The Apollo Space Program – A gigantic conspiracy?

Wolfgang G. Gasser

For the notation of numbers in dark-red see: A simplified and less confusing notation for numbers

Original discussion: International Skeptics Forum, Kinetic energy at atmospheric reentry from lunar mission

o    Kinetic energy at atmospheric reentry from lunar mission:  Posting #1

o    Essential postings on Apollo moon rock samples:  #172,  #184,  #420,  #435,  #533 and following

o    Slow Motion and other Techniques of Apollo Camera Work

#1  –  2016-08-02

Kinetic energy at atmospheric reentry from lunar mission

Escape velocity from Earth is 11.2 km/s = 4p112 m/s. A spacecraft leaving the gravitational field of the moon and falling on the Earth accelerates to a speed of 11 km/s = 4p11 m/s before reaching the atmosphere at an altitude of around 120 km = 5p12 meter (see: Atmospheric entry). The kinetic energy corresponding to 11 km/s = 4p11 m/s is around 60 Mega-Joule = 7p6 Joule for every kilogram of spacecraft and crew. For comparison: energy content of jet fuel, gasoline and diesel is only around 45 MJ per kg = 7p45 J/kg (source).

Figure 4.1.7-25 of
Returning from Space: Re-entry shows "Re-entry Profiles for the Shuttle Versus Gemini and Apollo":

"Notice Gemini and Apollo re-entered much more steeply than the Space Shuttle."

[This paragraph contains an instructive mistake, see next post:]

According to this figure, Apollo needed around 250 sec = 2p25 s to descent from altitude 120 km = 5p12 m to 60 km = 4p6 meter. This results in an average speed of v = 60 km / 250 s = 4p6 m / 2p25 s = 2p24 m/s = 240 m/s. To a speed of 240 m/s correspond only 28 kilo-Joule = 4p28 J per kg, which is a negligible quantity in comparison to the 60 Mega-Joule = 7p6 J corresponding to the entry speed of 11 km/s = 4p11 m/s. Thus in the order of 99.9% = 9n999 of the kinetic energy must have disappeared before falling to an altitude of 120 km = 5p12 m. Yet friction due to atmospheric pressure above 120 km = 5p12 m altitude is too weak to relevantly reduce speed.

Wikipedia on meteors:

"Meteors become visible between about 75 to 120 km above the Earth. They usually disintegrate at altitudes of 50 to 95 km."

The return capsule of Stardust (bringing down to Earth dust samples in 2006) reached its maximum deceleration of 34 g (!) at an altitude of only 55 km = 4p55 meter:

"On January 15, 2006, at 05:57:00 UTC, the Sample Return Capsule successfully separated from Stardust and re-entered the Earth's atmosphere at 09:57:00 UTC, at a velocity of 12.9 km/s, the fastest reentry speed into Earth's atmosphere ever achieved by a man-made object. The capsule followed a drastic reentry profile, going from a velocity of Mach 36 to subsonic speed within 110 seconds. Peak deceleration was 34 g, encountered 40 seconds into the reentry at an altitude of 55 km over Spring Creek, Nevada."

#14  –  2016-08-04

Dave Rogers in #8:

I would point out that velocity is a vector, not a scalar, and the graph of altitude vs. time can only be used to calculate the vertical component of that vector. It doesn't tell you anything at all about the horizontal component. Your numbers for the speed (i.e. the magnitude of the velocity vector) are therefore completely wrong; you'd need to know the angle of descent at every point on the re-entry profile to correct them.

Ok. My embarrassing calculation is based on two errors: First, "More steeply" doesn't mean steep. Second, the shallower the angle, the higher is resulting average speed because a longer path is needed to lower altitude. Mistakenly I assumed the opposite when writing the post. Nevertheless, the quoted figure cannot be correct. Quote from At what angle did Apollo 13 need to reenter?:

"All of the Apollo lunar returns were within about a tenth of a degree of -6.5° (halfway between -5.3° and -7.7°), except for Apollo 13, which was within a quarter of a degree."

As sin 6.5° ≈ 0.11 resp. sin 0p65° ≈ 9n11 , around 8.8 km = 3p88 meter are needed to lower altitude by one kilometer = 3p1 meter. This means, average speed during descent from 120 km = 5p12 m to 60 km = 4p6 m is not 240 m/s as I erroneously calculated, but 8.8 ∙ 240 m/s = (3p88 m / 3p1m) ∙ 2p24 m = 3p21 m = 2.1 km/s (assuming unchanged angle). As the Apollo curve is more or less a straight line in the 120 – 60 km range, the figure implies that Apollo missions moved at 2.1 km/s = 3p21 m/s when reaching altitude 120 km = 4p12 m with 6.5° = 0p65°.

To a speed v of 2.1 km/s = 3p21 m/s corresponds only ½ v2 = ½ ∙ 3p21 ∙ 3p21 J/kg ≈ 6p22 J/kg = 2.2 Mega-Joule per kg, which is a rather negligible quantity in comparison to the 60 MJ = 7p6 J corresponding to the entry speed of 11 km/s = 4p11 m/s. Thus around 96% = 9n96 of the kinetic energy would have disappeared before falling to an altitude of 120 km = 5p12 m. For an astronaut of 80 kg these "missing" 96% = 9n96 would result in 4.7 Tera-Joule = 9n47 Joule, enough to heat up ten tons (10000 kg = 4p1 kg) of steel by 1000° Kelvin.

#20  –  2016-08-08

Missing kinetic energy for one 80 kg astronaut is 4.7 Giga-Joule = 9p47 J and not 4.7 Tera-Joule = 12p47 Joule.

Nevertheless 4.7 GJ = 4.7 billion Joule = 4.7 ∙ 109 J = 9p47 Joule is actually enough to heat up ten tons (10000 kg = 4p1 kg) of steel by 1000 Kelvin = 3p1 Kelvin.

In the context of computers, the use of giga = 9p = nine-po of some years ago has been replaced more and more by tera = 1012 = 12p = twelve-po. Therefore I confused Tera-Joule with Giga-Joule. Yet my calculation is not affected by this mistake:

Specific heat capacity of steel is 470 Joule per kg per Kelvin = 2p47 J/kg/K = two.po four seven J/kg/K. With 470 J = 4.7 ∙ 102 J = 2p47 J we can heat up 1 kg by one degree Kelvin. For ten metric tons = 4p1 kg we need 4p1 kg ∙ 2p47 J/kg/K = 6p47 Joule/K = 4.7 MJ/Kelvin. In order heat up these ten tons = 4p1 kg of heat capacity 2p47 J/kg/K by 1000 Kelvin = 3p1 K, we need 9p47 Joules.

Summary: 2p47 Joule/kg/Kelvin ∙ 4p1 kg ∙ 3p1 Kelvin = 9p47 Joule

#25  –  2016-08-10

Gord_in_Toronto in #5:

Since it has been scientifically proven that re-entry is impossible, low Earth orbit must be crowded with Astronauts. QED.

The International Space Station is at an altitude of around 350 km = 5p35 m = five.po three five meter in the thermosphere. This is less than 3% = 8n3 = eight-ne three of the Earth's diameter. Due to atmospheric friction, the ISS must be regularly accelerated. Otherwise it would fall on the Earth.

The distance to the Moon with around 380,000 km = 8p38 meter is around 30 times = 1p3 the Earth's diameter and around 1000 times = 3p1 = 1p3 / 8n3 the distance from the Earth's surface to the ISS. This makes a huge difference!

It is true that kinetic and potential energy which has to be converted by atmospheric friction into thermal energy during re-entry is only twice as high in the lunar-mission case. But these additional 30 MJ/kg = 7p3 Joule per kg make a huge difference. From the ISS even a sailplane (glider aircraft) could safely reach the Earth, at least in principle.

#30  –  2016-08-15

abaddon in #26:

Did you have a point to make?

Jack by the hedge in #27:

No. I don't really know what the point was either.

Thor 2 in #29:

Can't see a point being made here Wolfgang, sorry.

Here again the sarcastic comment from Gord in post #5:

"Since it has been scientifically proven that re-entry is impossible, low Earth orbit must be crowded with Astronauts. QED."

Gord's comment implies that the many astronauts from the times of Gagarin to our days having visited the thermosphere could not have returned if a major difficulty existed in returning from a lunar mission. Thus, in #25 I explained that a huge difference prevails between returning from a stable orbit very close to the Earth's surface and coming back from an orbit around the moon.

The historic evolution of flights to orbits in the thermosphere is the normal one: After a first astronaut had visited this outer region of the atmosphere, this achievement has been repeated with increasing ease. Other example: After the first person(s) succeeded in climbing Mount Everest, more and more people have repeated this achievement with increasing ease. The same has been happening all the time with achievements in technology (e.g. automobile, airplane, computer), in science and in sports.

There is only one prominent exception to the rule that a first achievement can be repeated with increasing ease: the Apollo program! Interestingly, the increasing ease effect even worked normally during the Apollo program itself, culminating in driving a rover and singing on the moon. The big discontinuity only occurred after the Apollo program.

#44  –  2016-08-17

Dr.Sid in #9:

In reality, for each spacecraft, there is 'reentry angle'. If the spacecraft reentries at this angle, it will safely decelerate before diving too deep into denser air.

It is only a myth resp. unfounded belief that every spacecraft has a safe 'reentry angle'. The central problem is kinetic energy. If it substantially exceeds the energy of a stable orbit near terrestrial surface, then we have a problem. The speed of such a stable orbit at "reentry interface" altitude (120 km = 5p12 m) is 7.83 km/s = 3p783 m/s. The speed of the ISS is 7.7 km/s = 3p77 m/s. (If we add the potential energy of ISS with respect to "reentry interface", we get a theoretical speed of 7.94 km/s = 3p794 m/s.)

A spacecraft falling with 11 km/s = 4p11 m/s (even at optimal angle) resembles rather Fig. 4.1.7-9. Meteor Re-entering the Atmosphere:

"Recall, its velocity is pretty much constant initially, while it is high in the thin atmosphere. But then, it hits a wall as the atmosphere thickens and it slows rapidly."

Dr.Sid in #9:

Old style single use reentry modules … are built to have high drag coefficient (and high drag induced stability). Thus they will use much steeper reentry angle.

It is the other way round: The "steeper reentry angle" is the necessity resp. cause, and "high drag coefficient (and high drag induced stability)" is the solution resp. effect.

Dr.Sid in #9:

But old style reentry modules are not controlled at descent stage. So if they relied on shallow reentry angle, there would be danger that they slow down too much at high altitude, which will then change the reentry angle to deeper one, just to overheat at lower altitude.

Rather the opposite happens in reality: The steeper the reentry angle is, the faster a lower altitude is reached and the more acute is the overheat problem. The real problem with a shallow angle is that the spacecraft cannot slow down enough: Instead of changing the reentry angle to a deep enough one, the landing module goes off again on an elliptical tangent. And slowing down from more than 11 km/s = 4p11 m/s to around 7.8 km/s = 3p78 m/s needs to remove almost 30 Mega-Joule = 7p3 J of kinetic energy for each kilogram of the landing module. Also, the entry speed of 11 km/s = 4p11 m/s ≈ 1125 sec ∙ g  ≈ 3p1125 s ∙ 0p98 m/s2  is so high that the effect of the Earth's acceleration on further development of reentry angle is rather marginal.

Giordano in #11:

By the way - most of the meteors that burn up on reentry are very small - the size of dust or sand. No thermal mass, no place for the heat to go, and they are gone. … Even slightly larger meteors survive, hit the surface with no engineered solution, and are called meteorites. I have a small one at home. Larger ones survive and can create a lot of damage.

Meteors either burn up (almost) entirely, or they or their fragments still have huge kinetic energy per mass when reaching surface. In case of controlled reentry, all the kinetic energy must be converted by friction into heat, yet without burning the object being the primary cause of this friction.

#55  –  2016-08-18

wogoga in #44:

In case of controlled reentry, all the kinetic energy must be converted by friction into heat, yet without burning the object being the primary cause of this friction.

jadebox in #46:

No. Very little of the kinetic energy during reentry is lost to friction. Friction is not the main reason that heat is produced during reentry.

I used friction in the general sense of an interaction with the atmosphere leading to air drag, to air resistance. Whether a particle of the atmosphere directly interacts with the reentry module or indirectly via other particles having previously interacted directly or indirectly does not matter. At least in everyday language, the whole process can be called friction, i.e. kinetic energy of an object is transformed into heat by more or less direct contact.

