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Problem I - Cratering. 2. The Earth


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#501 indydave

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Posted 26 June 2016 - 03:07 PM

I should have mentioned this before, Pi, but best wishes for your 70th birthday!  And for making it through another school year! 



#502 indydave

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Posted 26 June 2016 - 05:39 PM

Pi>> keeping in mind that the only ones who can radiate to space are those exposed to space. The physicists have the expertise to know this.... the "BA in Bus Mgt" lacks that expertise.<<

As you know, your physicist experts never even mentioned radiation much less tried to estimate what amount of radiation over what amount of time would have taken place. Nor was any effort made to try to determine what would be the effect of having a layer of hot molecules at the very top of our atmosphere for a period of a few hours or days until all of the heat was radiated away to space. Again this topic involves craters on the earth and I believe I have proven the point I wanted to make here, which is that objects like we have discussed would indeed make craters on the moon without making craters on Earth.

#503 indydave

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Posted 27 June 2016 - 01:51 PM

Pi>>One of the reasons I was hesitant to accept the results of the lunar crater model you found was that I felt the parameters may have exceeded the limitations of the model and I wanted to hear from one of the researchers.  To some extent that was confirmed when I tried a density of 0.01 and the diameter of the resulting crater was smaller than the impactor but, IIRC, it was still pretty deep.  For that reason, there is still some concern the 0.1 density may exceed the useful parameters of the model .... but I can't "prove" it.>>

 

I found a users manual LINK  and a paper LINK  for the second crater size site LINK and found some equations. They did of course confirm that density of the impactor is a factor in determining size.  I could not tell where the density factor is used in the formulae, or if there is some limit for it.   I ran the program using .01 density and I did NOT get a crater smaller than the impactor, but it was equal in size for the final rim diameter.  I thought perhaps it would do that for any density below a certain amount so I checked it for .02 density and it was NOT equal.  A 10km diameter impactor made a 13.3km rim diameter.  That was using regolith for the target, but it suggested in a dialog box for larger impactors, hard soil/soft rock for the target.  Using that, the .02 impactor that was 10km diameter made a 17.2km final rim.  At .08 it was 28.4 km and at .1 it was 30.7km...with the final crater being 3x the diameter of the impactor.  All those seem quite reasonable.  This means that a .1 object that is 71km in diameter would indeed be able to make a 160km (100 mi) crater rim diameter on the moon...which is what I said before (I think).  If the target is regolith, then the impactor would have to be 82km.  BTW, if the density is half what we've discussed (.05) then you could still get a 160km/100 mi crater if the impactor was 93km diameter. 



#504 indydave

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Posted 07 July 2016 - 02:46 PM

indydave, on 11 Jun 2016 - 4:39 PM, said:snapback.png

However, in your calcs...found on pg 1 of A Rain of Fire and Brimstone LINK ...you fail to calculate how a hot and tiny particle at the very top of our atm would RADIATE AWAY its heat into space in a microsecond (just like tiny meteorites do)...with no bad effect on the earth below. AND you bragged about how you ran your calcs past some PhD's in Physics and they agreed with you and yet THEY ALSO failed to include radiation in their evaluation of what your wrote! I guess this BA in Bus Mgt could see things those PhD physicists could NOT!


Pi>>We recall that very differently. You claim that a molecule size particle at the top of the atmosphere will radiate away its heat to space. I pointed out (correctly) that the particle will radiate its heat in all directions cooler than the particle.... not just toward space as you claimed. The particle that enters the atmosphere will radiate its energy to the ones above, below, and to the sides of it. The only particle that is able to radiate its heat to space is the one at the top of the column and is exposed to space. Even then, it is able to radiate energy to space on only that side that is exposed to space. The remainder of the energy will be radiated to other molecules.>>

After you directly address the points I have recently made which show USING YOUR OWN SOURCES that an object like we have discussed COULD INDEED produce large craters on the moon w/o making craters on Earth....I would like to address (whether in this thread or another) your idea that heat would NOT be radiated into space at a faster rate than the objects which might fall back during the shut-off phase would fall. For YEARS now I have suggested that your cratering point was invalid and we need to GET ON WITH the job of addressing if adding heat to the upper atm would be lethal...so PLEASE HURRY UP to either concede or reply to my latest comments about cratering so we can look at the heat question.

