Forum Index Register Members List Events Mark Forums Read Help

 International Skeptics Forum Kinetic energy at atmospheric reentry from lunar mission

 Welcome to the International Skeptics Forum, where we discuss skepticism, critical thinking, the paranormal and science in a friendly but lively way. You are currently viewing the forum as a guest, which means you are missing out on discussing matters that are of interest to you. Please consider registering so you can gain full use of the forum features and interact with other Members. Registration is simple, fast and free! Click here to register today.
 2nd August 2016, 02:41 AM #1 wogoga Critical Thinker   Join Date: Apr 2007 Posts: 334 Kinetic energy at atmospheric reentry from lunar mission Escape velocity from Earth is 11.2 km/s. A spacecraft leaving the gravitational field of the moon and falling on the Earth accelerates to a speed of 11 km/s before reaching the atmosphere at an altitude of around 120 km (see: Atmospheric entry). The kinetic energy corresponding to 11 km/s is around 60 Mega-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 (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." According to this figure, Apollo needed around 250 sec to descent from altitude 120 km to 60 km. This results in an average speed of v = 60 km / 250 s = 240 m/s. To a speed of 240 m/s correspond only 28 kilo-Joule per kg, which is a negligible quantity in comparison to the 60 Mega-Joule corresponding to the entry speed of 11 km/s. Thus in the order of 99.9% of the kinetic energy must have disappeared before falling to an altitude of 120 km. Yet friction due to atmospheric pressure above 120 km 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: "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."So the question becomes even more acute: How could Apollo reduce its speed from 11 km/s to less than 1 km/s before or when entering the atmosphere (at an altitude of around 120 km)? Cheers, Wolfgang www.pandualism.com
 2nd August 2016, 03:11 AM #2 abaddon Penultimate Amazing     Join Date: Feb 2011 Location: Republic of Ireland Posts: 23,198 Originally Posted by wogoga Escape velocity from Earth is 11.2 km/s. A spacecraft leaving the gravitational field of the moon and falling on the Earth accelerates to a speed of 11 km/s before reaching the atmosphere at an altitude of around 120 km (see: Atmospheric entry). The kinetic energy corresponding to 11 km/s is around 60 Mega-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 (source). Figure 4.1.7-25 of Returning from Space: Re-entry shows "Re-entry Profiles for the Shuttle Versus Gemini and Apollo": http://www.pandualism.com/upload/apollo_reentry.JPG "Notice Gemini and Apollo re-entered much more steeply than the Space Shuttle." According to this figure, Apollo needed around 250 sec to descent from altitude 120 km to 60 km. This results in an average speed of v = 60 km / 250 s = 240 m/s. To a speed of 240 m/s correspond only 28 kilo-Joule per kg, which is a negligible quantity in comparison to the 60 Mega-Joule corresponding to the entry speed of 11 km/s. Thus in the order of 99.9% of the kinetic energy must have disappeared before falling to an altitude of 120 km. Yet friction due to atmospheric pressure above 120 km 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: "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."So the question becomes even more acute: How could Apollo reduce its speed from 11 km/s to less than 1 km/s before or when entering the atmosphere (at an altitude of around 120 km)? Cheers, Wolfgang www.pandualism.com Your math is not wrong. Your math is irretrievably broken. Your conceptualisation of the problem is similarly irretrievably broken. Your underlying assumptions are broken too. __________________ Who is General Failure? And why is he reading my hard drive? ...love and buttercakes...
 2nd August 2016, 03:12 AM #3 catsmate No longer the 1     Join Date: Apr 2007 Posts: 26,346 Originally Posted by abaddon Your math is not wrong. Your math is irretrievably broken. Your conceptualisation of the problem is similarly irretrievably broken. Your underlying assumptions are broken too. This.I suspect he's dabbling in Apollo denialism. I believe this may be of use. __________________ As human right is always something given, it always in reality reduces to the right which men give, "concede," to each other. If the right to existence is conceded to new-born children, then they have the right; if it is not conceded to them, as was the case among the Spartans and ancient Romans, then they do not have it. For only society can give or concede it to them; they themselves cannot take it, or give it to themselves.
