Has anyone ever done the calculations as to how much energy it takes to blast a crater from impact or explosives and compared that to the amount of energy delivered by the LM's descent engine on landing?
It seems to me that it wouldn't even be in the same ballpark, but I'm too math-challenged to find out.
"Straydog (Duane Daman) effectively managed to get himself banned from his own site." - Waspie_Dwarf
I don't think these crater forming mechanics would translate very well to a high velocity gas.
"Show us a photo of either the descent or the ascent rockets firing and the exhaust plume. And exactly how does a space craft fly when "The engine is NOT ON"" -Stelios at the DIF Google search this site
I messed around with some calculations years ago but I don't have anything formalized that would be much help.
There is a user at the BAUT forum named jrkeller who, as I recall, claimed to have done the calculations. I once knew jrkeller's full name and title, but I've forgotten it since I haven't been active at BAUT for several years. He's a thermal and fluids engineer or something impressive sounding like that. He and I had quite a bit of interaction several years ago when we teamed up to debunk some silly global warming related theory. I came away from that debate with a pretty healthy respect for the guy's knowledge and experience.
I don't know if he's still active at BAUT, but you could try sending him a PM to see if you get a response. If he still has those calculations (or can reproduce them), it would be a good thing to have for future reference.
The following is a list of reviews written by John R. Keller at Amazon.com. I'm pretty certain this is the guy.
If someone sends him a PM, he should get an email alert provided his BAUT profile is up to date with his current email address. That might get a reply from him even if he isn't regularly visiting BAUT anymore.
I suppose I ought to be the one to contact him since I'm the one who recalls him saying something about doing the calculations. He also might remember me from our previous discussions (even though it was several years ago). Unless someone else jumps up to volunteer, I'll try to PM him sometime before the end of the week (provided my account is still active at BAUT and I can remember my password).
Not much Apollo hoax stuff going on atm on BAUT, it's all UFO stuff. Yawn.
Well, if you're bored and happen to be in the mood to debunk some HB's claims, feel free to participate in this massive and on going 414 page thread at ATS. The current subject just happens to be about the lack of blast craters under the LM...
If John can produce the calculations in a presentable form, I'd like to add it to my web page for future reference. Perhaps you can suggest this to him and get his permission to reproduce the work. I can probably take whatever format he has it and make the necessary conversions to post it on the web.
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 30502 / 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 / 10002 = 179.35 kg
The bulk density of lunar soil at the surface is approximately 1.3 g/cm3 (Source, page 6). This increases to about 1.52 g/cm3 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/cm3 is,
179.35 x 1000 / 1.3 = 137,962 cm3
The 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 cm2. 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.
Brilliant. Why didn't someone do this earlier. I am so sick of the blast crater nonsense.
Just one thing, could you reference the source for the bulk density of the lunar soil for completeness sake.
Fine work sir.
Thanks. The method of figuring out the volume of lunar material needed to carry the amount of kinetic energy transferred to it from the exhaust gas didn't occur to me until this morning. It seemed like a pretty simple solution once the idea popped into my head.
The source for the soil bulk density is on page 6 of the following document (I've edited my previous post accordingly):
It seems a pretty elegant model to me and the worst case assumptions lead to a pretty small depression. I'd be surprised if it was 1/10th of that in reality. I'd hope most can follow the maths and it serves as an excellent starting point for a discussion on the topic when the matter is raised.
The exhaust is expelled from the engine traveling downward. It then impacts the surface, transfers some of its energy to the soil, and then flows radially outward. The soil particles flow outward along with the gas. If we assume the gas and soil particles speed outward at the same velocity, we can calculate the before and after energy of the gas and how much is transferred to the soil.
In my previous example, I assumed the outward flow had a velocity of 1,000 m/s. Therefore, if the gas velocity slows from 3,050 m/s at the nozzle exit to 1,000 m/s after impacting the surface, its kinetic energy reduces from 89.676 MJ to 9.640 MJ. In this case, the energy transferred to the soil is 80.036 MJ (89%). The mass of soil needed to carry this energy is 160.07 kg. This equates to a depth of 2.09 cm (0.82 inch) over the specified area.
Let’s now consider the gas and soil traveling at the high end of the velocity range, i.e. 2,400 m/s. In this case the energy of the gas reduces from 89.676 MJ to 55.526 MJ, transferring 34.150 MJ to the soil (38%). Not only is the energy of the soil significantly reduced, but also because it is traveling at a much higher velocity, the mass of soil needed carry this energy is dramatically reduced. The mass of soil is now only 11.86 kg, which equates to a depth of only 0.155 cm (0.061 inch).
The real answer is probably somewhere between these numbers. In even the worse case, we have only a minor amount of erosion with no obvious crater.
(The percent of energy transferred to the soil is higher than I would have guessed.)
Here's another entirely different way to go at it, as a sort of cross check. It occurred to me when you said that the DPS was burning about 8.5 lbm/sec of propellant during final approach.
Those hypergolic propellants have roughly the same energy density as TNT. (For representing the yields of nuclear weapons, 1 ton of TNT is considered to yield 1 million kcal or 4184 MJ.) So over your 5-second cavity-excavating period of interest, 42.5 lbm or about 19 kg of propellant was burned.
What if we just detonated 19 kg of TNT on the lunar surface? How big a crater would that make?
I am sure there must be detailed tables relating crater size to impactor size and velocity and hence impact energy, but I don't know where they are offhand.
I can say that your 90 MJ of total propellant energy would correlate to a ~32 kg object impacting the moon at its escape velocity of 2.38 km/s (the minimum velocity of any primary lunar impact) or only 200 grams hitting at 30 km/sec, the earth's orbital velocity around the sun. Depending on the original solar orbit of the impactor, the velocity could be anything up to over 60 km/sec.