Regarding heat...the duration of time before a particle would fall back could be MUCH longer than you or I have assumed. Something could remain within E's gravity influence (not beyond EV) and yet it could take a long time before it would return to the atm. For instance, the ISS is NOT beyond EV, and yet it is not reentering the atm. Yes, that is a well-controlled orbit, but objects that are NOT in controlled orbits at all can still remain aloft for long periods w/o returning to the atm. I read recently that one of the assent stages of Apollo 12 was intended to be slung (by Moon's gravity) to cause it to escape E's gravity but there was a failure and it remained within E's hold FOR YEARS and then something caused it to be released to orbit the sun and then recently it was detected returning! So it did not just go up, fail to reach EV and then immediately fall back. That illustrates that the amount of time available for a particle to radiate its heat is "up for grabs"...and that time period should be discussed. Also this time could be much longer because the power of gravity is less at greater heights than it is on the Earth's surface...so even IF it fell straight back, as the particle begins to fall back (from an average of 1/2 the EV altitude, whatever that is), it will very SLOWLY start to fall, until as it nears the Earth's surface THEN its acceleration is 9.8m/s/s. So it could take many hours or even a day or two...maybe more...to complete its fall IF it fell straight back rather than going into a type of orbit (that might eventually decay...maybe weeks or years later, as with Apollo 12's stage). The question then would be if (as you claim) the heat of a particle entering E's atm at (say) 100 miles up radiates upward to the molecule above it, and above that one etc...if it takes just a tiny part of ONE SECOND to radiate to the one directly above it (etc.) then how many of those molecules would there be up to whatever one half the EV altitude is? PLUS you must consider that the sphere/shell diameter would be far greater at that height, so there would be vastly more space BETWEEN particles than there would be at 100 miles alt. That means there would be fewer to radiate to, rather than radiating into space itself. BTW, although the particle would radiate downward, it would be at a VASTLY different rate than radiating upward due to the fact that those molecules above the topmost part of the atm would be at near absolute zero, while the temp of the clouds below would be far warmer. Stefan-Boltzmann considers the difference in temps of the two objects and the radiation is proportional to THE FOURTH POWER of that temperature difference. So there would be hardly any heat going downward compared to upward. One other point you miscalculated is that you figured a speed of an avg of 5.6km/s energy. However that figure (half of 11.2km/s) would not be close to correct if the heating of the particle took place at 100 mi height. It could be half that...or even less. That is because there would be a LOT of speeding up yet to be done in that last 100 miles of falling...and that would never take place before energy is radiated away. The launching would happen at the surface, but the final destination would NOT be at the surface, but rather 100 miles up, where maximum heating would happen.

I'm sure all your physics PhD's you consulted with had considered all of this 3 years or so ago when you started off "Fire and Brimstone"...but then, um, why didn't any of them MENTION these points??? The truth is, they (nor you) never CONSIDERED how quickly heat of reentry could be radiated away, NOR how much time there would be available to radiate it all, or other important factors affecting heating. This BA in Business Mgt had to point them out.

#505 indydave

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Posted 08 July 2016 - 01:51 PM

I went here to find gravitational acceleration for different altitudes above Earth.  I added the radius of the Earth so the first one would be an altitude of 206,400 km.  Here are the results. 

 

.009 m/s/s at 200,000 km.   .035 at 100,000   .126 at 50000   .574 at 20000   1.49 at 10,000   3.08  at 5000   7.31 at 1000   6.30 at 160 (100 mi.)  I stopped at 160 km because that is the place where the heating would happen and upward radiation would occur. 