 2nd August 2016, 04:33 AM #4 Crossbow Seeking Honesty and Sanity     Join Date: Oct 2001 Location: Charleston, WV Posts: 13,799 Originally Posted by wogoga ... snipped for brevity ... So the question becomes even more acute: How could Apollo reduce its speed from 11 km/s to less than 1 km/s before or when entering the atmosphere (at an altitude of around 120 km)? Cheers, Wolfgang www.pandualism.com They cheated. That is how they were able to slow down, by cheating. __________________ "But Presidents are not kings, and Plaintiff is not President." - Judge Chutkan On 15 FEB 2019 'BobTheCoward' said: "I constantly assert I am a fool." A man's best friend is his dogma.
 2nd August 2016, 06:35 AM #5 Gord_in_Toronto Penultimate Amazing     Join Date: Jul 2006 Posts: 21,882 Originally Posted by Crossbow They cheated. That is how they were able to slow down, by cheating. Since it has been scientifically proven that re-entry is impossible, low Earth orbit must be crowded with Astronauts. QED. __________________ "Reality is what's left when you cease to believe." Philip K. Dick
 2nd August 2016, 08:08 AM #6 3point14 Pi     Join Date: Nov 2005 Posts: 21,063 Originally Posted by Gord_in_Toronto Since it has been scientifically proven that re-entry is impossible, low Earth orbit must be crowded with Astronauts. QED. Jeb is. Well, that's LKO, but it amounts to the same thing. __________________ Up the River! Anyone that wraps themselves in the Union Flag and also lives in tax exile is a [redacted]
 2nd August 2016, 08:24 AM #7 The Don Penultimate Amazing     Join Date: Nov 2002 Location: Sir Fynwy Posts: 33,136 Originally Posted by Gord_in_Toronto Since it has been scientifically proven that re-entry is impossible, low Earth orbit must be crowded with Astronauts. QED. There is Abacha Tunde http://www.huffingtonpost.com/entry/...b0c3c550527f0b
 2nd August 2016, 09:07 AM #8 Dave Rogers Bandaged ice that stampedes inexpensively through a scribbled morning waving necessary ankles     Join Date: Jan 2007 Location: Cair Paravel, according to XKCD Posts: 33,049 I'm sure Jay Utah can answer the question in the OP much more comprehensively than I ever could, but just to be a little bit constructive 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. Since your starting numbers are wrong, the rest of your calculations aren't really any use. Dave __________________ There is truth and there are lies. - President Joseph R. Biden, January 20th, 2021
 2nd August 2016, 09:29 AM #9 Dr.Sid Illuminator   Join Date: Sep 2009 Location: Olomouc, Czech Republic Posts: 3,588 Reentry is all about angle. You can think about it as for every altitude there is 'safe' speed at which you can travel at for unlimited time, without overheating. Obviously this speed is lower at lower altitudes. The trick with reentry is to never be above safe speed at any given moment. You can think about it also as steps. You enter the very outskirts of atmosphere, and will stay at that altitude as until you are slow enough to descent another few hundred meters, again slow down enough, and again descent. Such approach would require lots of maneuvering, and it's not how it's done, it's just an example. 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. Thing is .. the angle is different for every spacecraft, and it depends on how much will the spacecraft decelerate due to the air friction. The Space shuttle was constructed to fly, and it uses the same configuration while going up. It's designed to have very low drag. Therefore it will slow down only very slowly .. so it needs very shallow reentry angle. Btw. Space shuttle also does high G horizontal turns to bleed speed faster, which is not part of your diagram. Old style single use reentry modules were designed differently. The are built to have high drag coefficient (and high drag induced stability). Thus they will use much steeper reentry angle. They will simply slow down fast enough. Space shuttle descent is controlled, so it can stay above ballistic course. It can do the shallow reentry in controllable fashion. 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. It's another reason why steep reentry angle is used. While there is huge difference between 5 and 15 degrees reentry angle, there is much less different between 60 and 70. That's why old style reentry module has different reentry angle. It will also be exposed to lot higher deceleration forces. Last edited by Dr.Sid; 2nd August 2016 at 09:32 AM.