Jrrarglblarg in #47:

Old literature may still refer to scrubbing velocity off as friction heat but it's been a couple of decades now that analysis of fallen space trash has revealed that it's compression of air.

Deceleration by interaction with air obviously leads to compression of the air in front of the decelerating object. I cannot imagine that this knowledge was not state of the art also "a couple of decades" ago.

By the way, rocket fuel with the highest energy content seems to be the combination of liquid hydrogen and liquid oxygen. Combustion energy of pure hydrogen is 140 MJ/Kg = 8p14 Joule per kg if oxygen is taken for granted. As we need 8 times more oxygen (atomic weight: 16) than hydrogen (atomic weight: 2 times 1), we get an energy content of only 15.5 MJ/kg = 7p155 J/kg = 8p14 ∙ 2/(16+1+1) for a fuel consisting of hydrogen and oxygen.

I'm actually astonished at this result which makes understandable that it is rather impractical to use fuels for reentry: In order to transport one kilogram of rocket fuel to an orbit around the moon, we need around 60 MJ
= 7p6 Joule. The energy content of 1 kg of fuel itself however is only around 15 MJ = 7p15 Joule and even cannot be transformed into kinetic energy without losses.

#68  –  2016-08-19

wogoga in #55:

At least in everyday language, the whole process can be called friction, i.e. kinetic energy of an object is transformed into heat by more or less direct contact.

jadebox in #57:

No, that is wrong. Drag and friction have little to do with the heat produced during reentry. Here is a link to some (mostly) good explanations:

Quotes from Why does a spacecraft heat up during reentry? concerning the use of "friction":

"When a spacecraft reenters, very little heat is generated by friction."

"Many people think that it's due to friction with air passing over the surface of the re-entering vehicle, but that's actually a relatively minor component. Most of the heating comes from the fact that, when you slam into a column of air really hard, really fast, it compresses; and when you compress air, it heats up. And that heat will be transferred, by direct conduction, to the object doing the compressing – i.e., the spacecraft."

"It has nothing to do with friction."

"This speed is reduced entirely through friction (drag) of the atmosphere. The thicker the atmosphere, the more the friction."

"The important thing is the above simulation is run without considering any friction (no viscous effects)."

"The compression friction heats the air."

"If the vehicle is slowed down by an ablating heat shield, it is friction with the air that heats up the shield."

A reasonable answer seems to me the second one:

"The reason why objects without a heat shield or entering the atmosphere uncontrolled incinerate is simply due to the fact air can't get out of their way fast enough. … As an object encounters the atmosphere, air cannot move out of the way due to the hypersonic velocity, and instead it piles up ahead of the object and compresses into a shock wave. The gases there are compressed rapidly enough to heat up to 20,000 degrees or more [Fahrenheit?], and because of that the radiant heat from the shock wave starts to heat up the incoming object. The heat will either melt or vaporize any material exposed to the shock wave, and since aluminum is used heavily in the construction of spacecraft, the spacecraft tears apart and the pieces melt or vaporize."

Yet the first answer (of a "NASA engineer that has worked spacecraft operations design"), seems rather illogical to me. Here two quotes:

"Through making the reentry vehicle blunt, air can't get out of the way quickly enough, and acts as an air cushion to push the shock wave and heated shock layer forward. Since most of the hot gases are no longer in direct contact with the vehicle, the heat energy would stay in the shocked gas and simply move around the vehicle to later dissipate into the atmosphere."

"Plasma, known as the fourth state of matter, does not conform to the gas laws of conventional thermodynamics, although it does share one familiar property, – a proportionality between pressure and temperature in a contained system. The formation of the pressure wave, therefore, also creates extreme temperatures. The plasma stream is electrostatically-charged too, and so it concentrates at acute surface contours. The resultant effect is particularly intense local heating at the airframe's leading edges.

If we have proportionality between pressure and temperature, then why isn't temperature highest, where pressure is highest? And pressure obviously is the higher, the closer to the heat shield, isn't it?

Dave Rogers in #56:

And there's another critical error in your thinking. It seems fairly clear to me that, because adiabatic compression takes place at greater distances from the falling object, the process of heat transfer into the falling object must be much less efficient than viscous drag. As a result, a much larger proportion of the energy loss goes into heating the air around the falling object, and a much smaller proportion into heating the object itself, than one would expect from viscous drag. Your unspoken suggestion that all the kinetic energy lost is converted into thermal energy in the falling body, as embodied in your calculation of what you expect its heating to be, is therefore completely specious.

Once again: All Apollo modules re-entered atmosphere with a speed quite close to escape velocity from Earth: v = 11.186 km/s ≈ 4p11 m/s, corresponding to a kinetic energy of Ekin = v2 /2 ≈ 62.6 Mega-Joule per kg ≈ 7p6 J/kg. If we assume an average heat capacity of 1000 J/kg/K (and 100% efficient heat transfer) then the heat released during atmospheric deceleration would be enough to heat up the return module of 5.5 tons = 3p55 kg by around 60 000 Kelvin = 4p6 K.

Thus, even if as much as 99.9% = 9n999 of the energy were used to heat up only the atmosphere and the vaporizing heat shield, the remaining 0.1% = 0p0001 = 9n001 = 8n01 = 7n1 would still be enough to heat the whole return module on average by 60 Kelvin = 1p6 K.

So the reproach that my argument is based on "assuming the entirety of the heat generated by atmospheric entry must be borne and rejected by the entrant" (#61) is rather unfounded.

Please imagine concretely: Every single kilogram of the return module releases as much heat as 4 kg rocket fuel do, if it is simply burned! And no action-at-a-distance can heat the atmosphere without affecting the landing module at the same time.


Nitpicking often results from lack of genuine objections

#75  –  2016-08-22

Forty years anniversary LUNA-24 returning 170 gram lunar rock on 22 Aug 1976

Luna 24 with launch date 9 Aug 1976 and recovery data 22 Aug 1976 was the last successful lunar sample return mission. According to Wikipedia, this was the third success in eleven attempts of the Soviets to bring down small amounts of moon rock to Earth. The task of returning 170 gram = 9n17 kg of moon rock is incomparably easier than of returning humans whose net weight alone is 200 kg = 2p2 kg or more. Rocks are not sensitive to temperature and high g's (decelerations). And we can take into consideration that the Soviets had been leading in space technology over many years.

In the four decades after Luna 24, only four "Far Earth" sample return missions have been completed, and only one single succeeded in safely landing an extremely small cargo (i.e. dust particles):

·         Stardust: It successfully reentered in 2006 with a peak deceleration of 34 g = 333 m/s2 = 2p333 m/s per sec = 1200 km/s per sec = 745 mph/s.

·         Genesis: "Genesis was launched on August 8, 2001, and crash-landed in Utah on September 8, 2004, after a design flaw prevented the deployment of its drogue parachute."

·         Hayabusa: "The reentry capsule and the spacecraft reentered Earth's atmosphere on 13 June 2010 at 13:51 UTC. The heat-shielded capsule made a parachute landing in the South Australian outback while the spacecraft broke up and incinerated in a large fireball."

·         Phobos-Grunt: "It was launched on 8 November 2011, 20:16 UTC from the Baikonur Cosmodrome, but subsequent rocket burns intended to set the craft on a course for Mars failed, leaving it stranded in low Earth orbit."

It is true that in case of return from "really" outside the Earth's gravitation, reentry velocity is higher than terrestrial escape velocity vesc = 11.2 km/s = 4p112 m/s, whereas in the lunar-mission case, it is normally slightly less than vesc. Nevertheless this difference in kinetic energy has not been substantial.

In meanwhile the Chinese have successfully landed their Chang'e 5 test return-vehicle after coming back from a short lunar orbit:


Xihua/Ren Junchuan

Interestingly, I did not hear about this Chinese success when it actually happened on Oct 31, 2014. Maybe, Western media were then primarily covering the launch and crash of Virgin Galactic's SpaceShipTwo happening on exactly the same day.

Unlike Apollo, no confusion prevails on how the Chinese succeeded in landing the return module on Earth (and in resolving other challenges, see). In comparison to Apollo, the Chinese had not only the advantage of 45 years of technological progress, but also of an economy much stronger than in the United States of the 1960s.

If US landing of humans on the moon actually were a hoax then it would have been an incredibly ingenious way to steal from the "evil communists" the honor of having first landed on a celestial body. By redefining "landing on the moon" by "landing men on the moon" with their propaganda machine, the leading capitalist nation then faked not only one landing, but several landings. The organizers would have dwarfed even Joseph Goebbels by making the falsehood so preposterously exaggerated that nobody could imagine such an audacity in deception.


In the long run TRUTH has the best chance

#89  –  2016-08-29

Giordano in #79:

The Apollo moon landings and returns were BY FAR the most documented explorations and returns of lunar samples (plus astronauts) of any that you cite. Far, far more through multiple independent documentations than any of the other missions you cite.

I would rather call it over-documented. One gets almost the impression that a reporter team was already on the Moon in order to document arrival and departure of Apollo 11. With modern technology it would probably be a child's play to analyze the original documentations (especially the films). Yet essential parts of the original documentations are unavailable for independent analysis, have disappeared, or have maybe even been preventively destroyed (Apollo 11 missing tapes).

Giordano in #79:

Also I have some trouble understanding your nomenclature: 170 grams is 0.17 E-3 kilograms, or 170 E-5 kilograms.

In past days I have become astonished at learning how many people do not even know that mega corresponds to million. At least according to what you write here, also you seem to get confused by orders of magnitude and measurement units with auxiliary numbers such as kilo. 170 gram = 0.17 kg is neither 0.17E-3 kg nor 170E-5 kg:

0.17E-3 = 0.17 ∙ 10-3 = 9n17 ∙ 7n1 = 6n17 = 0.00017 kg
170E-5 = 170 ∙ 10-5
= 2p17 = 5n1 = 7n17 = 0.0017 kg

Giordano in #79:

What is the 2n17 kilograms in your nomenclature?

2n17 = 10-10+2 ∙ 1.7 = 10-8 ∙ 1.7

abaddon in #80:

If you wish to use non-standard notation, fine. What do I care.
But you cannot even get that correct, so what does that tell us?
170 gram is plainly not 9n17 kg no matter how you slice it.

The sequence "100, 10, 1, 0.1, 0.01, 0.001" in A simplified and less confusing notation for numbers: "2p1, 1p1, 0p1, 9n1, 8n1, 7n1".

wogoga in #75:

Rocks are not sensitive to temperature and ...

abaddon in #80:

Care to visit a volcano or two?

Do you suggest that difference in sensitivity to temperature is not relevant between humans and rocks?

#95  –  2016-08-31

Dave Rogers in #56:

It seems fairly clear to me that, because adiabatic compression takes place at greater distances from the falling object, the process of heat transfer into the falling object must be much less efficient than viscous drag.

"The peak g stage will occur once it hits the thick part of the atmosphere at around 65 km altitude." (GPIS 6: Reentry)

Pressure at around 65 km = 4p65 m altitude: 10 Pascal = 1p1 Newton/m2.

Mean free path at 10 Pascal (medium vacuum): in the order of 1 mm = 0.001 m = 7n1 meter. For comparison mean free path at ambient pressure: 68 nm = 68 ∙ 1n1 m = 2n68 m ≈ 2n7 m ≈ 3n1 m ≈ 10-7 m.

"As an object encounters the atmosphere, air cannot move out of the way due to the hypersonic velocity, and instead it piles up ahead of the object and compresses into a shock wave." (#68).

An air particle colliding with the heat shield of the reentry module gets a huge impulse (off the shield). Yet the particle cannot leave the region close to the 12 m2 heat shield easily, because mean free path even in front of the highly compressed shock wave is only 1 mm = 7n1 meter.


Many prefer being further duped to admitting to themselves that they have been duped

#113  –  2016-09-04

wogoga in #75:

In the four decades after Luna 24, only four "Far Earth" sample return missions have been completed, and only one single succeeded in safely landing an extremely small cargo (i.e. dust particles).

John Nowak in #86:

Bonus points for following this with a list of four missions, only three of which left Earth orbit, (even the ESA wouldn't claim that a Mars probe stranded in LEO was "completed") and where samples were recovered from all three.