 

So if you assume that the EV altitude is the distance to the Moon (I have not yet found what that actually is...it is probably considerably more because the moon is NOT beyond E's gravity, traveling 1km/s) then half that would be 200,000 km.  That means if on average, stuff which did NOT achieve EV fell back from 200,000 then (as a rough guess) we could say that the average rate of fall (the rate from 100,000 km...conservatively***) might be .035m/s/s.  That means it accelerates 1 meter/second every 29 seconds on average during its entire fall.  I don't know how to do the precise calc, however it would take 29,000 seconds (8 hours) to reach a speed of 1 km/s and another 8 hours to reach 2km/s for an average of 1km/s for that 16 hours. In 16 more hours it would reach 4 km/s and average 2 km/s for 32 hours.  It would travel 230,400 km or a bit more than halfway to the moon.  So my estimate is about 30 hours to fall back from 200,000 km. 

 

That of course assumes that it fell back on a straight enough trajectory to hit the atm on the first pass and did NOT orbit around the Earth and then hit the atm on some later orbit...which is the most likely event. 

 

So this means that if a particle were to heat up and begin to radiate upward to hit the next highest particle above it and above it (etc) then at the LEAST there would be 30 hours available for the heat to reach the topmost particle and radiate to open space.  PLUS as the distance from Earth grows, the sphere above gains radius (i.e. 206400 km at half the orbit of the Moon, compared to 6560 km at the top of the atm) and that means the distance between particles would be much much greater, allowing that the heat would NOT hit another particle directly above it all the way upward, but at some lower altitude there would be NO particle directly above it. 

 

I am happy to discuss this heat estimation further AS SOON AS PI CAPITULATES regarding cratering!  Let's hear from you soon, Pi!  Stop the stalling!

 

===

 

*** the halfway point from a starting point of 200,000km (halfway to moon) in DISTANCE is 100,000km.  However the halfway point in TIME would be considerably further away.  Probably closer to 150,000km.  That means my estimate of the average rate of acceleration of .035km/s/s is likely high...so the time to fall is likely longer. 



#506 indydave

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Posted 08 July 2016 - 03:32 PM

I made a substantial edit to the prior post just now, so if you read it an hour ago, you should reread it...especially if you reply to it. 



#507 piasan

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Posted 08 July 2016 - 10:38 PM

OK.... with regard to the behavior of a 0.1 density impactor on the Earth or the moon....

There is another SL-9 paper here (Takata et. al), which says on p 7 that a 1000g/cm object (dense ice or water) which is 1km diameter traveling 60km/s would use all its energy by 100 bars.  (I cannot paste in the text...it is an image file)  The SL-9 paper we have been using (LINK) (Korycansky et. al.) says that 90% of the energy of a .6 density object going 61km/s would be expended (based on their fig. 3c) at about 15 bars, so 100% would be about 20 bars (based on my est. of the chart).  So in this case, an object that is 60% as dense travels only 20% as deep.  Using these two densities in the two SL-9 articles, it suggests there is a curved line and so if you drop density from .6 to .1 you would expect even LESS than 1/6th the depth of penetration.  This tends to confirm that my estimate of 1/6th the depth for 1/6th the density is more than fair.  It makes sense because there is a 1 to 1 relationship between mass (density, in this case) and energy, provided that there is no change in velocity.  The idea of it being even LESS than 1/1 ratio (density/depth) is likely due to the fact that there is greater surface area per unit of mass when the object is less dense, so there is more contact with the target atm.  This assumes a shape that remains FIXED.  BUT if the shape of the object is deformed more (and it WOULD be), then that WOULD cause it to penetrate LESS than 1/6th as far.  My 1/1 estimate was MORE than fair. 

The Takata link didn't work, but if an object penetrates to 100 bars on Jupiter, it would easily penetrate Earth's atmosphere where the pressure is only 1 bar.