 2nd August 2016, 09:38 AM #10 theprestige Penultimate Amazing   Join Date: Aug 2007 Location: The Antimemetics Division Posts: 55,852 Originally Posted by wogoga So the question becomes even more acute: How could Apollo reduce its speed from 11 km/s to less than 1 km/s before or when entering the atmosphere (at an altitude of around 120 km)? Do you even heat shield, bro? Get on my level.
 2nd August 2016, 09:42 AM #11 Giordano Penultimate Amazing   Join Date: Apr 2011 Posts: 17,646 Originally Posted by wogoga Escape velocity from Earth is 11.2 km/s. A spacecraft leaving the gravitational field of the moon and falling on the Earth accelerates to a speed of 11 km/s before reaching the atmosphere at an altitude of around 120 km (see: Atmospheric entry). The kinetic energy corresponding to 11 km/s is around 60 Mega-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 (source). Figure 4.1.7-25 of Returning from Space: Re-entry shows "Re-entry Profiles for the Shuttle Versus Gemini and Apollo": http://www.pandualism.com/upload/apollo_reentry.JPG "Notice Gemini and Apollo re-entered much more steeply than the Space Shuttle." According to this figure, Apollo needed around 250 sec to descent from altitude 120 km to 60 km. This results in an average speed of v = 60 km / 250 s = 240 m/s. To a speed of 240 m/s correspond only 28 kilo-Joule per kg, which is a negligible quantity in comparison to the 60 Mega-Joule corresponding to the entry speed of 11 km/s. Thus in the order of 99.9% of the kinetic energy must have disappeared before falling to an altitude of 120 km. Yet friction due to atmospheric pressure above 120 km 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: "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."So the question becomes even more acute: How could Apollo reduce its speed from 11 km/s to less than 1 km/s before or when entering the atmosphere (at an altitude of around 120 km)? Cheers, Wolfgang www.pandualism.com I will not bother to ask you to re-examine your own math, but just ask you to focus on the starting points you yourself provided: The unmanned Sample Return capsule entered the atmosphere at 12.9 km/s and braked to parachute speeds within 110 seconds using one of the most drastic re-entry profiles ever employed. This caused it to encounter a peak deceleration of 34 gs. Manned Apollo used a different re-entry profile but arrived at 11 km/s and required (over) 250 seconds to eventually slow down enough for parachute deployment. Due to the longer time (and you cut off events below 60 km altitude ) and the precise shape of the re-entry Apollo experienced 6 to over 7 gs decceleration (Mercury up to 11g!). The shuttles took even longer to break and maxed out around 3 g. Doesn't seem hard to understand why this works this way. If you look at the actual spacecraft in museums you will see that the re-entry heat nonetheless took quite on toll on them- either ablative shielding or thermal tiles. They certainly heated up enormously during re-entry and a lot of engineering went into preventing transfer of this heat into the cabins. 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. Many enter with a combined speed (adding Earth's movement) of 71 km/s. Versus multi-ton engineered spacecraft at 12 km/s. 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. Why do you still use velocity when you mean speed? Last edited by Giordano; 2nd August 2016 at 09:45 AM.
 2nd August 2016, 10:01 AM #12 3point14 Pi     Join Date: Nov 2005 Posts: 21,063 Originally Posted by Giordano The unmanned Sample Return capsule entered the atmosphere at 12.9 km/s and braked to parachute speeds within 110 seconds using one of the most drastic re-entry profiles ever employed. This caused it to encounter a peak deceleration of 34 gs. Holy McJesus, that's a braking maneuver. How hot did it get? __________________ Up the River! Anyone that wraps themselves in the Union Flag and also lives in tax exile is a [redacted]
 2nd August 2016, 10:04 AM #13 Jrrarglblarg Guest   Join Date: Nov 2010 Posts: 12,673 Oh, yeah, nobody gets launched and returned at the forces the early guys withstood. Apollo guys were around 11g peak and shuttles were around 3g IIRC. Soyuz reentries are apparently pretty rough - nominally 4g peaks but pretty variable from mission to mission apparently.