Other "Far Earth" sample missions may have been planned (maybe even started) and later cancelled in the forty years after Luna 24. In any case, only four of them got completed up to concrete realization (i.e. rocket launch). From these four "Far Earth" sample missions, only one was a full success (at least according to Sample return mission).

wogoga in #75:

Luna 24 with launch date 9 Aug 1976 and recovery data 22 Aug 1976 was the last successful lunar sample return mission. According to Wikipedia, this was the third success in eleven attempts of the Soviets to bring down small amounts of moon rock to Earth.

Reactor drone in #103:

If you don't have a problem with the Chinese mini-Soyuz sample return capsule being able to successfully return from the Moon do you think the Soviet Union faked the return of the full size Soyuz capsules that were sent into cislunar and circumlunar space as Zond probes? Some of those made very steep ballistic reentries, perhaps you imagine the Russians have some means of shedding energy that America didn't have.

Wikipedia on Skip reentry:

"The technique was used by the Zond series of circumlunar spacecraft, which planned for one skip before landing. Zond 6, Zond 7 and Zond 8 made successful skip entries, although Zond 5 did not. The Apollo Command Module, when returning from the moon, was capable of a one-skip entry. …The Chang'e 5-T1 [China, 2014] also used this technique."

Wikipedia on Zond 7:

"Like other Zond circumlunar craft, Zond 7 used a relatively uncommon technique called skip reentry to shed velocity upon returning to Earth."

Wikipedia on Zond 5:

Zond 5 launched on September 15 [1968] and became the first spacecraft to circle the Moon and return to land on Earth. … A biological payload of two Russian tortoises, wine flies, meal worms, plants, seeds, bacteria, and other living matter was included in the flight.

On September 22, the reentry capsule entered the Earth's atmosphere but could not perform a skip reentry due to a failure of the guidance system. Landing was supposed to occur in Kazakhstan, but instead Zond 5 splashed down in the Indian Ocean and was successfully recovered by the USSR recovery vessels Borovichy and Vasiliy Golovin. The USS McMorris was shadowing Soviet ships, collecting intelligence information. Photographs taken by the McMorris of the descent module bobbing in the ocean aroused concern at NASA that the Soviets were planning a manned circumlunar flight soon, especially since the US had been tracking Zond 5 for its entire flight, and may have been a catalyst for the decision to launch Apollo 8 to the Moon in December instead of its originally planned mission of testing the lunar module in high Earth orbit.

Although the ballistic reentry would have been bad for human occupants, it did not appear to affect the biological specimens, all of which were alive and well when the descent module was finally opened four days after landing. It was announced that the tortoises had lost about 10% of their body weight but remained active and showed no loss of appetite.

Let us assume that the survival of "tortoises and other biological specimens" was faked, and the Americans knew that. "The Chief Designer" Sergei Korlolev had died in 1966, and maybe in the aftermath of his death, less capable and honest persons (temporarily) took over.

If this or an analogue hypothesis were correct then Americans would have had a form of pressure on participants of the Soviet Space Program suspicious of the claimed successes of Apollo.

In any case, the majority of the Soviet Space Program has not been faked, such as e.g. the first photographs of the dark side of the moon by Luna 3.

sts60 in #82:

There is no confusion about how Apollo command modules landed on Earth.

See: Re-Entry Matters, A Detailed Investigation into Apollo Command Module Returns, Mary DM Bennett

#131  –  2016-09-07

international mile = …0003p1609344000… m = 3p1609.344 m ≈ 3p16 meter
foot = 0.3048 m = …9999n3048000… m = 9n3048 m ≈ 9n3 meter
nautical mile = 1852 m = …0003p1852000… m = 3p1852 m ≈ 3p185 meter

threadworm in #119:

Mary Bennett makes great play in her piece on 'Aulis' (which should in itself be a clue that it's all a complete crock) that Tom Stafford claimed in his autobiography that his re-entry speed was 28547 mph, which is faster than a return from Mars. Here are her exact words:

"Tom Stafford (Apollo 10 again) asserted in his own 2002 biography We Have Capture that his entry speed was 28,547mph, which works out at 41,869ft/sec – faster than a return from Mars!"

I have a copy of the book in front of me, and this is what it actually says:

Our trajectory was designed for the fastest return possible, allowing us to cover the 238,000 mile distance in 42 hours, rather than the usual 56 or more. This meant that at one point we reached a speed of 24,791.4 nautical miles (28547 statute miles) an hour, the fastest human beings have ever flown, or ever will fly, until astronauts return from a trip to Mars.

Wolfgang - do you see the subtle difference there between what Bennett claimed was written and what was actually written?

Statute mile: 5280 feet = 3p5280 ∙ 9n3048 m = 3p1609 meter = international mile

According to Mary Bennett, Apollo 10 astronaut Tom Stafford "claimed in his autobiography that his re-entry speed was 28547 mph". The unit mph = miles per hour is based on international mile (3p1609 meter). In order to convert the value attributed by Bennett to Stafford to International System of Units (i.e. to meter per second), we must at first multiply this value 4p28547 mph with factor 3p1609 m/s per mph. The result 7p4594 meter/hour = 107 ∙ 4.594 m/s must then be divided by 3600 sec/hour:

7p4594 m/h / 3p36 s/h = 4p1276 m/s = 12.76 km/s

According to those who call Bennett a liar, maximum speed was "24,791.4 nautical miles (28547 statute miles) an hour". So someone not being familiar with nautical mile will simply use the statute mile value 4p28547 which is what Mary Bennett seems to have done. And this value results in 4p1276 m/s = 12.76 km/s.

Obviously, such a speed v is highly unlikely. If we use unit meter, second and kg then we directly get Joule per kg (kinetic potential): potkin = ½ v2 = 7p8143 Joule/kg. Kinetic potential corresponding to terrestrial escape velocity is only potesc = ½ vesc2 = 7p6257 Joule/kg. To the difference Δpot = potkin - potesc = 7p1886 J/kg corresponds speed v = √(2 Δpot) = 3p614 m/s. This means that the spacecraft would have already had a velocity of around 6 km/s before falling back on the Earth.

The value "24,791.4 nautical miles" of Tom Stafford results in 4p1108 m/s = 11.08 km/s and is between the values given for Apollo 10 in Entry, Splashdown, and Recovery: 36,314 ft/sec = 4p1107 m/s = 11.07 km/s and 36,397 ft/sec = 4p1109 m/s = 11.09 km/s.

Such a reentry speed 4p1108 m/s may seem not exceptional enough to justify Stafford's statement "the fastest human beings have ever flown, or ever will fly, until astronauts return from a trip to Mars".

Why all this confusion with foot, mile, nautical mile and statute mile resp. international mile? In my opinion part of the answer is given by Mary Bennett:

"[O]ne might also conclude that the intention is to confuse any layman examining Apollo."

JayUtah in #117:

Bennett is a self-proclaimed psychic. She has no training at all in aerospace matters, and no experience. I wish I could count the number of times I've caught her in bald-faced lies and fabricated "research." Naturally neither she nor her sometimes co-author David Percy will debate me directly. They know they're lying.

The fact that Mary Bennett considers herself a psychic is as irrelevant as my belief in panpsychism. She also may have been misled into believing that Albert Einstein was a plagiarist (#122). I do believe in Einstein's honesty. Nevertheless I respect other opinions.

And the question whether Bennett has (officially recognized) training in aerospace matters seems less important to me than the question of whether she has common sense.

Please let me know what kind of dishonesty you are able to sense in Bennett's article.

#146  –  2016-09-09

Dave Rogers in #132:

Look up, wogoga. That thing flying over your head is the point.

Bennett claimed that Stafford said the re-entry speed was "faster than a return from Mars". Stafford actually said that the re-entry speed was "the fastest human beings have ever flown, or ever will fly, until astronauts return from a trip to Mars." Honest people can see the difference between these statements. Why can't you?

What is "honest people"? threadworm even succeeded in confusing me, because I could not imagine such "bald-faced lies", and therefore I erroneously assumed that he and those backing him up must have a legitimate point. But they do not!

Both reentry speed values presented by Tom Stafford in his book "24,791.4 nautical miles (28547 statute miles)" result in 4p1276 m/s = 12.76 km/s = 41869 ft/sec ≈ 4p42 ft/sec. And nobody can deny that this is "faster than a return from Mars". [Sorry, superficial calculations I have just performed seem to suggest that reentry speed from a Mars trip is or can be substantially higher than 12.76 km/s.]

Once again Mary Bennett's words:

"Tom Stafford (Apollo 10 again) asserted in his own 2002 biography We Have Capture that his entry speed was 28,547mph, which works out at 41,869ft/sec – faster than a return from Mars!"

JayUtah in #134:

People who make outrageous claims do not deserve the presumption of credibility you have given her.

According to my world view based on logical consistency and continuity, the claims made by the Apollo program are the ones being outrageous. I can honestly say that the Apollo program seems to me more a gigantic institutionalized lying program ala Richard Nixon than a gigantic technological achievement. History clearly shows that such huge conspiracies with corresponding disinformation programs are possible.

When around twenty years ago a former school mate told me that Apollo was faked, I could not believe and argued in favor of Apollo's reality. As an eight year old boy (then still very pious and religious) I had even prayed to God for the sake of the Apollo 13 astronauts. So it was a rather difficult process (over the last twenty years) to admit to myself that I had been fooled by a criminal government and collaborators.

Reality Check in #140:

We call Mary Bennett a liar because Mary Bennett lied as shown by threadworm in what you quoted!

JayUtah in #141:

And not the first time she has lied about and misrepresented her sources. In Dark Moon there's a photo purporting to be from NASA film/video footage. It depicts the Surveyor spacecraft purportedly as seen from Apollo 12 as they were in the process of landing. Bennett argued it disagreed with other photography taken of the Surveyor. But her photo and footage was to be found nowhere at any NASA source. Naturally her claim is valid only if she could prove her photo was actually published by NASA. After a lengthy period of stalling and insisting her source was NASA, Bennett finally admitted it had instead come from a German documentary and she had just "assumed" it was official NASA footage. It was instead, as we discovered, a reconstruction produced by the German documentarians.

What you address here is obviously a problem. A huge number of more or less officially manipulated or "reconstructed" documents exist (e.g. Earth on photo, 2013, 2015). This creates additional confusion which helps to defend the Apollo myths.

Quite similar is the situation with the moon rocks distributed all over the world (mainly as donations). Even here in Liechtenstein we have one in the Treasure Chamber of the National Museum:

"Further important objects include lunar rocks from the Apollo 11 and Apollo 17 missions as well as Liechtenstein national flags which were carried on the first and last manned flights to the Moon. These were given to the Principality as a token of gratitude for the help provided to NASA by the Liechtenstein-based company Balzers AG, which at the time specialised in vacuum technology and supplied protective coatings for the space rockets."

When such a "moon rock" after having been examined turns out to be undoubtedly terrestrial, then a story that the original rock must have been stolen and replaced with a less valuable stone, or a similar tale can easily be served.


Our Sun is a huge, open nuclear reactor

#150  –  2016-09-11

wogoga in #75:

The task of returning 170 gram = 9n17 kg of moon rock is incomparably easier than of returning humans whose net weight alone is 200 kg = 2p2 kg or more. Rocks are not sensitive to temperature and high g's (decelerations).

Giordano in #79:

The weights of the samples/human beings are not relevant in terms of heat/energy to be dissipated - it is the weight of the total re-entry vehicle. Your use of the weights of samples vs human beings incorrectly greatly exaggerates the difference (for some reason).

Apart from Apollo, "Far Earth" reentry capsules have carried cargos only ranging from dust particles to a maximum of 170 gram of rock. In such a case, the only or at least the main purpose of the heat-shielded reentry capsule is to protect the small but valuable quantity of dust or stones. In this respect, the performance requirements for a safe Apollo landing have a different quality.

Only by concrete calculations we can get a feel for the orders of magnitude involved (→ notation). Let us assume that all the kinetic energy of a car with a weight of one metric ton = 3p1 kg is released in the automotive brake as heat. If the car of our thought experiment stops from a speed v = 22.37 mph = 36 km/h = 10 m/s = 1p1 m/s by applying the brake, then the brake must absorb or conduct away thermal energy of E = mass ∙ v2 / 2 = 3p1 kg ∙ 1p1 m/s ∙ 1p1 m/s / 2 = 5p1 / 2 kg m2/s2 = 4p5 Joule = 50 kilo-Joule.

