077  75.504  2,106.6  136,767,677  163.0679  9.1510 
016:00:00  7/17/69, 05:32:00  158.5629  75.759  2,036.0  143,989,147  163.7286  8.7711 
017:00:00  7/17/69, 06:32:00  159.2470  75.983  1,972.0  150,983,728  164.3260  8.4260 
018:00:00  7/17/69, 07:32:00  159.8713  76.180  1,913.6  157,771,041  164.8703  8.1102 
019:00:00  7/17/69, 08:32:00  160.4448  76.354  1,859.8  164,367,793  165.3695  7.8195 
020:00:00  7/17/69, 09:32:00  160.9747  76.510  1,810.2  170,788,358  165.8300  7.5505 
021:00:00  7/17/69, 10:32:00  161.4666  76.649  1,764.0  177,045,222  166.2571  7.3002 
022:00:00  7/17/69, 11:32:00  161.9253  76.773  1,720.9  183,149,320  166.6549  7.0665 
023:00:00  7/17/69, 12:32:00  162.3547  76.884  1,680.6  189,110,296  167.0270  6.8474 
024:00:00  7/17/69, 13:32:00  162.7582  76.984  1,642.7  194,936,715  167.3762  6.6413 
026:00:00  7/17/69, 15:32:00  163.4980  77.155  1,573.1  206,215,699  168.0159  6.2627 
028:00:00  7/17/69, 17:32:00  164.1625  77.293  1,510.6  217,038,742  168.5896  5.9220 
030:00:00  7/17/69, 19:32:00  164.7650  77.403  1,453.9  227,448,909  169.1092  5.6126 
032:00:00  7/17/69, 21:32:00  165.3156  77.491  1,402.1  237,482,107  169.5836  5.3294 
034:00:00  7/17/69, 23:3200  165.8224  77.559  1,354.3  247,168,686  170.0198  5.0684 
036:00:00  7/18/69, 01:32:00  166.2915  77.609  1,310.1  256,534,595  170.4233  4.8266 
038:00:00  7/18/69, 03:32:00  166.7280  77.645  1,269.0  265,602,250  170.7985  4.6014 
040:00:00  7/18/69, 05:32:00  167.1362  77.667  1,230.5  274,391,185  171.1492  4.3906 
042:00:00  7/18/69, 07:32:00  167.5194  77.677  1,194.4  282,918,557  171.4781  4.1926 
044:00:00  7/18/69, 09:32:00  167.8805  77.677  1,160.4  291,199,543  171.7880  4.0058 
046:00:00  7/18/69, 11:32:00  168.2218  77.666  1,128.2  299,247,650  172.0808  3.8292 
048:00:00  7/18/69, 13:32:00  168.5456  77.646  1,097.7  307,074,968  172.3583  3.6616 
051:00:00  7/18/69, 16:32:00  169.0018  77.600  1,054.7  318,425,471  172.7493  3.4252 
054:00:00  7/18/69, 19:32:00  169.4273  77.536  1,014.7  329,335,876  173.1138  3.2046 
057:00:00  7/18/69, 22:32:00  169.8260  77.456  977.1  339,834,316  173.4551  2.9978 
060:00:00  7/19/69, 01:32:00  170.2012  77.360  941.8  349,945,393  173.7762  2.8031 
063:00:00  7/19/69, 04:32:00  170.5556  77.249  908.4  359,690,771  174.0795  2.6191 
066:00:00  7/19/69, 07:32:00  170.8917  77.124  876.8  369,089,655  174.3670  2.4445 
069:00:00  7/19/69, 10:32:00  171.2114  76.984  846.8  378,159,169  174.6404  2.2784 
072:00:00  7/19/69, 13:32:00  171.5164  76.830  818.1  386,914,660  174.9011  2.1199 
075:00:00  7/19/69, 16:32:00  171.8081  76.662  790.7  395,369,948  175.1505  1.9683 
Mapping the Trajectory
Data like that shown in the above table allows for the mapping of the trajectory. By converting the polar coordinates to rectangular (X,Y,Z) coordinates, the trajectory can be plotted graphically and shown from different perspectives. Before we do that, however, let's first take a look at what the trajectory looks like from an Earthcentered view:
This is an insideout view showing the track of Apollo 11 across Earth's sky, indicated by the dark blue line. The light blue markers show the location of Apollo 11 in fourhour increments. The yellow disc is the position of the Sun at TLI, the gray circle is the Moon at TLI, and the gray disc is the position of the Moon at lunarorbit insertion (075:49:50 GET). The coordinates are shown in units of right ascension and declination, which are the normal coordinates used for locating objects on the celestial sphere in the equatorial system. Declination is equivalent to latitude and right ascension is analogous to longitude with one hour equaling 15 degrees. (Note that right ascension increases righttoleft because of the insideout perspective.) Since the viewing perceptive is geocentric, the apparent position of the spacecraft will be different for an observer on the surface of Earth due to parallax. The apparent motion of the spacecraft decreases dramatically after the first several hours due to (1) the spacecraft is slowing down and (2) it is moving mostly away from us rather than across our field of view.
