I’ve decided not to wait for Mr. Keller and devised my own way of analyzing this problem.
Let’s use Apollo 11 for an example, which is also the mission for which we have the best photos of the engine’s effect on the lunar surface. At 26 seconds prior to engine shutdown, Aldrin reported an altitude of 40 feet announced that they were “picking up some dust.” 18 seconds before shutdown, Aldrin reported an altitude of 20 feet and a forward velocity of 4 ft/s. Contact light (altitude of approximately 5 feet) occurred 3 seconds before shutdown.
Since the engine exhaust rapidly expanded outward after leaving the nozzle, as the LM descended and neared the surface, the exhaust was concentrated into a smaller and smaller area. While still 20 feet or more above the ground, the surface area impacted by the exhaust was quite large. It wasn’t until the final seconds that the exhaust stream was concentrated into a small enough area that we should concern ourselves with the possibility of it blasting out a crater. I think a reasonable assumption is to say the exhaust was concentrated into an area twice the diameter of the nozzle exit for a period of about five seconds. We'll ignore the effect of horizontal velocity. Since the LM was drifting horizontally, the exhaust was continually sweeping over new surface, thereby diminishing its effect and lessening the chance of a crater forming.
The Apollo 11 LM had an initial launch mass of 33,278 lbm. During descent 17,414 lbm of propellant was burned, making its landing mass approximately 15,864 lbm. In 1/6th gravity the LM’s weight on the Moon was about 2,644 lbf. To hover over the surface, the descent engine had to counterbalance the vehicle’s weight, thus the engine’s thrust near the surface was approximately 2,644 lbf.
The LM’s descent engine had a specific impulse of 311 seconds. Therefore, the amount of propellant burned each second was 2,644 / 311 = 8.50 lbm. In the 5-second window that we’re concerned with, about 42.5 lbm (19.28 kg) of propellant was burned.
The effective exhaust gas velocity of the engine is the specific impulse times standard gravity, or 311 x 9.80664 = 3,050 m/s. The actual velocity would be less than this because effective exhaust gas velocity includes the contribution of pressure thrust. Nonetheless, let’s us the higher number because this is conservative and will favor the HB argument.
In the 5-second window that I say we have a chance of a crater forming, the total kinetic energy of the exhaust gas expelled is,
KE = 19.28 x 3050
2 / 2 = 89,676,000 J
The big question now is how much of this energy is transferred to the lunar soil. Let’s make the totally absurd and ultra conservative assumption that 100% is transferred to the soil.
It has been estimated that the dust blown away by the engine exhaust attained velocities of about 0.6 to 1.5 miles per seconds (
Source), or about 1,000 to 2,400 m/s. Let’s assume the lower end of this range, which is again conservative and favors the HB argument.
If 100% of the exhaust energy goes into propelling lunar soil at a velocity of 1,000 m/s, then the mass of soil needed to carry this energy is,
m = 89,676,000 x 2 / 1000
2 = 179.35 kg
The bulk density of lunar soil at the surface is approximately 1.3 g/cm
3 (
Source, page 6). This increases to about 1.52 g/cm
3 at a depth of 10 cm, but we’re not going to get this deep so let’s use the lighter surface density. The volume occupied by 179.35 kg of material at a density of 1.3 g/cm
3 is,
179.35 x 1000 / 1.3 = 137,962 cm
3The assumption was that the exhaust stream was concentrated into an area twice the diameter of the engine nozzle exit. The descent engine’s nozzle had a diameter of about 137 cm (54 inches). We are, therefore, forming our crater over a zone with a diameter of 274 cm, giving us a surface area of 58,965 cm
2. Thus, the depth of material removed is,
Depth = 137,962 / 58,965 = 2.34 cm (0.92 inch)
That’s not much of a crater. And this was assuming all of the gas energy goes into excavating it, which we know is certainly not true. The gas continued to carry a large amount of the energy as the gas itself sped away at high velocity. The amount absorbed by the soil was probably actually quite small.
So in conclusion, no appreciable crater is to be expected.