If we increase speed by factor 10 then we get v = 223.7 mph = 360 km/h = 100 m/s = 2p1 m/s with corresponding E = mass ∙ v2 / 2 = 3p1 kg ∙ 2p1 m/s ∙ 2p1 m/s / 2 = 7p1 / 2 kg m2/s2 = 6p5 Joule = 5 Mega-Joule. This is an increase by factor 100!

In order to get a speed close to Apollo reentry speed 4p11 m/s, we must multiply the previous speed by factor 110. We get 24600 mph = 39600 km/h = 11000 m/s = 4p11 m/s with corresponding heat energy E = mass ∙ v2 / 2 = 3p1 kg ∙ 4p11 m/s ∙ 4p11 m/s / 2 = 11p121 / 2 kg m2/s2 = 10p605 Joule = 60.5 Giga-Joule. The speed increase by factor 110 leads to an energy increase by factor 12100 = 2p11 ∙ 2p11 = 4p121! The resulting theoretical brake energy has a TNT equivalent of 4p1446 kg = 14.46 tons TNT.

If we load up such a car with an additional ton, i.e. from 3p1 kg to 3p2 kg, then the break must absorb or conduct away twice the heat of the empty car. In the situation of the previous paragraph, the same automotive brake would have to absorb or conduct away the energy equivalent of 28.92 tons = 4p2892 kg TNT instead of 14.46 tons = 4p1446 kg TNT.

What is relevant is the ratio between total weight and brake capacity. In the case of Apollo return module, the relevant ratio is between surface and weight of the heat shield on the one hand, and total weight of the module on the other hand.


9/11 would not have been possible if the Apollo Program had not shown how easy it is to make honest enlighteners look like "conspiracy nuts" in the eyes of the naïve majority

#164  –  2016-09-13

JayUtah in #71:

Using the total spacecraft mass at entry is a red herring. Some parts of the ship (i.e., the ablatant) got very hot while others stayed relatively cool. ... Automotive brakes convert kinetic energy to heat as well, but you don't use the whole mass of the car to estimate a temperature. The shoes and rotor heat up -- quite a lot, if you've ever accidentally touched a recently driven car's brake rotor. But the whole car doesn't.

wogoga in #150:

Let us assume that all the kinetic energy of a car with a weight of one metric ton = 3p1 kg is released in the automotive brake as heat.

JayUtah in #151:

Assume all you want, but what you're suggesting by assumption has nothing to do with how spacecraft actually manage the energy of re-entry.

Dave Rogers in #156:

This assumption is the key error. The kinetic energy of a spacecraft on re-entry is not released as heat within the spacecraft. It is released as heat in the atmosphere below the heat shield, much of which is then dissipated without transfer to the spacecraft. So the amount of energy you're claiming must be absorbed by the spacecraft is greatly over-estimated.

Do you despite #68 still honestly assume that my argument depends on the premise that the whole energy equivalent of as much as 80 tons TNT = 4p8 kg TNT has to be absorbed by the Apollo return capsule (with a weight of 5.5 ton = 3p55 kg)?

The energy content of objects falling on Earth such as meteors can be expressed as
TNT equivalent, e.g. in case of the Chelyabinsk meteor:

The bulk of the object's energy was absorbed by the atmosphere, with a total kinetic energy before atmospheric impact equivalent to approximately 500 kilotons = 8p5 kg of TNT (about 1.8 Petajoule = 15p18 Joule), 20–30 times more energy than was released from the atomic bomb detonated at Hiroshima.

In comparison with the Chelyabinsk meteor, a total kinetic energy before atmospheric impact equivalent to approximately 80 tons TNT = 4p8 kg TNT seems petty. Apollo reentry energy is not even 1% = 8n1 = one of Little Boy having laid Hiroshima in ashes "with an energy of approximately 15 kilotons = 7p15 kg of TNT (63 Terajoule = 13p63 Joule)". Yet on the other hand, 190 Apollo reentries correspond energetically to one Hiroshima bomb, since 7p15 kg / 4p8 kg = 2p1875 ≈ 190.

Lance Armstrong seems to be representative of those having seized the control of the United States, whereas Neil Armstrong who died three days after the irretrievable fall on 2012-10-22 of Lance was rather a tragic hero and liar against will

#172  –  2016-09-15

wogoga in #146:

When such a "moon rock" after having been examined turns out to be undoubtedly terrestrial, then a story that the original rock must have been stolen and replaced with a less valuable stone, or a similar tale can easily be served.

threadworm in #171:

The statement in the last paragraphs never happened. It is a complete fabrication.

Wikipedia on Goodwill Moon rocks:

In 1970, US president Richard Nixon gave presentation samples of Moon rock brought back by Apollo 11 as gifts to 135 countries and 50 US states.

Near the end of their third and final moonwalk, and what would be the last moonwalk of the Apollo program, Apollo 17 astronauts Eugene Cernan and Harrison Schmitt "picked up a very significant rock, typical of what we have here in the valley of Taurus-Littrow... composed of many fragments, of many sizes, and many shapes, probably from all parts of the Moon, perhaps billions of years old" and made a special dedication to the young people of Earth. This rock was later labeled sample 70017. President Nixon ordered the distribution of fragments of the rock to 135 foreign heads of state and the 50 U.S. states. These gifts were distributed in 1973. The fragments were presented encased in an acrylic sphere, mounted on a wood plaque which included the recipients' flag which had also flown aboard Apollo 17.

Many of the presentation Moon rocks are now unaccounted for, having been stolen or lost.

"Attempt at subtle bribery rocks" would probably have been a better name for Richard Nixons "Goodwill Moon rocks".

Quote from 'Moon rock' given to Holland by Neil Armstrong and Buzz Aldrin is fake:

"Curators at Amsterdam's Rijksmuseum, where the rock has attracted tens of thousands of visitors each year, discovered that the "lunar rock", valued at £308,000, was in fact petrified wood.

Xandra van Gelder, who oversaw the investigation, said the museum would continue to keep the stone as a curiosity."

By the way, the Fake Moon Rock Hypothesis could resolve the Lunar Isotopic Crisis. Abstract from New approaches to the Moon's isotopic crisis, 2014:

"Recent comparisons of the isotopic compositions of the Earth and the Moon show that, unlike nearly every other body known in the Solar System, our satellite's isotopic ratios are nearly identical to the Earth's for nearly every isotopic system. The Moon's chemical make-up, however, differs from the Earth's in its low volatile content and perhaps in the elevated abundance of oxidized iron. This surprising situation is not readily explained by current impact models of the Moon's origin and offers a major clue to the Moon's formation, if we only could understand it properly. Current ideas to explain this similarity range from assuming an impactor with the same isotopic composition as the Earth to postulating a pure ice impactor that completely vaporized upon impact. Several recent proposals follow from the suggestion that the Earth–Moon system may have lost a great deal of angular momentum during early resonant interactions. The isotopic constraint may be the most stringent test yet for theories of the Moon's origin."

#184  –  2016-09-17

The "NASA-Baked Moon-Rocks" Hypothesis

wogoga in #172:

By the way, the Fake Moon Rock Hypothesis could resolve the Lunar Isotopic Crisis. Abstract from New approaches to the Moon's isotopic crisis, 2014:

"Recent comparisons of the isotopic compositions of the Earth and the Moon show that, unlike nearly every other body known in the Solar System, our satellite's isotopic ratios are nearly identical to the Earth's for nearly every isotopic system. The Moon's chemical make-up, however, differs from the Earth's in its low volatile content and perhaps in the elevated abundance of oxidized iron.

Reality Check in #182:

The second sentence that you did not highlight debunks your delusion that rocks recorded as being collected from the Moon did not come from the Moon, i.e. are terrestrial!

Relevant quotes from the New approaches to the Moon's isotopic crisis:

The returned Apollo samples … initiated a period of deep confusion about the Moon's origin lasting from about 1969 to 1984.

It is now clear that the Moon is a differentiated body with a bulk chemical composition generally similar to that of the Earth's mantle, but compared with the Earth, the Moon is strongly depleted in volatiles such as Na, K, Rb and especially water.

The volatile element depletions, however, remain as strong chemical differences between the Earth and Moon, in spite of more recent discoveries suggesting that the Moon's interior is not as depleted in water as initially supposed.

The radical idea that the Moon might have been born during a planetary size impact late in Earth's accretion history was first vetted at a conference in 1974, … These apparently outrageous suggestions were not taken seriously by most of the lunar science community until a conference on the origin of the Moon in 1984 pointed up the difficulties of the classic origins.

Given these constraints, all of these previous simulations found that the projectile [Theia], while contributing about 10% to the mass of the Earth …, contributed between 70 and 90% to the mass of the Moon-forming disc …

There were signs of an isotopic problem with the then-understood giant impact hypothesis as early as 2001.

This can, of course, be reconciled if the actual impactor is isotopically nearly identical to the Earth …, but as error bars shrank, the O-isotopic similarity between the Earth and Moon came to seem anomalous.

Adding to the eerie similarity between oxygen isotopes of the Earth and Moon, many other isotopic systems show a near identity of the two bodies.

The similarity in the W system indicates that the impactor and proto-Earth must have had a nearly identical history of iron–silicate differentiation, a situation that is hard to envision in the context of current models for terrestrial planet formation.

Some isotopes do show differences between the Earth and Moon. For example, Paniello et al. find an enrichment of heavy isotopes of Zn in lunar rocks which they attribute to the depletion of the volatile element Zn in a hot proto-lunar disc.

The obvious conclusion of all these isotopic similarities is that the Moon is almost entirely derived from the Earth's mantle.

We have the peculiar circumstance that the Earth and Moon, while apparently nearly identical in most isotope ratios, are substantially different in their bulk chemistry.

Maybe the following hypothesis could lead to simpler explanations of our Moon's origin:

The majority of so-called "moon rocks" has been baked or cooked by NASA from a mixture of terrestrial rocks peppered with some meteorites at high temperatures and vacuum, leading by vaporization to depletion of volatile elements and enrichment of heavy isotopes.

In any case, such baking of moon-stones would have been a much easier and cheaper task than fetching home genuine stones from the Moon.

A strong argument against moon-rock-baking by NASA would be the absence of biological molecules in the interior of so-called "lunar rocks". Quotes from New NASA Study Reveals Origin of Organic Matter in Apollo Lunar Samples:

A team of NASA-funded scientists has solved an enduring mystery from the Apollo missions to the moon – the origin of organic matter found in lunar samples returned to Earth. Samples of the lunar soil brought back by the Apollo astronauts contain low levels of organic matter in the form of amino acids. Certain amino acids are the building blocks of proteins, essential molecules used by life to build structures like hair and skin and to regulate chemical reactions.

Since the lunar surface is completely inhospitable for known forms of life, scientists don't think the organic matter came from life on the moon. Instead, they think the amino acids could have come from four possible sources.

"People knew amino acids were in the lunar samples, but they didn't know where they came from," said Jamie Elsila of NASA's Goddard Space Flight Center in Greenbelt, Maryland. "The scientists in the 1970s knew the right questions to ask and they tried pretty hard to answer them, but they were limited by the analytical capabilities of the time. We have the technology now, and we've determined that most of the amino acids came from terrestrial contamination, with perhaps a small contribution from meteorite impacts."

The team analyzed seven samples taken during the Apollo missions and stored in a NASA curation facility since return to Earth, and found amino acids in all of them at very low concentrations (105 to 1,910 parts-per-billion). One of the key new capabilities of the Goddard Astrobiology Analytical Laboratory was instrumentation with high enough sensitivity to determine the isotopic composition of an amino acid molecule, according to Elsila. This capability enabled the team to say terrestrial contamination was the primary source of the lunar amino acids.

Also, if the solar wind were responsible for the amino acids, then samples taken from near the lunar surface, which had the highest exposure to the solar wind, should have a greater abundance of amino acids than samples taken from deeper beneath the surface. This is the opposite of what was found – the deepest samples, which were the most sheltered from the solar wind, produced the most amino acids.

However, the team found that a sample taken from 6.5 kilometers (four miles) away had similar amino acid abundances to the one taken beneath the module.

The ability to determine the orientation of an amino acid molecule was another significant new capability of the Goddard lab that enabled them to discover the origin of the lunar amino acids... Since life uses the left-handed versions, this suggests terrestrial life as the source of these amino acids.

By the way, concentrations of organic matter up to "1,910 parts-per-billion" are not "very low". These concentrations have even been high enough to determine isotopic composition! 1910 per billion = 3p191 / 9p1 = 4n191 ≈ 4n2 = 0.000,002. If we take e.g. world population of around 7 billion = 9p7 then "1,910 parts-per-billion" resp. 4n2 result in around 9p7 ∙ 4n2 = 4p14 = 14 thousand humans.