Let's now examine the flight path in three dimensions. For this exercise I've defined three axes, as follows:
These axes were so defined to depict the trajectory in the correct orientation, i.e. north up and orbit counterclockwise.
The first illustration is a "Top View" looking down on Earth's north pole along the Zaxis. The path of Apollo 11 is represented by the blue line and starts at 002:50:13 GET (TLI) and ends at 075:49:50 GET (LOI). The aqua disc is Earth and the gray disc is the Moon as it was positioned at lunar orbit insertion. Both axes are drawn to the same scale with the units in kilometers.
The second illustration is a "Side View" looking along the Yaxis in the direction of 270^{o} longitude. In this view, Earth's axis is straight upanddown (the Zaxis) and we're viewing the equatorial plane edgeon (the Xaxis). The drawing scale is the same as the Top View.
The third illustration is an "End View" looking along the Xaxis in the direction of 180^{o} longitude. Like the Side View, Earth's axis is vertical and the equator is seen edgeon. The scale here is larger than the Top and Side Views.
Together, these drawings show the translunar trajectory in all three dimensions. With some spatial visualization, it should not be too hard to see the flight path Apollo 11 followed en route to the Moon. Generally, it is an elongated ellipse inclined along the minor axis.
Something you likely noticed in these illustrations is that the trajectory misses the Moon. How is that possible? There are two reasons. First, Apollo 11 made a midcourse correction, which has been omitted from my calculations. Second, and most significantly, the gravity of the Moon has been ignored. As Apollo 11 traveled away from Earth, it started to be pulled toward the Moon, thus deflecting the trajectory. In these illustrations we see the path Apollo 11 would have followed had the Moon not been there.
To illustrate how the Moon influences the flight path, the following is a simulation of a freereturn trajectory. This is just a generic illustration and does not represent any specific mission. Note how the trajectory begins to noticeably bend as the spacecraft nears the 350,000 km mark.
FreeReturn Trajectory
Van Allen Radiation Belts
A common claim of the moon landing conspiracy theorists is that Apollo was impossible because the Van Allen Radiation Belts (VARB) form an impenetrable barrier to human space flight beyond low Earth orbit. Aside from the fact that the man after which the VARB are named, Dr. James Van Allen, has specifically repudiated the claim, there are several things wrong with this theory. One reason is specifically relevant to this web page  the Apollo missions didn't fly straight through the teeth of the VARB, they mostly went around them.
The Van Allen Radiation Belts are a torus of energetic charged particles circling Earth around its magnetic equator and held in place by Earth's magnetic field. The VARB are split into two distinct belts, with energetic electrons forming the outer belt and a combination of protons and electrons forming the inner belt. The energy and density of the particles varies by many orders of magnitude depending on where inside the VARB one is located.As we have seen from the previous illustrations, Apollo 11's translunar trajectory was inclined, allowing the spacecraft to rise rapidly above the equatorial plane as it headed away from Earth. (Although we've been dealing specifically with Apollo 11, all Apollo missions flew similar trajectories.) Since we know Apollo 11's equatorial latitude and radial distance from Earth, we can easily calculate the spacecraft's horizontal and vertical distances from Earth relative to the geographic equator and axis of rotation. Apollo 11's distance from Earth in the equatorial plane is R×cos(d), and the altitude above the equator is R×sin(d). The following graph shows Apollo 11's altitude above the equatorial plane relative to its horizontal distance from Earth:
Unlike the previous illustrations, which attempted to accurately depict what the orbit looked like from various perspectives, the above does not represent any specific viewing angle. The radial movement of the spacecraft is depicted in a flattened out profile view. Motion tangential to Earth's surface is in the third dimension not visible. The red markers indicate the spacecraft position in 10minute increments, with four hours total being shown.
The above is helpful, but it doesn't really tell us what we need to know, since the VARB circle the geomagnetic equator rather than the geographic equator. We need to transform the coordinates to yet another reference plane. To do so, we must know the inclination and ascending node of the geomagnetic plane in relation to geographic equator. In 1969, the north geomagnetic pole was located at approximately 78.6^{o} N and 70.2^{o} W. The inclination angle is, therefore, 90  78.6 = 11.4 degrees. The longitude of the ascending node is 90^{o} ahead of the north geomagnetic pole, or 70.2^{o} + 90 = 19.8^{o} E.