 4th August 2016, 02:29 AM #14 wogoga Critical Thinker   Join Date: Apr 2007 Posts: 334 Originally Posted by wogoga in #1 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." According to this figure, Apollo needed around 250 sec to descent from altitude 120 km to 60 km. This results in an average speed of v = 60 km / 250 s = 240 m/s. Originally Posted by 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, around 8.8 km are needed to lower altitude by one kilometer. This means, average speed during descent from 120 km to 60 km is not 240 m/s as I erroneously calculated, but 8.8 ∙ 240 m/s = 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 when reaching altitude 120 km with 6.5°. To a speed of 2.1 km/s corresponds only 2.2 Mega-Joule per kg, which is a rather negligible quantity in comparison to the 60 MJ corresponding to the entry speed of 11 km/s. Thus around 96% of the kinetic energy would have disappeared before falling to an altitude of 120 km. For an astronaut of 80 kg these "missing" 96% would result in 4.7 Tera-Joule, enough to heat up ten tons (10000 kg) of steel by 1000° Kelvin. Cheers, Wolfgang
 4th August 2016, 03:19 AM #15 abaddon Penultimate Amazing     Join Date: Feb 2011 Location: Republic of Ireland Posts: 23,198 Originally Posted by wogoga 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, around 8.8 km are needed to lower altitude by one kilometer. This means, average speed during descent from 120 km to 60 km is not 240 m/s as I erroneously calculated, but 8.8 ∙ 240 m/s = 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 when reaching altitude 120 km with 6.5°. To a speed of 2.1 km/s corresponds only 2.2 Mega-Joule per kg, which is a rather negligible quantity in comparison to the 60 MJ corresponding to the entry speed of 11 km/s. Thus around 96% of the kinetic energy would have disappeared before falling to an altitude of 120 km. For an astronaut of 80 kg these "missing" 96% would result in 4.7 Tera-Joule, enough to heat up ten tons (10000 kg) of steel by 1000° Kelvin. Cheers, Wolfgang Still fractally wrong. __________________ Who is General Failure? And why is he reading my hard drive? ...love and buttercakes...
 4th August 2016, 03:38 AM #16 Jack by the hedge Safely Ignored     Join Date: Oct 2009 Posts: 12,828 I don't think you are starting with an accurate source for the Apollo reentry profile, as an initial Google image search for the term seems to suggest.
 4th August 2016, 06:35 AM #17 abaddon Penultimate Amazing     Join Date: Feb 2011 Location: Republic of Ireland Posts: 23,198 Originally Posted by wogoga 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, around 8.8 km are needed to lower altitude by one kilometer. This means, average speed during descent from 120 km to 60 km is not 240 m/s as I erroneously calculated, but 8.8 ∙ 240 m/s = 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 when reaching altitude 120 km with 6.5°. To a speed of 2.1 km/s corresponds only 2.2 Mega-Joule per kg, which is a rather negligible quantity in comparison to the 60 MJ corresponding to the entry speed of 11 km/s. Thus around 96% of the kinetic energy would have disappeared before falling to an altitude of 120 km. For an astronaut of 80 kg these "missing" 96% would result in 4.7 Tera-Joule, enough to heat up ten tons (10000 kg) of steel by 1000° Kelvin. Cheers, Wolfgang Straight out of the box:Pretty much none of the relationships involved is linear, as you assume. Ablative heat shield Apollo capsule is an aerodynamic lifting body and actually ascends during part of re-entry Invalid use of average speed Just a partial list of how wrong this whole proposal is for starters. __________________ Who is General Failure? And why is he reading my hard drive? ...love and buttercakes...
 4th August 2016, 08:13 AM #18 WhatRoughBeast Graduate Poster   Join Date: Apr 2011 Posts: 1,480 Originally Posted by wogoga 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. Cheers, Wolfgang It's nice to see that you're rethinking your assumptions, and here's another which you need to rethink The attitude of the Apollo is the angle of the vehicle's path. Nope. It's the angle of the vehicle's axis with respect to the current velocity vector. It has nothing to do with the angle of that velocity vector with respect to local horizontal. Your revised calculation is getting better, but the path angle was actually (in the early stages of reentry) much shallower than you realize. Consequently, the velocity (and the kinetic energy) were much greater.