Even if the NASA-baked moon-rocks hypothesis should eventually turn out to be untenable, it remains a legitimate hypothesis for me as long as I do not know of serious objections.


Genuine scientists try out the simplest hypotheses regardless of prejudices and scientific taboos

#196  –  2016-10-01

wogoga in #14:

To a speed v of 2.1 km/s = 3p21 m/s corresponds only ½ v2 = ½ ∙ 3p21 ∙ 3p21 J/kg ≈ 6p22 J/kg = 2.2 Mega-Joule per kg, which is a rather negligible quantity in comparison to the 60 MJ = 7p6 J corresponding to the entry speed of 11 km/s = 4p11 m/s. Thus around 96% = 9n96 of the kinetic energy would have disappeared before falling to an altitude of 120 km = 5p12 m. For an astronaut of 80 kg these "missing" 96% = 9n96 would result in 4.7 Giga-Joule = 9n47 Joule, enough to heat up ten tons (10000 kg = 4p1 kg) of steel by 1000° Kelvin.

WhatRoughBeast in #23:

It would help if you paid attention to your own links and figures. The Apollo CM did not start slowing appreciably until it reached the atmosphere.

The problem is that the figure (see #1) of Returning from Space: Re-entry constitutes disinformation insofar as it implies an entry speed of only 2100 m/s instead of 11000 m/s = 4p11 m/s.

WhatRoughBeast in #23:

The period of greatest energy shedding is clearly the portion of the flight which is temperature limited, from roughly 300 to 1000 seconds. At 5500 kg, the kinetic energy of the CM at reentry is about 340 GJ. Shedding this energy over 700 seconds requires a power level of about 500 MW, more or less.

[500 MW per 12 m2 heat shield → 40 MW/m2 = 7p4 W/m2]

According to NASA Technical Note D-5399 of Apollo 4, Table I, peak g of around 7 g ≈ 70 m/s2 is already reached 77 seconds after "entry interface" at an altitude of ~5p12 meter, and 46 sec after "0.05 g interface" (when the spacecraft reaches a deceleration of 0.05 g = 8n5 g ≈ 9n5 m/s2).

The power of heat and radiation production is P = E / t = F ∙ s / t = m ∙ a ∙ v. For the "0.05 g interface" this results in 3p54 kg ∙ 8n5 ∙ 0p98 m/s2 ∙ 4p11 m/s ≈ 7p29 kg m2/s3 = 29 Megawatt, resp. 6p24 W/m2 = 2.4 Megawatt per square meter of the heat shield of 12 m2.

During the next 46 sec until peak g, speed decreases to around 10 km/s = 4p1 m/s, and deceleration power increases to around 3p54 kg ∙ 0p7 ∙ 0p98 m/s2 ∙ 4p1 m/s ≈ 9p37 kg m2/s3 = 3.7 Gigawatt, resp. 8p31 W/m2 = 310 Megawatt per m2 heat shield. For comparison, radiation at Sun's surface (black body radiation of 5780°K) is only 63 MW per m2 = 7p63 W/m2.

WhatRoughBeast in #23:

Radiation: Although minor, you might want to do the calculations. The Apollo CM had a diameter of 3.9 meters, for a heat shield area of about 12 m2. Blackbody radiation from such a surface at 1500 K runs about 7 MW.

A correct calculation according to the Stefan–Boltzmann law results in only half of your value. Yet in case of a 100 percent efficient heat shield (i.e. with 100% emissivity), a temperature of 1790°K resp. 1520°C would be enough to radiate away your 7 MW = 6p7 Watt on a surface of 12 m2, i.e. ~580 kW/m2 = 5p58 W/m2.

WhatRoughBeast in #23:

Ablation: You've made much of the temperature rise if the available energy were transferred to, for instance, steel. This is a perfectly valid concern, and the response was to make sure that that didn't happen. As I 'm sure you're aware, phase changes can absorb far more energy than a simple temperature change - look at the difference between heating water from 99 to 100 degrees, and from 100 degrees to 101 degrees. For Apollo, the heat shield actually burned off due to pyrolysis, limiting the maximum temperature.

According to NASA Technical Note D-5399 of Apollo 4, Table II, mass of the return module reduces during reentry from 369.7 slugs (→ 5395 kg) to 362.1 slugs (→ 5284 kg). The mass loss of 7.6 slugs results in 111 kg. What is the maximum "cooling by vaporization" we can achieve with a mass of 111 kg?

The element with the highest vaporization energy is boron with 500 kJ/mol = 5p5 Joule/mole (boiling point at 4200°K). Taking also into account molar heat capacity and heat of fusion, we calculate that around 600 kJ = 5p6 Joule are needed in order to vaporize one mole of boron. Since the mass of 1 mole is only 10.8 gram = 8n108 kg, we get as total vaporization energy 5p6 Joule / 8n108 kg ≈ 7p55 J/kg = 55 Mega-Joule per kg.

Thus, even if all the 111 kg having disappeared during reentry of Apollo 4 consisted of boron, vaporization of it could have absorbed only 2p111 kg ∙ 7p55 J/kg ≈ 9p6 Joule = 6 GJ. This is not even 2% = 8n2 of total heat and radiation released during reentry, which is around 330 GJ = 11p33 Joule! Because high vaporization resp. pyrolysis energy is only one of several other constraints (such as stability and low thermal conductivity), the real heat shield of Apollo 4 could probably not even have absorbed 1% = 8n1 of reentry energy, or do I miss something?

Let us admit the truths to ourselves, better late than never!

#291  –  2016-10-13

wogoga in #196:

The problem is that the figure (see #1) of Returning from Space: Re-entry constitutes disinformation insofar as it implies an entry speed of only 2100 m/s instead of 11000 m/s = 4p11 m/s.

Reality Check in #205:

It is a lie that Figure 4.1.7-25 of Returning from Space: Re-entry (PDF) has any entry speeds!
It is a lie that anything in that section of the document "implies an entry speed of only 2100 m/s". Figure 4.1.7-24 explicitly has the Space Shuttle with a speed of 7300 m/s at 80 km.

Space Shuttle is different from Apollo. Here again Returning from Space – Re-entry, Figure 4.1.7-25:


If you elongate straightforward the upper left part of the green Apollo line, you reach zero altitude at around 500 sec. This means a decrease in altitude of ~240 m/s at the beginning of reentry. Taking into account a reentry angle of 6.5°, reentry speed results in ~2100 m/s (see #14).

In the meanwhile, I found a hypothesis why the textbook authors may have assumed such a low altitude-decrease at entry interface (apart from heat shield calculations). Quote fromNASA Technical Note D-5399 of Apollo 4:

After the CM separated from the SM [Service Module] and prior to reaching the entry interface at 400 000 feet [120 km = 5p12 m], the spacecraft was oriented in pitch with its stability axis along the AGC-estimated relative wind-velocity vector … Pitch and yaw attitude control was maintained until 0. 05 g deceleration was reached. … At the entry interface, the initial roll program of the INITIAL ENTRY phase was in command. … When 0.05 g was sensed (0. 05 g interface), the AGC [Apollo guidance computer] automatically began the entry computations... A decision was made to continue the flight with the lift vector up. When thermodynamic deceleration level exceeded 0.2 g and the altitude was decreasing at a rate less than 700 ft/sec, control was transferred to the HUNTEST phase.

"Deceleration exceeding 0.2 g and altitude decreasing at less than 700 ft/sec = 210 m/s" may be somehow misleading. So the textbook authors could erroneously have interpreted this as "when deceleration first exceeded 0.2 g, altitude-decrease became less than ~210 m/s". And with a lift-to-drag ratio of ~0.4 they may have calculated their altitude-decrease of ~240 m/s at entry interface.

Further quote from the Apollo 4 Note:

The principal objectives of the Apollo 4 mission were to demonstrate the structural and thermal integrity of the space vehicle and to verify the adequacy of the heat shield when subjected to entry at lunar return flight conditions. To reach lunar return flight conditions, the service propulsion system (SPS) engines were fired, and an entry velocity of 36 545 ft/sec was obtained.

To 36 545 ft/sec correspond 11.139 km/s ≈ 4p111 m/s. "Entry velocity" with respect to what? My previous posts have been written under the premise that entry velocity is speed relative to the ground below the entering spacecraft, since this speed is relevant to "thermodynamic deceleration", resp. heating due to kinetic energy. Yet at the latest in NASA Technical Note D-7564 of 1974, there is a clear distinction between "inertial velocity at 400 000 ft" and "relative velocity at 400 000 ft". The latter is given as 35 220 ft/sec = 10.735 km/s ≈ 4p107 m/s, a value which cannot be found in the original Apollo 4 note (maybe the value is hidden as a Mach number, where the speed-unit Mach depends on pressure and temperature, i.e. on altitude and weather).

It seems that "inertial velocity" means relative to the Earth's center, and "relative velocity" relative to the surface below the spacecraft. Nevertheless, Figure 7b, Earth relative velocity of the Apollo 4 note starts with a value that seems to be closer to 36 545 than to 35 220 feet.

In any case, the difference in "thermodynamic deceleration" energy due to the two different entry-speed values is not relevant in this discussion (58 MJ/kg = 7p58 J/kg versus 62 MJ/kg= 7p62 Joule per kg).

In Figure 6, Entry trajectory control programs, ground distance from 'entry interface' to '0.05 g interface' is around 2180 - 2035 = 145 nautical miles = 265 km = 5p265 m. As 30.5 sec= 1p305 s are needed (TABLE I.), we calculate a speed of 265 km / 30.5 s = 5p265 / 1p305 m/s ≈ 3p87 m/s = 8.7 km/s, which is incompatible with an entry speed close to 11 km/s = 4p11 m/s and an entry angle of ~7°. So we can be quite sure that this "best-estimate trajectory" as presented in Figure 6 of the Apollo 4 note is not (fully) consistent.

Here a scaled graphic of reentry interface and angle of Apollo 4 (based on data from Figure 6):


Point with dotted line represents entry point with entry direction. The red surface part ranges from '0.05 g interface' (on the right side) to 'peak g' (resp. 'HUNTEST', see Figure 6). The left end of the surface corresponds to the landing site of Apollo 4 at a distance of 2180 nautical miles = 3980 km ≈ 6p4 meter from 'entry interface' (see Figure 6).

wogoga in #196:

Because high vaporization resp. pyrolysis energy is only one of several other constraints (such as stability and low thermal conductivity), the real heat shield of Apollo 4 could probably not even have absorbed 1% = 8p1 of reentry energy, or do I miss something?

threadworm in #199:

Yes, you missed the undeniable fact that Apollo 4 (if that is the example you are going to use) took off, orbited and returned to Earth. [...] and this youtube video of mine shows all the orbital images strung together as a movie:

Where does the wiggling in your video come from? Did the engine run when the photos were taken? In purely inertial motion, the Apollo 4 camera position should not move at all or only smoothly and slowly. And at what distance where the photos taken?

threadworm in #199:

So if Apollo 4 launched and orbited but then burned up, how did we get those photos?

The photos are not clear and sharp enough to exclude a fake. And even if they are genuine, they could have survived in an additionally protected container within the disintegrating spacecraft.

#308  –  2016-10-19

wogoga in #95:

An air particle colliding with the heat shield of the reentry module gets a huge impulse (off the shield). Yet the particle cannot leave the region close to the 12 m2 heat shield easily, because mean free path even in front of the highly compressed shock wave is only 1 mm = 7n1 meter.

Dave Rogers in #97:

"An air particle colliding with the heat shield" is therefore an event that can only occur for air particles within that mean free path once the shock wave forms, and much of the momentum transfer is therefore due to collisions between atoms elsewhere, ...

Such "collisions elsewhere" can only slow down the spacecraft, if momentum ultimately gets transferred to the heat shield (via collisions with particles situated closer and closer to the heat shield). Without such momentum transfer, a deceleration of the spacecraft is impossible.

The mean free path in the shock-compressed air just before the heat shield is even substantially lower than 1 mm (here assumed as free path outside the compressed shock wave), since this region has much higher density due to bombardment by new air particles (with e.g. Mach 30 relative to heat shield). Particles jammed between the 12 m2 heat shield and the newly arriving high-speed air particles need many collisions in order to escape. Where could all this kinetic resp. thermodynamic energy disappear without affecting the heat shield?