 

Using data from Figure 7 of the SL-9 paper we have been using (LINK), I get the following "eyeball" values: 

Density   Penetration (km)

 2.7              120

1.76             105

0.917            80

0.6                60

Note:  Penetration is below the point where the atmospheric pressure = 1 bar.  This is the pressure at the surface of the Earth.  In other words, all of these would impact the surface with energy to spare.

 

When I did an Excel plot of the data (sorry, couldn't copy-paste it), the "best fit" line shows a penetration well in excess of 50 Km for a density of 0.  However, the plot does have a downward trend (most notable in the interval from density 0.917 and 0.6 that strongly suggests it's not a linear trend.  The best fit line at a density of 0 shouldn't be above 50 km.  In fact, at density of 0, there should be a negative penetration since the "0" line is a basically atmospheric pressure at Earth's surface.  Basically, the data shows that above a density of 1.0, the plot follows a fairly straight line, but below that, it curves sharply toward 0.  The problem is there is no data below a density of 0.6 so any proposal we may make is no more than guesswork.

 

Your "1/6 the mass means 1/6 the penetration" claim is completely misguided. 

1)  Atmospheric resistance doesn't follow a linear function so most of the energy is dissipated at lower altitudes.  (Note: the pressure scales are logarithmic.) 

2) Examining Figure 1 shows energy release doesn't peak until about 10 bars.  One sixth of that is 1.67 bars.  You still have surface impact.

3)  The abstract states 50% of the energy reaches the -70km level.  Using Figure 1 for conversion, that's a pressure of about 8 bars.  See (2).

4)  The paper suggests "pancaking" is limited to about 2x the radius.  That would translate to 4x the area, not 6x.  This means deeper penetration.  See (2).

5)  The abstract points out that depth of penetration is "weakly" increased by grater mass.  Table 8 shows this relationship is also a curve.  Using the crater calculator (link) to get the mass of a 0.1 density 70km diameter bolide, it comes up 1.8e16kg or 1.8e19g.  Notice, the logarithmic (horizontal) scale stops at 1e16g.  This takes the line another 3 full scale steps to the right.  The downward trend can be expected to continue at an increasing rate as the chart is extended to the right.  See (2). 

6)  The paper specifically considers ablation. (See statement to the right of Fig. 8)

 

... I think we should limit comments about that and try to reach a point of resolution about crater formation on the Moon and on Earth. I don't think we should have to wait much longer for your concession about that. You are not of course giving away the whole store by conceding this because you could still question the matter of too much heat or even whether anything could ever be launched at all. But this topic deserves to be put to bed and I believe unless you have some argument to make you should give me your concession at this time, at least tentatively. Of course if you find new information you can retract your concession.

Based on the data presented in the SL-9 paper we've been using and other factors concerning the crater calculators, here are my conclusions.

First: we have both experienced issues with the calculators when low densities are used.  From things stated by both of them and the models reported by the SL-9 paper, it is clear there is a lower bound to their reliability.  It appears, from the data in Fig. 7 that they are not useful with densities of only 0.1... especially if an atmosphere is involved.

Second:  On very careful review of the SL-9 paper I am convinced a 0.1 density bolide with a diameter of 70 km (43 mile) will reach the surface of the earth and release most of its energy there .... resulting in a significant crater that would be, at a minimum, tens of miles across.

Third:  There is a point at which an incoming projectile would be of low enough density it wouldn't reach the surface.... but the proposed object isn't it.

 

There is no need for a concession on my part.

 

I agree this topic needs to be put to bed.  We're more than 500 posts into it and haven't come to agreement on even the most simple things, such as how many events with these objects we should expect Earth to encounter.

 

Faulkner has already addressed the launch process and finds it doubtful Brown's process can penetrate the atmosphere as cleanly as claimed.  And that's for a one time blowout.... the atmospheric penetration, and backpressure, would need to be dealt with each time Brown's "flutter" cycle repeats.

 

We've already had a considerable discussion (here) about the very formation of these objects with no agreement.

 

We've discussed the orbital mechanics with no agreement.

 

So, let's see... we've disagreed on:

1)  The number of encounters Earth should expect to have with the proposed impactors.