 8th August 2016, 05:37 AM #20 wogoga Critical Thinker   Join Date: Apr 2007 Posts: 334 Originally Posted by wogoga in #14 As sin 6.5° ≈ 0.11, around 8.8 km are needed to lower altitude by one kilometer. This means, average speed during descent from 120 km to 60 km is not 240 m/s as I erroneously calculated, but 8.8 ∙ 240 m/s = 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 when reaching altitude 120 km with 6.5°. To a speed of 2.1 km/s corresponds only 2.2 Mega-Joule per kg, which is a rather negligible quantity in comparison to the 60 MJ corresponding to the entry speed of 11 km/s. Thus around 96% of the kinetic energy would have disappeared before falling to an altitude of 120 km. For an astronaut of 80 kg these "missing" 96% would result in 4.7 Tera-Joule, enough to heat up ten tons (10000 kg) of steel by 1000° Kelvin. Originally Posted by abaddon in #15 Still fractally wrong. You are right: Missing kinetic energy for one 80 kg astronaut is 4.7 Giga-Joule and not 4.7 Tera-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 = 7p1 gram) 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 0.47 Joule per gram per Kelvin = 9n47 J/g/K = nine-ne four seven J/g/K. With 0.47 kJ = 4.7 ∙ 102 J = 2p47 J = two-po four seven Joule we can heat up 1 kg = 3p1 gram by one degree Kelvin. For ten metric tons = 7p1 g we need 7p1 g ∙ 9n47 J/g/K = 6p47 Joule/Kelvin. In order heat up these ten tons (7p1 gram) of heat capacity 9n47 J/g/K by 1000 Kelvin (3p1 K), we need 9p47 Joules. Summary: 9n47 J/g/K ∙ 7p1 g ∙ 3p1 K = 9p47 J Cheers, Wolfgang Efficient Concept-Formation for Languages of the Future
 8th August 2016, 06:04 AM #21 abaddon Penultimate Amazing     Join Date: Feb 2011 Location: Republic of Ireland Posts: 23,198 Originally Posted by wogoga You are right: Missing kinetic energy for one 80 kg astronaut is 4.7 Giga-Joule and not 4.7 Tera-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 = 7p1 gram) 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 0.47 Joule per gram per Kelvin = 9n47 J/g/K = nine-ne four seven J/g/K. With 0.47 kJ = 4.7 ∙ 102 J = 2p47 J = two-po four seven Joule we can heat up 1 kg = 3p1 gram by one degree Kelvin. For ten metric tons = 7p1 g we need 7p1 g ∙ 9n47 J/g/K = 6p47 Joule/Kelvin. In order heat up these ten tons (7p1 gram) of heat capacity 9n47 J/g/K by 1000 Kelvin (3p1 K), we need 9p47 Joules. Summary: 9n47 J/g/K ∙ 7p1 g ∙ 3p1 K = 9p47 J Cheers, Wolfgang Efficient Concept-Formation for Languages of the Future Swing and a miss. __________________ Who is General Failure? And why is he reading my hard drive? ...love and buttercakes...
 8th August 2016, 06:55 AM #22 abaddon Penultimate Amazing     Join Date: Feb 2011 Location: Republic of Ireland Posts: 23,198 Originally Posted by wogoga You are right: Missing kinetic energy for one 80 kg astronaut is 4.7 Giga-Joule and not 4.7 Tera-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 = 7p1 gram) 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 0.47 Joule per gram per Kelvin = 9n47 J/g/K = nine-ne four seven J/g/K. With 0.47 kJ = 4.7 ∙ 102 J = 2p47 J = two-po four seven Joule we can heat up 1 kg = 3p1 gram by one degree Kelvin. For ten metric tons = 7p1 g we need 7p1 g ∙ 9n47 J/g/K = 6p47 Joule/Kelvin. In order heat up these ten tons (7p1 gram) of heat capacity 9n47 J/g/K by 1000 Kelvin (3p1 K), we need 9p47 Joules. Summary: 9n47 J/g/K ∙ 7p1 g ∙ 3p1 K = 9p47 J Cheers, Wolfgang Efficient Concept-Formation for Languages of the Future Put it this way. Even if you eventually correct your calculations, 4.7x109 Joules is sufficient to heat 18076923 cups of coffee to 90 C from room temperature. Did you have a point? __________________ Who is General Failure? And why is he reading my hard drive? ...love and buttercakes...