Almost half of the radiation emerging in this region is directed to the heat shield.


High speed reentry velocities lead to a strong shock wave upstream of the vehicle which strongly modifies the state of the fluid behind it. The internal thermodynamic energy of the free-stream fluid particles is small when compared with the kinetic energy of the free stream. As the Mach number is increased, the flow density increases progressively, so the shock layer created by the strong bow shock over the blunt nose becomes thinner; the detachment distance between the wall boundary and the shock wave is really small.

As aforementioned, the collisions between free stream particles and molecules of the dense shock layer excite the rotational, vibrational and electronic modes of the molecules, and give rise to a non-equilibrium between the temperatures and energies corresponding to these modes. Generally, translational and rotational energy modes require a few collisions to equilibrate, so they are usually considered in equilibrium even for Hypersonic flow conditions, for the sake of time-saving. On the other hand, chemical dissociation and vibrational excitation require more collisions to reach their equilibrium state and non-equilibrium effects may be important for hypersonic flow conditions, especially at high altitudes.

In any case, the only reasonable hypothesis for Apollo reentry being possible seems this one:

The by far biggest part (substantially more than 90% = 9n9) of the energy emerging due to transfer of momentum from the spacecraft to the atmosphere is used to break and ionize air molecules or to induce energy in form of rotational, vibrational or electronic "excitation" not transmittable to the heat shield. All these energy forms are only released again when the molecules or their fragments have left the region close to heat shield.

Since a reentry similar to a Apollo reentry has never been successfully repeated, I prefer the simple and clear arguments suggesting physical impossibility to the complicated and opaque arguments suggesting possibility.

#320  –  2016-10-20

wogoga in #164:

Lance Armstrong seems to be representative of those having seized the control of the United States, whereas Neil Armstrong who died three days after the irretrievable fall on 2012-10-22 of Lance was rather a tragic hero and liar against will.

Reality Check in #165:

A lie of calling Neil Armstrong a "tragic hero and liar against will" is even worse. Neil Armstrong was a hero before Apollo (a test pilot) and a hero during Apollo because he was the first man on the Moon. Neil Armstrong did not lie.

Please watch depressed Neil Armstrong speaking (especially at 12:11):

"Today we have with us a group of students, among America's best. To you we say we have only completed a beginning. We leave you much that is undone. There are great ideas undiscovered, breakthroughs available to those who can remove one of the truth's protective layers. There are many places to go beyond belief. Those challenges are yours – in many fields, not the least of which is space, because there lies human destiny." (Transcript)

A reasonable video on Apollo with some quite convincing arguments: Conspiracy Theory: Did We Land on the Moon?

#331  –  2016-10-22

wogoga in #308:

Almost half of the radiation emerging in this region is directed to the heat shield.

abaddon in #311:

Evidence for this claim please.

Take a location close to the heat shield. How much of the world is hidden behind the heat shield? An infinitely extended or infinitely close plane would cover half of the world. Since thermal radiation goes in all directions, the probability that a photon emitted by a particle close to the heat shield travels in direction to the shield is not far below 50%.

Elagabalus in #301:

This one shows the heat shield honeycomb being fitted, filled (all 370,000) machined and tested @ 11:29.

@ 20:20 shows heat shield failing a drop test.


The picture is from your video at 20:22 when first water splashes become visible. Even this low speed water drop test resulted in "failure of the heat shield supporting structure", and water started penetrating into the capsule. In any case, only persons with no idea of orders of physical magnitudes can think that such tests and arguments are relevant to this discussion.

#336  –  2016-10-24

Jrrarglblarg in #313:

I've attempted to tell this upthread but it was apparently overlooked: air particles do not impact the heat shield.

The heat shield is ablative, meaning it burns away during use, like brake pads, which are made of similar organic resin binders. The resin decomposes at 200-something degrees C, about the temp for baking muffins in an oven. As it decomposes, the gasses outflow. It's this boundary layer of gas that actual air molecules are impacting, and it's these gasses flowing out of the heat shield that carry the heat away. The plume of reentry gasses are very hot.

The interesting point here is not that somebody makes a fool of himself in trying to defend what in reality is only huge hoax. The interesting point is that the (other) forum pundits do not correct such obvious disinformation resp. nonsense.

A material decomposing "at 200-something degrees C, about the temp for baking muffins in an oven" is assumed to protect a heat-shield assumed to withstand temperatures up to 5000°F resp. 3000°K (Figure 2, NASA TN D-7564)!

Jrrarglblarg in #313:

The bottom of the heat shield is like the bottom of a pot of beans. As long as the water is boiling off it's able to keep the pot bottom at the boiling temp of water. The flame is very hot, but the beans don't burn because hot gasses on the bottom of the pot form a temperature boundary layer, and carry incoming heat away.

Since water vaporization energy is 2257 kJ/kg = 6p2257 Joule per kg, "cooling" capacity of water is not unlimited. The same is valid for the heat shield. In #196 I calculate an upper limit for heat absorption if all the 111 kg mass loss during reentry of Apollo 4 was due to vaporization.

Jrrarglblarg in #313:

The total energy dissipation budget of Apollo reentry cannot be discussed accurately until you acknowledge the energy consumption of the ablative material decomposing, and realize that these gasses prevent ANY air from impacting the surface of the reentry module.

Momentum in the order of 5400 kg ∙ 11 km/s = 3p54 kg ∙ 4p11 m/s ≈ 7p6 kg m/s must be taken away from the reentering spacecraft. And you claim that this happens without air particles impacting "the surface of reentry module". In case of Apollo 4, the total mass loss during reentry was 111 kg. In order to slow down the spacecraft from 11 km/s, these 111 kg would have to move away from the heat shield at a speed of around 11 km/s ∙ 5300 kg / 111 kg ≈ 530 km/s = 1,900,000 km/h = 1,200,000 mph = 5p53 m/s. Even the most efficient rocket engines reach exhaust gas velocity of only 4.5 km/s = 16,200 km/h = 10,100 mph = 3p45 m/s.

As long we are able to change our minds we are mentally young

#375  –  2016-10-26

jaydeehess in #253:

To this non-engineer one thing is quite obvious, the air below the capsule is not static, it heats up and flows away to be replaced by cooler air.

JayUtah in #254:

Yes, there is a convective component to the model.

This exemplifies one central error of my opponents.

Hot air and/or gaseous heat-shield particles are assumed to be replaced by "cooler air". Yet in the decisive deceleration phase of the Apollo module, such air particles have a typical speed of 10 km/s = 4p1 m/s relative to the heat shield. Averaged thermal velocity of nitrogen is ~470 m/s = 2p47 m/s at room temperature (293°K). Temperature is essentially proportional to the square of particle speed. This roughly means that e.g. an increase in particle speed by factor 10 leads to a temperature increase by factor 100. Since our typical interaction speed is around 20 times higher than the nitrogen thermal speed at 300°K (10 km/s versus 500 m/s resp. 4p1 m/s versus 2p5 m/s), we can attribute a theoretical temperature of 300°K ∙ 20 ∙ 20 = 2p3 ∙ 1p2 ∙ 1p2 Kelvin = 5p12°K = 120 thousand degrees Kelvin to the nitrogen molecules interacting (directly or indirectly) with the heat shield. Up to an altitude of around 100 km = 5p1 m, around 80% = 9n8 of the atmosphere consists of nitrogen (source; see also stagnation enthalpy).

Thus, the assumption that bombardment of the spacecraft by high-speed air particles represents "cooler air" for the heat shield is completely absurd.

It is true that we can cool something by blowing air on it. The main reason is that air heated by the object is replaced by still unheated air. Convective heat transfer is simply more efficient than purely conductive. Yet such a cooling process is only possible if the kinetic energy of the air particles is lower than the thermodynamic energy of the surface particles of the object. Both energy and momentum are simply redistributed between colliding particles, and neither energy nor momentum can disappear in the process.

JayUtah in #254:

There is also a radiant component whereby the heated atmospheric gas applies heat to the heat shield via infrared. That's part of the brilliance of the design: the gases produced in the ablation process are substantially opaque to infrared. This principle is also applied in ablative linings for rocket nozzles.

Two devastating objections can be raised against this disinformation:

Since air particles colliding with the shock layer in front of the heat shield (see #308) have theoretical temperatures in the order of tens of thousands degrees Kelvin (see above), their collisions with shock layer particles lead in the first place rather to ultraviolet than to infrared radiation.

The Apollo heat shield is assumed to generally work substantially below 5000 Fahrenheit resp. 3000°K, and thermal radiation in this temperature range lies primarily in the infrared (see Wien's displacement law). So "gases produced in the ablation process" which are "substantially opaque to infrared" would prevent the heat shield from performing [one of] its main task, namely from radiating off heat.

The believer (e.g. in Apollo or 9/11), instead of questioning his faith with arguments, feels entitled to judge arguments with his faith

#403  –  2016-10-28

wogoga in #375:

So "gases produced in the ablation process" which are "substantially opaque to infrared" would prevent the heat shield from performing its main task, namely from radiating off heat.

JayUtah in #388:

No, it isn't. The main task of an ablation shield is to ablate. Its own heat rejection via radiative means is a third-order term in the problem.

The Stefan–Boltzmann law states that the total energy radiated per unit surface area of a black body is directly proportional to the fourth power of temperature. The constant of proportionality is 5.67 x 10-8 W/m2/K4 = 2n567 W/m2/K4. In case of 1000 Kelvin, (ideal) radiation results in 2n567 ∙ 3p14 = 2n567 ∙ 12p1 = 4p567 W/m2 = 4560 Watt per square meter. In case of 3000 Kelvin however, we get 6p46 W/m2 = 4.6 MW/m2 (i.e. increase by factor 81).

Thus the question whether "heat rejection via radiative means is a third-order term" depends on temperature.

Jrrarglblarg in #283:

I've been researching the chemistry of the ablative heat shield itself and I'm finding this paper quite interesting:

JayUtah in #286:

It is, and that's the one that I thought had the quantitative analysis of the design.


"The Apollo spacecraft flew on numerous earth-orbital and lunar-landing missions, and there were no problems or anomalies associated with the TPS [thermal protection subsystem]. The success of the system can be attributed to the somewhat conservative design philosophy that was adopted and to the rigorous analytical and test certification requirements that were imposed."


"In space, the temperature requirements for the ablator were limited to a cold temperature of -260° F [-162°C] and a hot temperature of 250° F [121°C]. The maximum initial bondline temperature at the start of entry was specified as 250° F [121°C], and the limiting temperatures at splashdown were 600°F [316°C] at the interface of the ablator and the stainless-steel honeycomb structure and 200°F [93°C] for the aluminum-honeycomb pressure-vessel structure."

A blind eye is turned to the crucial problem of reentry heat. Yet, as also our discussion vividly demonstrates: Most people are credulous enough to believe in APOLLO even in the absence of hard evidence. All what could easily be faked with the technology of 1960s does not count as hard evidence. Ask e.g. James Randi how easy it is to mislead others.

The only relevant information for our discussion is contained in TABLE II. – APOLLO TPS FLIGHT VERIFICATION:

Total reference heating load, Btu/ft2 is given as 37 522 for Apollo 4. Since 1 Btu = 1055 Joule, this value translates into 4p37522 Btu/ft2 / (9n3048 m/ft)2 ∙ 3p1055 J/Btu ≈ 8p426 J/m2 = 0.43 Gigajoule/m2

However, total kinetic energy of reentry per square meter heat shield is E ≈ ½ m ∙ v2 / 12m29n5 ∙ 3p54 kg ∙ 4p11 m/s ∙ 4p11 m/s / 1p12 / m210p27 J/m2 = 77 Gigajoule/m2. This more than 100 times more!

Maximum heating rate, Btu/ft2-sec is given as 425. This translates into 425 Btu/ft2/s / (9n3048 m/ft)2 ∙ 1055 J/Btu ≈ 6p48 J/s /m2 = 4.8 Megawatt/m2. This is a reasonable result insofar as blackbody radiation of a body of 5000°F = 3033°K also results in 2n567 ∙ 30334 W/m2 ≈ 6p48 W/m2 = 4.8 MW/m2.

However, maximum thermodynamic heating (at peak g) due to momentum transfer between spacecraft and atmosphere is 8p31 W/m2 = 310 Megawatt/m2 (see #196). This is around 65 times more than "maximum heating rate" of the heat shield.

