Robert A. Braeunig
Despite overwhelming evidence that American astronauts landed on the Moon during the period 1969-1972, there are many conspiracy theorists that dispute this. One of their arguments point to the fact there is no evidence of a pronounced crater beneath the engine nozzle of the lunar module (LM). They insist a crater should have been formed by the erosive action of the engine's exhaust stream. The typical response to this is that the exhaust stream was not strong enough or concentrated enough to excavate a crater, but can this be proven scientifically? This is the purpose of this web page. We will show that the transfer of kinetic energy from the engine exhaust to the lunar soil was insufficient to produce an obvious crater.
Although any of the Apollo missions could be used for the purpose of our analysis, we'll study Apollo 11 since the action of this LM's exhaust on the lunar surface was most closely examined and documented. The first thing we must do is to recreate the path the LM took to the lunar surface during the final seconds before landing. Although we lack the data to know this precisely, we can create an approximation from the radio communications of the Apollo 11 astronauts. During the final landing phase, lunar module pilot Buzz Aldrin frequently reported the LM's altitude, vertical velocity, and horizontal velocity. Aldrin first reported "picking up some dust" at GET (ground elapsed time) 102:45:17, so this is a good place to start. From the mission transcript, we have the following data:
|102:45:40||"Contact Light" (altitude approx. 5 ft)|
|102:45:43||"Shutdown" (contact with surface)|
This isn't much data to work with, but we can fill in the blanks with educated guesses. Figure 1 below is an approximation of Apollo 11's approach during the final 26 seconds leading up to touchdown. It is assumed the LM's descent during the final three seconds is vertical, which, conservatively, results in the greatest concentration of the exhaust stream. By "conservative" it is meant that the assumptions are worse than reality. For instance, if the LM were still drifting forward during the final seconds prior to touchdown, as it likely was, the exhaust stream would be sweeping across the surface rather than being concentrating on a single spot. Concentrating the exhaust into a small area produces a deeper excavation that reveals itself more conspicuously. In all cases conservative assumptions, i.e. the worst than can be, will be made to favor the argument of the conspiracy theorists. If this conservative approach shows that no conspicuous crater is produced, we can be sure that reality will be even less.
Exhaust Gas Power
Determining the power of the exhaust gas requires that we know the velocity and mass flow rate of the exhaust stream. Exhaust velocity can be calculated from the engine specific impulse, which for the LM's descent engine was 311 seconds at full throttle. In our case the engine was operating in a throttled down condition. Throttling the propellant flow to the engine reduces the combustion chamber pressure, which results in a lower exhaust gas velocity, and consequently, a lower specific impulse. The mathematics to determine this are complex and more than I'd like to present in this article, but my calculations have estimated the specific impulse to be about 305 s in the throttle range of interest.
Normally we would obtain the gas velocity by multiplying the specific impulse by standard gravity (32.174 ft/s2), but this method gives the effective exhaust gas velocity, which includes the effect of unbalanced pressure forces at the nozzle exit. We're interest only in the thrust that comes from the momentum of the exhaust gases. For a large expansion ratio nozzle, pressure thrust is typically very small. Again the mathematics are complex, but it's been determined that in our particular case the pressure thrust is about 3.7% of the total. Therefore, the exhaust gas velocity at the nozzle exit is,
Ve = 305 × 0.963 × 32.174 = 9,450 ft/s (2,880 m/s)
Before we can derive the mass flow rate, we must determine the thrust of the engine. This can best be achieved by calculating the thrust required to perform the maneuvers shown in Figure 1. We obtain the mass of the LM from Selected Mission Weights, where we see the mass at lunar landing was 16,153.2 lbm. By performing a simple simulation, it's been found that the thrust settings shown below in Table 2 replicate the movements illustrated in Figure 1. It was assumed that the abrupt change in vertical velocity centered on T-18 s was attained by a two-second long increase in thrust. Once we have the thrust, the mass flow rate, , is easily calculated by dividing the thrust by the specific impulse. Power is then calculated using the equation v2/2. Below is a sample calculation when the thrust is 2,652 lbf,
= 2,652 / 305 = 8.695 lbm/s (3.944 kg/s)
P = 3.944 × 2,8802 / 2 = 16.36×106 W
Table 2 summarizes the changes in thrust, exhaust gas mass flow rate, and exhaust gas power for the 26-second period leading up to touchdown. It should be noted that the power is that at the nozzle exit.