 10th August 2016, 05:18 AM #25 wogoga Critical Thinker   Join Date: Apr 2007 Posts: 334 Originally Posted by 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 (1p3 / 8n3 = 3p1) 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. Cheers, Wolfgang Last edited by wogoga; 10th August 2016 at 05:39 AM.
 10th August 2016, 08:13 AM #26 abaddon Penultimate Amazing     Join Date: Feb 2011 Location: Republic of Ireland Posts: 23,198 Originally Posted by wogoga 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 (1p3 / 8n3 = 3p1) 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. Cheers, Wolfgang Did you have a point to make? Different energies are different. And........ __________________ Who is General Failure? And why is he reading my hard drive? ...love and buttercakes...
 10th August 2016, 08:21 AM #27 Jack by the hedge Safely Ignored     Join Date: Oct 2009 Posts: 12,828 Originally Posted by abaddon Did you have a point to make? Different energies are different. And........ Well, yes but compare dropping a brick on your head from 50 feet versus 50,000 feet. Hugely different yet somehow the same. No. I don't really know what the point was either.
 10th August 2016, 08:38 AM #28 Giordano Penultimate Amazing   Join Date: Apr 2011 Posts: 17,646 Originally Posted by wogoga 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 (1p3 / 8n3 = 3p1) 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. Cheers, Wolfgang Here is a good article for you to begin to re-channel your thinking into a better understanding of how it works: https://en.wikipedia.org/wiki/Atmospheric_entry Best regards!
 10th August 2016, 04:21 PM #29 Thor 2 Philosopher     Join Date: May 2016 Location: Tiny town west of Brisbane. Posts: 6,922 Originally Posted by wogoga 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 (1p3 / 8n3 = 3p1) 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. Cheers, Wolfgang Can't see a point being made here Wolfgang, sorry.
 15th August 2016, 09:26 AM #30 wogoga Critical Thinker   Join Date: Apr 2007 Posts: 334 Originally Posted by abaddon in #26 Did you have a point to make? Originally Posted by Jack by the hedge in #27 No. I don't really know what the point was either. Originally Posted by 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. Cheers, Wolfgang Last edited by wogoga; 15th August 2016 at 09:29 AM.
 15th August 2016, 09:28 AM #31 Dave Rogers Bandaged ice that stampedes inexpensively through a scribbled morning waving necessary ankles     Join Date: Jan 2007 Location: Cair Paravel, according to XKCD Posts: 33,049 Still not seeing a point here. Dave ETA: I'll see your Apollo and raise you Concorde. __________________ There is truth and there are lies. - President Joseph R. Biden, January 20th, 2021
 15th August 2016, 09:35 AM #32 phunk Illuminator     Join Date: Aug 2007 Posts: 4,127 Originally Posted by wogoga The big discontinuity only occurred after the Apollo program. Because no one wanted to pay for it anymore.
 15th August 2016, 09:47 AM #33 Giordano Penultimate Amazing   Join Date: Apr 2011 Posts: 17,646 Originally Posted by wogoga [snip] 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. [snip] Cheers, Wolfgang Perhaps you would be willing to present as numbers this "huge" difference in energy that is represented by the higher re-entry speeds from a lunar mission versus an orbital one? I believe you have the information as to the re-entry speeds. There is no doubt that the former is higher than the latter, but I find the term "huge" to not quite fit my calculations, which suggest a less than two-fold difference and why, by slightly modifying the same basic approaches, the former would be impossible whereas the latter you acknowledge has been achieved routinely for decades.