It cannot be denied that the whole momentum of around 5400 kg times 11 km/s = 3p54 kg ∙ 4p11 m/s ≈ 7p6 kg m/s = 60 Mio. kg m/s is transferred via contact interaction from the spacecraft to the atmosphere. Momentum change of the entering spacecraft is exactly the same as corresponding momentum change of the atmosphere (actio = reactio). Why should only in the order of one percent of the energy emerging during this mutual momentum transfer affect the heat shield? And even this calculation (heat shield removing in the order of one percent of total emerging heat) is based on the quite unlikely hypothesis that the heat shield works fine up to a temperature of 5000°F resp. 3000°K.

#420 –  2016-11-05

wogoga in #184:

Maybe the following hypothesis could lead to simpler explanations of our Moon's origin:

The majority of so-called "moon rocks" has been baked or cooked by NASA from a mixture of terrestrial rocks peppered with some meteorites at high temperatures and vacuum, leading by vaporization to depletion of volatile elements and enrichment of heavy isotopes.

What I wrote here is wrong. They probably used an artificial atmosphere with at least radiogenic 40Argon in order to mislead radioactive dating techniques.

A quote from Age Determinations and Isotopic Abundance Measurements on Lunar Samples, 1970:

The presence of an anomalously large quantity of 40Ar, in a sample of type C breccia precluded the calculation of its K-Ar age.

A simple hypothesis for such an anomalously large quantity of 40Argon is an error in the process of "moon rock" baking. The NASA bakers would have mistakenly added too much of the ingredient 40Argon.

Logically consistent creationists are pleased with lunar rocks: The Age of the Moon

It would be quite easy for an unprejudiced scientific research to decide whether Apollo moon rocks are faked, and if yes, to exactly determine how the Apollo moon rock samples were created. Some "moon rocks" may simply be special meteorites (or maybe even very special terrestrial stones) which are now called "lunar meteorites". An interesting list of lunar meteorites: List

Secondary meteorites originating from a primary impact on the moon enter our atmosphere with a speed as high as Apollo reentry speed! Thus the high proportion of "lunar" meteorites among all meteorites becomes rather unrealistic. Small pieces catapulted out of the lunar surface and accelerated to more than the moon's escape velocity of 1620 m/s would mostly vaporize in terrestrial atmosphere. Reaching regions close to the poles such as Antarctica is even more improbable. So how is it possible that "0.11% of all meteorite stones found in Antarctica are of lunar origin" (Source)?

Centuries ago, those being knowledgeable and courageous enough to cast doubt on the legitimacy of the Inquisition were hated by those having naïve faith in the Holy Church

#435  –  2016-11-10

wogoga in #184:

Even if the NASA-baked moon-rocks hypothesis should eventually turn out to be untenable, it remains a legitimate hypothesis for me as long as I do not know of serious objections.

Jrrarglblarg in #187:

Here are several serious objections:

1) Hundreds of geologists around the world have studied and continue to study the lunar rocks. Not one has ever raised any objection.

"Friendly" researches receive carefully selected, maybe even freshly doctored small samples from NASA, normally in order to deal with one single question. It may be unlikely that no one of the receivers has become suspicious, yet outing suspicions representing scientific taboos is not everyone's hobby.

Take e.g. 9/11. Every unprejudiced scientist with common sense comes to the conclusion that the twin towers and WTC7 have been brought down by explosives, whereas the rest of the WTC buildings such as the Marriott hotel were destroyed in a way expected for steel-frame buildings in the absence of explosives. Nevertheless, if you rely on peer reviewed articles and the mainstream media you get a completely different picture (i.e. a fully disinformative one).

Jrrarglblarg in #187:

2) If you want to study lunar rocks you can. Nothing is preventing you from getting a geology degree and sending in your request for a specific sample to study.

Quote from The Age of the Moon,, 2008:

Even before the Apollo 11 astronauts brought rocks back from the moon, scientists from all over the world were clamoring to get the chance to analyze them. Therefore, NASA gave many scientists the opportunity to write proposals telling how they would analyze the moon rocks if they were given the opportunity. Based on the merit of the proposals and the qualifications of the scientists, they allowed a few select scientists access to the samples.


In the four decades since the Apollo Moon landings, scientists have studied only about ten percent of the lunar rocks and dust collected by astronauts from 1969 to 1972.

Jrrarglblarg in #187:

3) Volcanically created rocks have internal gas pockets. These gas pockets in lunar rocks have physical characteristics which differ greatly from earth rocks of similar chemistry, because they cooled in a different gravity than earth rocks.

"Different gravity" at cooling is quite irrelevant. Relevant are pressure (weight above) and atmospheric composition. I do not doubt that Apollo moon samples on the one hand and genuine moon rocks on the other hand are different from terrestrial rocks. Apollo samples are assumed to be "the result of magmatic degassing during the eruption of lunar magmas into the near-vacuum of the Moon’s surface" leading to "rapidly quenched lunar volcanic glasses".

Quotes from High Pre-Eruptive Water Contents Preserved in Lunar Melt Inclusions, Jul 2011:

Melt inclusions are small samples of magma trapped within crystals that grow in the magma before eruption. By virtue of their enclosure within their host crystals, melt inclusions are protected from loss of volatiles by degassing during magma eruption [or moon-rock baking]. … The most important aspect of our volatile data on lunar melt inclusions is their similarity to melt inclusions from primitive samples of terrestrial mid-ocean ridge basalts, like those recovered from spreading centers located within transform faults; the melt inclusions from [Apollo sample] 74220 are markedly similar to melt inclusions from the Siqueiros Fracture Zone on the East Pacific Rise, some of the most primitive mid-ocean ridge magmas that have been measured.

Thus we should not apriori exclude a further explanation for "the paradox of identical isotopic compositions of Earth and the Moon" (Science, Jun 2014), namely that Apollo samples are in fact terrestrial. Scientific taboos can do a lot of harm to science!

#452  –  2016-11-25

JayUtah in #387:

Again, this [insulating the spacecraft against radiative heat transfer] was not new technology for Apollo. This was developed in the 1950s for ICBMs.

In case of intercontinental ballistic missiles (ICBM), the objective is to reduce atmospheric friction in order not to lose speed. In case of a reentry, the objective is to get rid of kinetic energy by means of atmospheric friction. The goals are almost as different as "efficient braking" versus "reduction of frictional losses" in case of cars.

I must admit that the values of a reentry from ISS with a Soyuz capsule are closer to the alleged Apollo reentry values than I would have thought. But in case of a return from ISS, heating rate can easily be made as low as needed by increasing reentry time, and current efficiency resp. short reentry time can result from experience and optimization.

In any case, I do not think that lunar reentry is impossible for a spacecraft with form and total weight of the Apollo Command Module. I only think that it is beyond NASA's technology of the 1960s.

We humans tend to believe what we want to be true, and we want to be true what we already believe

#471  –  2016-12-02

wogoga in #403:

Thus the question whether "heat rejection via radiative means is a third-order term" depends on temperature.

JayUtah in #408:

No. You don't understand what is meant by "the problem." The problem is the design of a heat shield. The solution involves several factors working in concert. Rejection of heat by radiative heat transfer is a third-order term in the quantitative solution. You're trying to claim the gaseous ablation products cannot be suitably opaque as I claimed because then the heat shield wouldn't be able to radiate away heat. That's a knee-jerk response based on what you desperately want to be true, not what is true.

In the textbook provoking this thread (#1), the authors deal with three approaches to thermal-protection systems:

If you are right and "radiative heat transfer is a third-order term in the quantitative solution" then thermal protection in case of Apollo must almost fully rely on ablation, since involved thermal energy is far too enormous for being relevantly absorbed by heat sinks of the reentry capsule (#68).

Jrrarglblarg in #458:

Does the Orion/constellation stuff use substantially different technology for reentry shielding? The capsule is bigger, and designed for larger payloads up and down, does it still use slow-charring epoxy for an ablative reentry shield? I've not really followed the thing because it seems like NASA was designing a tool for a job that doesn't exist.

JayUtah in #459:

It's almost a carbon copy of Apollo's heat shield. No pun intended.

There is a huge difference between a heat shield losing roughly 100 kg of "slow-charring epoxy" and a heat shield losing e.g. 1000 kg carbon (#196#336). Carbon has a sublimation temperature of 3915°K and a vaporization enthalpy of 355.8 KJ/mol ≈ 5p356 Joule per mole. Taking into account heat capacity, this results in a theoretical heat absorption capacity of more than 30 MJ/kg = 7p3 Joule per kg. Thus, by vaporizing 1000 kg = 3p1 kg, the heat shield can remove 30 GJ = 10p3 Joule of reentry heat. This would already result in around 10% = 9n1 = n1 of total heat of 330 Gigajoule = 11p33 J emerging during an Apollo reentry #196).

And 3915°K also renders radiative cooling very effective: around 13 Megawatt/m2 = 7p13 Watt per square meter (#403), resp. 160 MW = 8p16 Joule/sec for the whole heat shield area. At this vaporization temperature, 2000 seconds would be enough to radiate away a quantity equal to total reentry heat.

Here some relevant NASA documentations concerning the suspicious material of the Apollo heat shields:




[4] Thermal / Pyrolysis Gas Flow-Analysis of Carbon Phenolic Material, 2002

According to [3], Table B-3 – Normal Thermo-chemical Properties for the Apollo Heat Shield Material, density is 32 lbm/ft = 512 kg/m3, and effective heat of pyrolysis is 250 Btu/lbm = 0.51 MJ/kg = 5p51 Joule per kg.

If the whole 111 kg mass loss of Apollo 4 was due to vaporization of this Apollo heat shield material, these 111 kg could have removed not even 0.06 GJ = 7p6 Joule of total reentry heat of ~ 330 GJ = 11p33 Joule (#196), i.e. a portion of not even 0.02% = 6n2.

At least the order of magnitude of the value given in [3], Table B-3 is also confirmed by [1], TABLE 1. – ENTHALPY OF SIX CHARRING ABLATORS AND TWO CHARS. Enthalpy resp. thermodynamic energy with respect to 0°C = 273°K has been measured for the ablative material AVCOAT 5026/39. The highest temperate mentioned is 662°K = 389°C with an enthalpy of 0.6364 MJ/kg ≈ 5p64 Joule/kg and a mass loss of 27.5% = n275.

I conclude that all this stuff concerning labor-intensive processes, where each individual cell of the Apollo heat shield was filled by hand with a magic material was primarily a propaganda bluff.

Quote from NASA Applies Insights for Manufacturing of Orion Spacecraft Heat Shield, 2015:

The heat shield was composed of a titanium skeleton and carbon fiber skin that gave the crew module its circular shape on the bottom and provided structural support, on top of which a fiberglass-phenolic honeycomb structure was placed. The honeycomb structure had 320,000 tiny cells that were individually filled by hand with an ablative material called Avcoat designed to wear away as Orion returned to Earth through the atmosphere. During the labor-intensive process, each individual cell was filled by hand as part of a serial process, cured in a large oven, X-rayed and then robotically machined to meet precise thickness requirements.

However, during the manufacture of the heat shield for Orion's flight test, engineers determined that the strength of the Avcoat/honeycomb structure was below expectations.

And we all know where such wrong expectations come from: They stem from the belief that such a technology worked without problems already more than 40 years ago.

#487  –  2016-12-18

Reality Check in #416:

Ablative heat shields work because the ablation at lower temperatures as the shockwave begins to form creates a layer of cooler gas that protects them from the shockwave temperatures.

Jrrarglblarg in #475:

Among the functions of the ablative heat shield is to generate a coolant fluid to carry away heat. Decomposition of the shield material isn't the entire energy budget. It's just one stage in the process of shedding the incoming heat. The gasses generated by decomposition absorb heat from the shockwave and carry them away, hotter than they were when cooked off.

How could ~ 100 kg = 2p1 kg of such a "coolant fluid" vaporizing already at a relatively low temperature be responsible for carrying away relevant parts of total reentry heat which is ~ 300 Gigajoule = 11p3 Joule? Gases emerging at a rather low temperature from such a coolant fluid cannot have a relevant cooling effect, since heat capacity of gases cannot be high enough.

How should such gases protect the heat shield? They are rapidly heated by a continuous high-speed bombardment with atmospheric particles. Imagine concretely these magic gas particles in front of the heat shield being bombarded by atmospheric particles! (#308, #336)

If we do not believe in miracles then such a decomposing "cooling fluid" also cannot absorb relevantly more energy by breaking apart or by ionizing than normal (parts of) atmospheric molecules.