Unlike the ascending node of the orbital plane, which is fixed in relation to the stars, the ascending node of the geomagnetic plane is fixed to Earth. As Earth rotates, the node moves through 360 degrees of celestial longitude each day, and is, therefore, time dependent. The celestial longitude of the ascending node at the time of TLI is equal to the local sidereal time: LST @ 19.8^{o} E, 16Jul69, 16:22:13 GMT = 13^{h} 18' 43" = 199.68^{o}. The node advances at a rate of 360 / 86,164.1 = 0.00417807^{o}/sec.
Locating Apollo 11 in relation to the geomagnetic plane is similar to the method employed earlier: (1) determine the equatorial coordinates, (2) calculate the longitude of the geomagnetic plane's ascending node, and (3) transform the coordinates using the previously given equations. After deriving the spacecraft's latitude in the geomagnetic coordinate system, we can plot Apollo 11's horizontal and vertical distances from Earth relative to the geomagnetic equator and axis, as follows:
Let's now overlay the trajectory on flux maps of the VARB (resizing to match scale). The first map shows the electron flux and the next two maps show the proton flux at two energy levels. As you can see, Apollo 11 easily avoided the areas of highest flux, thereby minimizing the exposure. As before, the red markers indicate the time in 10minute increments. The far edge of the electron belt was reached in about 90 minutes, the inner zone was traversed in about 30 minutes, and the region of the most energetic particles was skirted in just about 10 minutes.
From AP8 Trapped Proton Environment for Solar Maximum and Solar Minimum, National Space Science Data Center, December 1976.
From AP8 Trapped Proton Environment for Solar Maximum and Solar Minimum, National Space Science Data Center, December 1976.
Below is another VARB map showing the radiation exposure rate in rads per second. We see clearly that Apollo 11 bypassed the most intense zones, instead skirting along the edges. Furthermore, in agreement with the electron belt overlay, we see Apollo 11 leaving the vicinity of the VARB after about 90 minutes.
Of course, the outbound leg was only half the journey; Apollo 11 had to also fly past the VARB on the return trip. Applying to the transearth trajectory the same computational methods used to analyze the translunar trajectory, we can determine Apollo 11's position relative to the VARB (see below). Like before, the trajectory was optimized to bypass the VARB to the greatest extent possible. This time, however, Apollo 11 approached from the south and was traveling at an even greater inclination and velocity than it had on its way to the Moon. The spacecraft contacted the fringes of the VARB for a period of about 60 minutes.
From the above we can estimate the theoretical radiation dose received; however, we must do some extrapolation because, for a time, Apollo 11's flight path strayed outside the mapped area. I've done my best to estimate how long Apollo 11 was in each of the colored zones, thereby allowing a calculation of the radiation dose. Be advised, however, that the above map is just an average representation of the VARB. In fact, the VARB are not uniformly distributed around Earth, being compressed on the sunward side and elongated on the opposite side. Furthermore, the radiation intensity varies depending on solar activity. The actual geometry and intensity of the VARB at the time and direction of Apollo 11's flight may be different than that shown.
Estimated Radiation Dose  
Zone  Duration (min)  Rate (Rad/s) 
Dose (Rads) 

Outbound  Inbound  
Blue  53  33  0.0001  0.52 
Green  17  15  0.001  1.92 
Yellow  10  10  0.005  6.00 
Orange  5  0  0.01  3.00 
TOTAL  11.44 
Rad is a unit of absorbed dose, while rem is a unit of dose equivalent and its biological effects in man. The dose equivalent in rems is equal to the absorbed dose in rads multiplied by a quality factor, Q, for the specific type of radiation. For beta radiation (electrons) Q = 1, and for highenergy protons Q = 10. The VARB consists of both electrons and protons, with electrons forming the outer belt and a combination of protons and electrons forming the inner belt. As can be seen from the above, most of the dose is received from the outer belt, i.e. from beta radiation**. It is impossible to accurately determine the dose equivalent from the limited data provided here, but a reasonable guess might place the value at somewhere around 2025 rem (Q ≈ 2). A person will experience radiation sickness with a dose of 100200 rem, and death with a dose of 300+ rem.
This is, of course, the radiation dose that an unprotected astronaut, i.e. one outside the spacecraft, would receive. The Apollo 11 astronauts where safely inside a spacecraft having a multilayer hull with a total thickness of several centimeters (see below) and a shielding rating of 7 to 8 g/cm^{2}. This hull provided excellent shielding from the particulate radiation of the VARB, as it was constructed from materials ideal at stopping this type of radiation. The actual radiation dose received by the astronauts throughout the entirety of the Apollo 11 mission was less than one rem, as measured by dosimeters.
(** Please note that this is true only for the case of the unprotected astronaut, as beta radiation lacked the energy necessary to penetrate the spacecraft shielding. It was the highest energy protons of the inner radiation belt that posed the only real concern in trajectory planning and, as we have seen, Apollo was beyond this region in as little as ten minutes.)
From Virtual Apollo, Apogee Books, Scott P. Sullivan.