|-26 to -19||2,707||8.875||16.70|
|-19 to -17||3,205||10.508||19.77|
|-17 to 0||2,652||8.695||16.36|
If you look at Figure 1 you'll see that the exhaust stream is shown to diverge into a conical shaped stream after exiting the nozzle. This is typical of a rocket operating in a vacuum or near-vacuum environment. Since there is no external air pressure to confine the stream, the gas will immediately expand into the surrounding vacuum. The photo below is a good example showing the expanding exhaust plume of a rocket traveling through a low-pressure environment. So how quickly does the stream expand? This is a difficult question to answer, but a conservation assumption is to use a 15-degree divergent cone half-angle as shown in Figure 1. Fifteen degrees is selected because this is the standard angle used in the design and construction of conical-shaped nozzles. We'll assume the gas continues to expand at this standard angle after exiting the nozzle, though in reality it will most assuredly expand at an even faster rate. Using this small angle keeps the exhaust concentrated into a narrow crater-forming stream. It is also important to note that the diameter of the nozzle exit is 63 inches (1.60 meters). (Added 21-Aug-15 — The nozzle exit diameter on Apollo 11 was 59 inches. It was not until Apollo 15 that the nozzle was upgraded to 63 inches. This error is insignificant in the overall analysis.)
As the exhaust gas expands its pressure and temperature decreases, thus so does its enthalpy. Enthalpy is the thermodynamic potential of the system. Most of this potential is used to produce thrust by expanding the gas inside the nozzle; however, as the gas continues to expand outside the nozzle, some of the remaining enthalpy is converted to kinetic energy. Since we've assumed the gas expands in a cone with a half-angle of 15-degrees, it's possible to calculate the enthalpy of the gas at a given distance from the nozzle exit. The calculations are complex, but it has been determined that the following equation approximates the enthalpy at the nozzle exit less the enthalpy at some distance "D" (in feet) aft of the nozzle. The answer, therefore, represents the increase in kinetic energy of the gas in the exhaust stream. The units are J/kg.
Hexit– H = -13640 LN(D)3 + 114500 LN(D)2 - 8160 LN(D) + 109900
We can now calculate the total power of the exhaust gas as it impacts the lunar surface. For example, let's examine the situation at T-3 s (see Figure 1). At this time the power of the exhaust as it exits the nozzle is 16.36 MW (from Table 2). The bottom of the nozzle is about two feet above the landing pads, therefore the distance of the nozzle exit from the lunar surface is 7 feet. We add to the power at the nozzle exit the increase in power resulting from the expansion of the gas,
Ptotal = 16.36×106 + 3.944 × (-13640 LN(7)3 + 114500 LN(7)2 - 8160 LN(7) + 109900) = 18.04×106 W
You may have noticed that we've used a mix of U.S. and metric units. This is because NASA typically uses the FPS system while thermodynamic data is usually given in the MKS system. From here on all work will be performed using metric units.
Mapping the Exhaust Stream
To better illustrate the LM's flight path, Figure 2 below is a to-scale plot of the LM's altitude and range as it approached the lunar surface. Markers placed one second apart indicate time. It is during the final twelve seconds that most of the crater-forming action takes place.
We now have the data necessary to determine how the exhaust gas mass is distributed over the lunar surface as the stream sweeps across the ground. To show how this is done we'll choose an example. Let's select a point lying on the centerline of the LM's flight path and 13 meters up range of the final landing spot. If we define the landing point as X=0, then our point can be defined as X-13 meters. We're concerned only with the period of time that X-13 lies within the exhaust stream. First contact occurs when the leading edge of the exhaust stream touches X-13, mid-point occurs when the centerline of the exhaust stream contacts X-13, and final contact occurs when the trailing edge of the exhaust stream touches X-13. Figure 1 below illustrates these events along with pertinent dimensional data and the time of occurrence.
Note that the LM is descending as it passes over X-13. The distance and time from A-to-B is greater than the distance and time from B-to-C because the footprint diameter of the exhaust stream reduces as the LM descends closer to the surface. Also note that the intensity of the exhaust stream increases as the footprint shrinks because the gas is being concentrated into a smaller and smaller area. We can determine the intensity of the stream at the surface by dividing the mass flow by the footprint area. Figure 4 below is a graph of the exhaust gas mass flow per square meter versus time for the three events illustrated in Figure 3. The three points have been fitted with a curve, for which we've derived an equation.
The total mass deposited per unit area at point X-13 is obtained by integrating the mass flow equation from time T-24.36 to T-18.88, as follows:
By repeating the above operation thousands of times, a map can be generated showing the distribution and intensity of the exhaust gas at the lunar surface. Figure 5 below shows the distribution of mass across the LM's flight path. The actual computer program used to generate Figure 5 used a non-calculus approximation of the solution demonstrated above. The greatest intensity of gas is 6.332 kg/m2 at the position indicated.