On the other hand, low vaporization temperature of such a coolant fluid has a serious drawback: It reduces working temperature of the heat shield, relevant to radiative cooling. A heat shield of 3500°K = 5 ∙ 700°K radiates away 5 ∙ 5 ∙ 5 ∙ 5 = 625 times more energy than an analogous heat shield of only 700°K. (#403)

The NASA movie Apollo 4: The first flight of a Saturn V launch vehicle claims (at 4:42) that the reentering spacecraft was "cooling as it climbed back towards space" after a first heating phase. According to Figure 7c of NASA TN D5399: Entry aerodynamics at lunar return conditions obtained from the flight of Apollo 4, deceleration was around 0.3 g ≈ 3 m/s2 during cooling maximum (deceleration minimum) at t ≈ 30240 sec. According to Figure 7b, "Earth relative velocity" at t ≈ 30240 was ~ 22000 ft/s = 6.6 km/s = 3p66 m/s. Thus, heating power due to deceleration results in as much as P = E / t = F ∙ s / t = m ∙ a ∙ s/t = m ∙ a ∙ v ≈ 3p54 kg ∙ 3 m/s2 ∙ 3p66 m/s ≈ 8p1 kg m2/s3 = 8p1 Watt = 100 Megajoule/sec even during so-called "cooling" maximum.

Since vaporization of the heat-shield epoxy AVCOAT 5026-39 could have absorbed maximally ~ 7p6 Watt-second = 60 Megajoule (#471), this coolant fluid could not even have absorbed the heat emerging during only one second of the "cooling phase".

Momentum-change per second during this "cooling phase" is p = m ∙ a ∙ t ≈ 5400 kg ∙ 3 m/s. From a spacecraft speed of ~ 6600 m/s relative to the atmosphere we can calculate* an atmospheric mass of 5400 kg ∙ 3/6600 ≈ 2.5 kg = 0p25 kg bombarding every second the "protective layer" of the heat shield with a speed of ~ 3p66 m/s = 6600 m/s ≈ 24000 km/h ≈ 15000 mph.

* Simplifying assumption: Via inelastic collision, atmospheric mass is accelerated to speed of decelerating spacecraft. Then atmospheric mass leaves region before heat shield laterally, i.e. without influencing spacecraft momentum.

Thus, the assumption that gas particles stemming from ~ 100 kg of a magic coolant fluid (vaporizing during ~ 600 sec deceleration time) can somehow create a protective layer between heat shield and atmospheric bombardment, cannot be much more than wishful thinking resp. disinformation.

Declaring to have debunked is not the same as debunking

#492  –  2017-01-04

wogoga in #44:

Meteors either burn up (almost) entirely, or they or their fragments still have huge kinetic energy per mass when reaching surface.

phunk in #48:

That's not true at all. Most meteorites that reach the surface do so by falling at terminal velocity, not with "huge kinetic energy". They don't hit any harder than they would if you tossed them off a tall building.

My statement of #44 may be prone to misunderstanding, but what I wanted to say is that on average only quite small mass portions of meteors reach the Earth' surface with "terminal velocity".

Collectable meteorites are the result of a selection. Meteors and parts of disintegrating meteors or asteroids normally do not become collectable meteorites if they still have "huge kinetic energy" at impact on the ground. They get destroyed instead.

Small pieces entering the atmosphere mostly vaporize. They arrive with at least terrestrial escape velocity of 11.2 km/s = 4p112 m/s and therefore with a kinetic energy of at least ½ v ∙ v ≈ 9n5 ∙ 4p112 m/s ∙ 4p112 m/s ≈ 7p63 Joule/kg = 63 MJ/kg. For comparison, vaporization energy of iron is only 6p609 Joule/kg = 6.09 MJ/kg.

The Chelyabinsk meteor of 2013 is probably the best documented meteoritic impact. Total known weight (TKW) of all collected pieces is approximately 1000 kg = 3p1 kg. Yet initial mass is estimated to be more than 10000 metric tons = 7p1 kg. Thus initial mass is around four decimal powers = 4p1 times higher than collected mass.

This example clearly shows: The assumption that collectable meteorites are representative of the initial objects entering the atmosphere is simply wrong. The by far biggest mass portion of the Chelyabinsk meteorite dissolved into a thermal debris cloud.

Giordano in #51:

As to meteorites: it depends on the size, of course, but the average pea sized meteorite one might collect probably had a terminal velocity of 200 mph or somewhat higher at impact, the same speed as if someone threw it off of a tall building (terminal velocity). Essentially the same speed as the Apollo capsule after its atmospheric braking and just prior to the release of its drogue parachutes (200 to 300 mph). So they both shed most of their velocity through atmospheric braking, released as heat.

Correct insofar as Apollo's reentry would have been quite similar to a normal meteor (suffering substantial vaporization and disintegration) due to its flimsy heat shield (#471, #487).

#533  –  2017-02-21

Terrestrial magnetic remanence in Apollo moon rock samples

If we take into account that every kilogram of cargo produces during reentry as much heat as 4 kg of rocket fuel (#55) then the claim that the Apollo missions brought back 381 kg of moon rock becomes quite suspicious, especially if we take into account that the strength of the Apollo heat shield is primarily based on propaganda and public relations (#471, #487, #492).

Apart from the astonishing fact the Apollo rocks have the same isotopic fingerprint as terrestrial rocks (
#172) and contain traces of terrestrial life (#184), there is a further problem: Remanent magnetism of moon rocks. Quotes from On the natural remanent magnetism of certain mare basalts, 1979:

"After ten years of research directed at understanding lunar magnetism, it remains an enigma. Neither the NRM [Natural Remanent Magnetism] of the returned samples, nor the remanence which gives rise to crustal anomalies is properly understood. Consequently, the implications of lunar magnetism are unclear, and the important developments in lunar science have taken place with little or no contribution from paleomagnetism. Nevertheless, the observation of relatively stable and strong NRM in certain returned samples, and of surface magnetic anomalies, defined intriguing phenomena to be explained."

"These detailed studies of the NRM of a small group of mare basalts epitomize the problem of the interpretation of the lunar sample magnetization. To each simple interpretation, there are clear, if not insurmountable objections, and we are left with inelegant and in somewhat ad hoc explanations of the magnetization."

According to the "NASA-baked moon-rocks" hypothesis (#184), the magnetic remanence of the Apollo rocks is simply terrestrial remanence. Wikipedia on the Earth's magnetic field: Its magnitude at the Earth's surface ranges from 25 to 65 microteslas [5n25 – 5n65 Tesla]. For comparison, Surface vector mapping of magnetic anomalies over the Moon using Kaguya and Lunar Prospector observations, 2015, is "showing the highest intensity of 718 nT [~ 3n7 Tesla] in the Crisium antipodal region."

Magnetic remanence of Apollo rocks comes close to terrestrial remanence. MAGNETISM OF THE MOON AND METEORITES: IS IT ASSOCIATED WITH CRATERING AND COLLISIONAL PROCESSES?, 1991:

"However, some authors have indicated that the deduced maximum field strength of about 50 - 100 μT [5n5 – 6n1 Tesla] is too large to be plausibly generated by a dynamo in a small lunar core, and considerable attention has been directed towards another possible source, impact-generated, transient magnetic fields associated with a TRM or shock magnetization."

Is there real evidence suggesting that meteoritic or other impacts generate magnetic fields, which are even durable enough to entail "natural remanent magnetism" in surrounding solidifying rocks? I don't think so. And how probable is such a scenario? Isn't the obvious hypothesis that NASA moon rock bakers did not properly take into consideration terrestrial magnetism much simpler?

The capability of changing one's mind as an adult has been neglected by human evolution

#540  –  2017-03-01

JayUtah in #538:

Remnant magnetism in lunar surface samples is no mystery. While the Moon lacks a global magnetic field, it features several local magnetic fields. Remnant magnetism in samples we think came from deep under the lunar surface are a "mystery" only because the Moon cannot support a thermal dynamo. That's typically the reason large planets have magnetic fields. But it's not the only form of dynamism, and in the years following 1979 we discovered the tidal dynamo.

Because the Moon's orbit and its rotational axis are not coplanar, back when the Moon did have a molten center, different portions of it rotated along different axes. The Earth's gravity preferentially affected the rotation according to layers, and — more importantly — made the boundary non-spherical. That means you get a mixing of layers and a dynamo. The Moon had a magnetic field until it cooled to the point where fluid effects gave way to mere plasticity, which doesn't allow a meaningful dynamo. Researchers who tried to attribute the remnant magnetism to a thermal dynamo rightly were unable to get it to fit.

You probably mean Ancient lunar dynamo may explain magnetized moon rocks, 2011. Quotes:

The presence of magnetized rocks on the surface of the moon, which has no global magnetic field, has been a mystery since the days of the Apollo program. Now a team of scientists has proposed a novel mechanism that could have generated a magnetic field on the moon early in its history.

[The authors of November 10 issue of Nature] describe how an ancient lunar dynamo could have arisen from stirring of the moon's liquid core driven by the motion of the solid mantle above it.

Dwyer and her coauthors calculated the effects of differential motion between the moon's core and mantle. Early in its history, the moon orbited the Earth at a much closer distance than it does today, and it continues to gradually recede from the Earth. At close distances, tidal interactions between the Earth and the moon caused the moon's mantle to rotate slightly differently than the core. This differential motion of the mantle relative to the core stirred the liquid core, creating fluid motions that, in theory, could give rise to a magnetic dynamo.

"The moon wobbles a bit as it spins--that's called precession--but the core is liquid, and it doesn't do exactly the same precession. So the mantle is moving back and forth across the core, and that stirs up the core," explained Nimmo, a professor of Earth and planetary sciences at UCSC.

The researchers found that a lunar dynamo could have operated in this way for at least a billion years. Eventually, however, it would have stopped working as the moon got farther away from the Earth. "The further out the moon moves, the slower the stirring, and at a certain point the lunar dynamo shuts off," Dwyer said.

This is a highly speculative theory which probably cannot be backed up by independent research. It reminds me of the former belief in differential rotation of the sun below the convective zone:

Until the advent of helioseismology, the study of wave oscillations in the Sun, very little was known about the internal rotation of the Sun. The differential profile of the surface was thought to extend into the solar interior as rotating cylinders of constant angular momentum. Through helioseismology this is now known not to be the case and the rotation profile of the Sun has been found. On the surface the Sun rotates slowly at the poles and quickly at the equator. This profile extends on roughly radial lines through the solar convection zone to the interior. At the tachocline the rotation abruptly changes to solid-body rotation in the solar radiation zone.

Wouldn't friction between different lunar layers rotating at different speeds immediately weaken these differences in rotation speed? In any case, the hypothesis that these Apollo rocks have been created under terrestrial magnetism is more straightforward than such ad-hoc hypotheses concerning the moon.

#544  –  2017-03-03

Reality Check in #542:

Just published in 2017: Further evidence for early lunar magnetism from troctolite 76535

This is one mechanism proposed for an early Moon magnetic field - also see: A long-lived lunar dynamo powered by core crystallization

Quote from the abstract of Further evidence for early lunar magnetism from troctolite 76535, 2017:

The lunar sample with the oldest known paleomagnetic record is the 4.25 billion year old troctolite 76535. Previous studies of unoriented subsamples of 76535 found evidence for a dynamo field with a paleointensity of several tens of microteslas [i.e. ~ 5n5 Tesla]. … We infer a field paleointensity of approximately 20-40 μT [5n2 - 5n4 Tesla], supporting the previous conclusion that a lunar dynamo existed at 4.25 Ga [9p425 years ago].

Quotes from the abstract of A long-lived lunar dynamo powered by core crystallization, 2014:

For reasonable initial conditions, a solid inner core between 100 and 200 km is always produced. During its growth, a surface magnetic field of about 0.3 μT [3n3 Tesla] is generated and is predicted to last several billion years.

A hypothetical magnetic field of ~ 3n3 T = 0.000,000,3 Tesla obviously cannot explain field intensities of ~ 5n3 T = 0.000,03 Tesla. Thus, paleo-intensities derived from the Apollo 17 sample in the one paper are two orders of magnitude stronger than the intensity explained by the other paper.

However, all mentioned paleo-intensity values for troctolite 76535 (from 5n2 to several 5n1 Tesla) come suspiciously close to terrestrial magnetism (5n25 - 5n65 Tesla).

© – No rights reserved – – 2017-04-25