2)  The effects of those encounters if and/or when they may occur.

3)  The behavior of whatever ejecta is in near Earth orbit on near misses by the Earth/moon.

4)  The formation of the proposed impactors in the first place.

 

I think that about covers it ......



#508 indydave

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Posted 09 July 2016 - 09:04 AM

Pi>> 2) Examining Figure 1 shows energy release doesn't peak until about 10 bars. One sixth of that is 1.67 bars. You still have surface impact.<<

That might be true of an object traveling 61 km/s. Why did you conveniently neglect to notice that a drop in speed to 17 km/s would mean 1/13 the energy and therefore as I have said the energy would be used up high in Earth's atmosphere and cause no cratering? It's hard to believe you would have accidentally forgotten the speed component. What is your reply about that? Let me make it easy for you . What is 1.67 divided by 13?

>> I think that about covers it ....<

No, I don't believe you have covered anything until you address the much slower speed of the object we are discussing. Please don't take two months this time to address this component and put this to bed with your concession rather than your bloviating.

#509 indydave

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Posted 09 July 2016 - 11:58 AM

The Takata SL-9 paper is here:  https://www.google.c...XUhwDKw&cad=rja

 

I am not sure why links are not working when I use the link function here.  I wanted to do it that way so it would not crash the site...and also so the long google link was not given.  I will also email you a copy.



#510 indydave

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Posted 09 July 2016 - 12:25 PM

>>

Using data from Figure 7 of the SL-9 paper we have been using (LINK), I get the following "eyeball" values: 

Density   Penetration (km)

 2.7              120

1.76             105

0.917            80

0.6                60

Note:  Penetration is below the point where the atmospheric pressure = 1 bar.  This is the pressure at the surface of the Earth.  In other words, all of these would impact the surface with energy to spare.

 

When I did an Excel plot of the data (sorry, couldn't copy-paste it), the "best fit" line shows a penetration well in excess of 50 Km for a density of 0.  However, the plot does have a downward trend (most notable in the interval from density 0.917 and 0.6 that strongly suggests it's not a linear trend.  The best fit line at a density of 0 shouldn't be above 50 km.  In fact, at density of 0, there should be a negative penetration since the "0" line is a basically atmospheric pressure at Earth's surface. >>

I think you did something wrong using your Excel plot...since a density of .6 travels 60km below 1 atm and you say a 0 density is well in excess of 50km.  Please recheck that.

 

Also remember that you are looking at various densities but ALL are at the speed of SL-9...which is 61km/s.  We have been discussing speeds of 17km/s and that is 13x less energy.

 

>>The Takata link didn't work, but if an object penetrates to 100 bars on Jupiter, it would easily penetrate Earth's atmosphere where the pressure is only 1 bar.>

Making a bare assertion like that proves nothing.  Show your calcs...like I have.  When you consider BOTH the lower density AND the lower speed, the object never gets below high up in the atm. before all the energy is used. 

 

I'd like to also point out that in K's fig 7, the depth in atm for a 2.7 basalt is about 90 bars for the middle peak.  For a .6 object (about 1/4 as dense) it is about 8 bars...less than 1 tenth the pressure.  So if that same relationship applies below .6, then an object which is 1/6 that density (i.e. .1) could be expected to travel 1/10th (or less) as deep.  So by that measure, the middle peak would be at about .8 bars.  You said 1.67 which is more than twice what is most likely, given the chart.  But even then, you STILL MUST FACTOR IN THE SLOWER SPEED and so far you have NOT.  If you divide 1.67 by 13 that gives you .128 bars which is far up in the atm. 



