Robert A. Braeunig
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 not-to-scale 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.
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):
KSC Date: 16-Jul-1969
GMT Date: 16-Jul-1969
KSC Time: 12:22:13 PM
Time Zone: EDT
GMT Time: 16:22:13
Altitude (ft): 1,097,229
Altitude (n mi): 180.581
Earth-Fixed Velocity (ft/sec): 34,195.6
Space-Fixed 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
C3 (ft2/sec2): -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 7.367o 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 9.9204o 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 × 10,834.32) = 1.011851
(Rp/r) = (-C + SQRT[ C2 - 4 × (1 - C) × -cos2(7.367) ] ) / (2 × (1 - C)) = 0.9833646
Rp = 0.9833646 × 6,711,964 = 6,600,308 m
(Ra/r) = (-C - SQRT[ C2 - 4 × (1 - C) × -cos2(7.367) ] ) / (2 × (1 - C)) = 84.39725
Ra = 84.39725 × 6,711,964 = 566,471,300 m
e = SQRT[ (6,711,964 × 10,834.32 / 3.986005E+14 - 1)2 × cos2(7.367) + sin2(7.367) ] = 0.976965
n = arctan[ (6,711,964 × 10,834.32 / 3.986005E+14) × cos(7.367) × sin(7.367) / ( (6,711,964 × 10,834.32 / 3.986005E+14) × cos2(7.367) - 1 ) ] = 14.9087o
The semi-major 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 31.383o and 9.9204o N respectively. From this we calculate l and a as follows:
l = arcsin[ sin(9.9204) / sin(31.383) ] = 19.3189o
a = arctan[ tan(19.3189) × cos(31.383) ] = 16.6614o
The argument of perigee, w, is equal to l of the launch point minus the true anomaly:
w = 19.3189 - 14.9087 = 4.4102o
The geographic longitude of the ascending node equals the geographic longitude of the launch point, -164.8373o E, minus the launch point a:
Geographic longitude of ascending node = -164.8373 - 16.6614 + 360 = 178.5013o 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 @ 178.5013o E, 16-Jul-69, 16:22:13 GMT = 23h 53' 31" = 358.380o
Apollo by the Numbers gives the longitude of the descending node as 121.847o, 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 56.604o, the celestial longitude of the descending node is 56.604 + 121.847 = 178.451. The ascending node is 180o opposed at 358.451o. This checks very closely with my calculated value of 358.380o. 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 / 286,535,8003 ] = 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 16-Jul-69
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 three-dimensions. 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 space-fixed 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)) ] = 121.765o
Radius, r = 286,535,800 × (1 - 0.9769652) / (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)) ] = 59.684o
Velocity, v = SQRT[ 3.986005e+14 × 286,535,800 × (1 - 0.9769652) ] / (26,866,200 × cos(59.684)) = 5,318 m/s
Orbit longitude, l (from ascending node) = 121.765 + 4.4102 = 126.175o
Equatorial longitude, a (from ascending node) = arctan[ tan(126.175) × cos(31.383) ] + 180* = 130.581o
Equatorial longitude, a (celestial) = 358.380 + 130.581 - 360 = 128.961o
Equatorial latitude, d (celestial) = arcsin[ sin(126.175) × sin(31.383) ] = 24.858o
(* The arctan function is problematic because calculators and computers cannot distinguish the quadrant of the angle. For instance, arctan(1) = 45o and 225o. 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 two-body 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|
|Greenwich Mean Time
|Equatorial Coordinates (Celestial)|
|Longitude (o)||Latitude (o)|
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 Earth-centered view:
This is an inside-out 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 four-hour 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 lunar-orbit 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 right-to-left because of the inside-out 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 Z-axis. 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 Y-axis in the direction of 270o longitude. In this view, Earth's axis is straight up-and-down (the Z-axis) and we're viewing the equatorial plane edge-on (the X-axis). The drawing scale is the same as the Top View.
The third illustration is an "End View" looking along the X-axis in the direction of 180o longitude. Like the Side View, Earth's axis is vertical and the equator is seen edge-on. 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 mid-course 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 free-return 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.
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 figure is a simple two-dimensional plot of height above the equator versus distance from Earth. It is important to note that the spacecraft was swinging around Earth at the same time it was moving up and away. The red markers indicate the spacecraft position in 10-minute 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 78.6o N and 70.2o W. The inclination angle is, therefore, 90 - 78.6 = 11.4 degrees. The longitude of the ascending node is 90o ahead of the north geomagnetic pole, or -70.2o + 90 = 19.8o 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 @ 19.8o E, 16-Jul-69, 16:22:13 GMT = 13h 18' 43" = 199.68o. The node advances at a rate of 360 / 86,164.1 = 0.00417807o/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 10-minute 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. Again note that the trajectory as shown is a simple X-Y data plot that is not intended to depict the actual 3D shape or appearance of the orbit.
From AP-8 Trapped Proton Environment for Solar Maximum and Solar Minimum, National Space Science Data Center, December 1976.
From AP-8 Trapped Proton Environment for Solar Maximum and Solar Minimum, National Space Science Data Center, December 1976.
Estimating the Radiation Dose
Knowing the trajectory we can estimate the radiation dose received by the astronauts. For this analysis please see my companion article,
Apollo and the Van Allen Belts (an estimate of the radiation dose received)