 15th August 2016, 10:42 AM #34 abaddon Penultimate Amazing     Join Date: Feb 2011 Location: Republic of Ireland Posts: 23,198 Originally Posted by wogoga 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. Cheers, Wolfgang First, quoting post # is pointless. Violations get shuffled of to AAH and whatnot, threads get merged and variously divide up. Citing post numbers without links is either ignorance of how things actually work, or an intentional tactic. Pick one. Re-entry is not disproven, it is proven. we do it all the time. You do not understand orbits, nor the velocities involved. You don't understand atmospheric layers and the meaning thereof. You don't understand that climbing Everest is a different challenge to travelling to the Moon. You don't understand what the thermosphere actually is. Frankly, at this point, I am sort of wondering what it might be that you actually understand? If anything? __________________ Who is General Failure? And why is he reading my hard drive? ...love and buttercakes...
 15th August 2016, 11:15 AM #35 sts60 Illuminator     Join Date: Jul 2007 Posts: 3,770 Originally Posted by wogoga 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. There was no discontinuity in "ease". The Apllo program ended. It ended early, because there was no political \$upport to keep it up. In fact, the Apollo workforce was already ramping down well before the first landing. So what, exactly, is your point? Last edited by sts60; 15th August 2016 at 11:20 AM. Reason: Corrected to "[i]no[/i] political..."
 15th August 2016, 11:19 AM #36 sts60 Illuminator     Join Date: Jul 2007 Posts: 3,770 Bonus question: Suppose manned lunar exploration was officially set to be an active goal of the U.S. civil space program. Do you expect a monotonic decrease in "ease" to apply, nearly half a decade after the last active manned exploration program ended? Why, or why not?
 15th August 2016, 06:13 PM #37 ferd burfle Graduate Poster     Join Date: Jan 2006 Location: Just short of Zeta II Reticuli Posts: 1,551 Originally Posted by Giordano Here is a good article for you to begin to re-channel your thinking into a better understanding of how it works: https://en.wikipedia.org/wiki/Atmospheric_entry Best regards! Originally Posted by wogoga 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. Cheers, Wolfgang One great big chunk of irrelevance. If you want to discuss the actual science, start by following Giordano's link above and focus on the heat transfer discussions. Until you do, any further discussion will be pointless. Sent from my iPad using Tapatalk __________________ What are the odds something bad will happen to a submarine in a Godzilla movie?-Brandon
 15th August 2016, 09:09 PM #38 Ranb Penultimate Amazing     Join Date: Jul 2003 Location: WA USA Posts: 10,445 Originally Posted by wogoga From the ISS even a sailplane (glider aircraft) could safely reach the Earth, at least in principle. You mean like the Space Shuttle?
 16th August 2016, 06:32 AM #39 Jrrarglblarg Guest   Join Date: Nov 2010 Posts: 12,673 Originally Posted by Ranb You mean like the Space Shuttle? Exactly like the space shuttle. Which, I will point out, was never flown to the moon. The question of why has been answered over the years in a variety of ways, but the energy required to decelerate to earth orbit and reentry speeds exceeds the amount of fuel that a shuttle can carry.
 16th August 2016, 06:43 AM #40 3point14 Pi     Join Date: Nov 2005 Posts: 21,063 Originally Posted by Jrrarglblarg Exactly like the space shuttle. Which, I will point out, was never flown to the moon. The question of why has been answered over the years in a variety of ways, but the energy required to decelerate to earth orbit and reentry speeds exceeds the amount of fuel that a shuttle can carry. I've often wondered about that - even if you filled the cargo bay with as much hypergolic propellant as it could take and piped it through to the oms? I mean, I've always guessed the answer was no and that re-entry frrom a lunar trajectory would be, erm, bad, but that cargo bay could hold a lot of fuel. ETA: I am aware that there are about a million other technical reasons that make this impractical. __________________ Up the River! Anyone that wraps themselves in the Union Flag and also lives in tax exile is a [redacted] Last edited by 3point14; 16th August 2016 at 06:50 AM.

International Skeptics Forum