Likewise, Figure 6 below is a map showing the distribution and intensity of the exhaust gas kinetic energy at the lunar surface. The method used to generate Figure 6 is similar to that described for Figure 5. This is the energy of the gas as it impinges on the surface, not the energy transferred to the lunar soil. The greatest intensity of kinetic energy is 29.08 MJ/m2 at the position indicated.
Digging a Crater
As the exhaust gas impacts the ground, a portion of its kinetic energy is transferred to the soil. The kinetic energy transferred to the soil causes the soil particles to fly away at high velocity, leaving behind a void where the soil once was. From this news article we find that the velocity of the lunar dust has been estimated to be between 0.6 and 1.5 miles per second (about 1,000 to 2,400 m/s). It is logical to assume the gas velocity (after losing energy to the soil) is the same as the dust. That is, the dust particles are propelled by the gas and are carried along with the gas flow at the same speed. We'll assume the low end of the velocity range, i.e. 1,000 m/s, because this is ultra-conservative. The lower the final gas velocity, the more energy it gives up to the soil. Furthermore, the lower the soil velocity, the more mass is required to carry the energy. All this adds up to a bigger crater.
To be further conservative, we'll assume no loss of kinetic energy. In reality, the kinetic energy at the end of the process will be less than that before because some of the energy will be used to perform other actions, such as dislodging or detaching soil particles. There will also be some dissipation of energy due to friction, turbulence, etc.
From the preceding section, we know the mass and kinetic energy of the exhaust gas as it contacts the ground. After colliding with the ground and losing energy to the soil, the gas will slow to 1,000 m/s. Knowing the final velocity of the gas we can calculate its final kinetic energy. Using the maximum mass value from Figure 5, we have
KEgas = 6.332 × 10002 / 2 = 3.17×106 J/m2
The amount of energy transferred to the soil is equal to the amount given up by the gas (assuming no loses). Using the maximum kinetic energy from Figure 6, we have the following. Note that in this scenario 89% of the gas kinetic energy is transferred to the soil.
KEsoil = 29.08×106 – 3.17×106 = 25.91×106 J/m2
Now that we know the kinetic energy and velocity of the soil, we can calculate the mass of soil needed to carry this energy,
m = 2 × 25.91×106 / 10002 = 51.82 kg/m2
To convert the above to a depth, we simply divide by the soil density. From Geotechnical Properties of Lunar Soil (Page 6), we read "On average, the bulk density, r, is approximately 1.30 g/cm3 at the surface, increases rapidly to 1.52 g/cm3 at a depth of 10 cm, then more gradually to 1.83 g/cm3 at a depth of 100 cm." Since we're dealing essentially with surface material, we'll use the low number of 1.30 g/cm3 (1,300 kg/m3), which is also conservative because a lower density means a larger volume of material is excavated. We have,
Depth = 51.82 / 1300 = 0.0399 m, or 39.9 mm (1.57 inches)
This is the maximum depth taken at the point where we have the greatest concentration of exhaust gas mass and kinetic energy. Since the mass and energy is distributed unevenly over the lunar surface, the depth will vary. Figure 7 below shows the depth of soil removed from the area near where the LM came to rest.
Despite our conservative assumptions, this is a very small amount of soil. Figure 8 below is a sectional view cut through the center of the crater and drawn with equal horizontal and vertical scales. Although some soil is removed, the hole left behind would be hardly noticeable among all the naturally occurring ground undulations. There is clearly no large and obvious crater as claimed by the conspiracy theorists.
Of all our assumptions, perhaps the one that most dramatically affects the results is the final velocity of the gas and soil. For instance, if we used the middle of the velocity range, i.e. 1,700 m/s, rather than 1,000 m/s, we get a maximum crater depth of just 10.6 mm (0.42 inch). In reality, the exhaust stream blew away very little soil because there wasn't enough energy in the gas to move a large volume of surface material.
Many conspiracy theorists claim not only that a crater should have been formed, but also that there is no evidence whatsoever of the exhaust's interaction with the ground. This claim is patently false. Below is a photograph showing the area beneath the Apollo 11 lunar module. Here we see that much of the fine powdery surface dust observed in other Apollo photos has been blown away. We also see radial streaks and other features produced by the erosive action of the exhaust gas. Generally, the photo looks much like what we'd expect given the results of the above analysis.
This analysis is nothing but a theoretical construct designed to form a correlation between the energy of the exhaust gas and the volume of soil displaced. The actual interaction between the exhaust gas and the lunar soil is far more complex than has been taken into consideration here. Despite the limitations of this analysis, the conclusion is inescapable: no pronounced crater is formed. No more soil can be removed than there is energy available to detach, entrain, and transport it away. The energy of the exhaust gas has been determined to a reasonably high degree of confidence, and the energy require to produce a large crater is simply not present.