#511 indydave

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Posted 09 July 2016 - 06:13 PM

Pi >> 5) The abstract points out that depth of penetration is "weakly" increased by grater mass. Table 8 shows this relationship is also a curve. Using the crater calculator (link) to get the mass of a 0.1 density 70km diameter bolide, it comes up 1.8e16kg or 1.8e19g. Notice, the logarithmic (horizontal) scale stops at 1e16g. This takes the line another 3 full scale steps to the right. The downward trend can be expected to continue at an increasing rate as the chart is extended to the right.<<

You really need to read more carefully. That chart is showing 5 different masses however it is also showing 5 different diameters. For instance the chart has triangles at the top. There are 5 triangles and the 5th one is 5 times the diameter of the second one. That means it is 125 times greater volume and greater mass. That explains why that is a curved line. The five triangles show the various depths where 10% of the energy is expended for different sizes. The open squares are 5 different sizes of objects where 90% of the energy is expended. If you look at that you see that the second square is a one kilometer object and 5th square is a five kilometer object which is 125 times greater in mass. It is 5 times larger in diameter but it travels 10 times deeper because it is 125 times heavier. The mass is not directly proportional to the cross sectional area of the object as it increases in size.

If you want to try to make a point about density then that chart in figure 8 is not going to help you because I believe those densities are unchanged however the total mass is increased due to the size (diameter) of the bolides.

#512 indydave

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Posted 10 July 2016 - 06:50 PM

Pi>> Faulkner has already addressed the launch process and finds it doubtful Brown's process can penetrate the atmosphere as cleanly as claimed. And that's for a one time blowout.... the atmospheric penetration, and backpressure, would need to be dealt with each time Brown's "flutter" cycle repeats.<<

All the sudden you want to change the subject from craters! What you apparently don't seem to appreciate is that although Faulkner so far has opposed the Brown model because he believes there is too much heat, in private discussions with me he has indicated that he believes the low density objects striking the moon to cause craters and not harm the Earth is a good idea. After writing my heat article for CRSQ it is my intention to write another one about craters and perhaps have Danny Faulkner co-author it with me. Of course if that were to happen all of the sudden he would not be someone that you want to try to refer to positively in this forum, eh?

#513 piasan

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Posted 11 July 2016 - 03:52 AM

The Takata link didn't work, but if an object penetrates to 100 bars on Jupiter, it would easily penetrate Earth's atmosphere where the pressure is only 1 bar.

Making a bare assertion like that proves nothing.  Show your calcs...like I have.  When you consider BOTH the lower density AND the lower speed, the object never gets below high up in the atm. before all the energy is used. 

OK.... went to the Takata paper and reviewed it.  There wasn't even a suggestion of anything that would help your case regarding the 0.1 density, 70km bolide you propose.

 

What calculations are needed to conclude that an object penetrating to an atmospheric pressure of 100 bars on Jupiter would penetrate to the surface of the Earth?  Your whole argument is based on the claim that a proposed bolide with a density of 0.1 and a diameter of 70km will not penetrate Earth's atmosphere.  The surface pressure of Earth's atmosphere is 1 bar.  If the object survives to that pressure, it has done the equivalent of passing through Earth's atmosphere are reaching the surface.  A pressure of 100 bars is 100x Earth's atmospheric pressure.

 

I think if anyone needs to show calcs on that point, it would be you showing how an object can penetrate a planet's atmosphere until the pressure is nearly 1500 psi, but will vaporize before reaching Earth's surface where the pressure is only 14.7 psi.

 

BTW, the Takata paper was published and revised before the SL-9 impact with Jupiter.  The intention was to predict the results of the event.  The actual fireworks far exceeded expectations.

 

Pi>> 2) Examining Figure 1 shows energy release doesn't peak until about 10 bars. One sixth of that is 1.67 bars. You still have surface impact.<<

That might be true of an object traveling 61 km/s. Why did you conveniently neglect to notice that a drop in speed to 17 km/s would mean 1/13 the energy and therefore as I have said the energy would be used up high in Earth's atmosphere and cause no cratering? It's hard to believe you would have accidentally forgotten the speed component. What is your reply about that? Let me make it easy for you . What is 1.67 divided by 13?

First:  It's another non-linear relationship so you can't just divide by 13.

Second:  IIRC, when I pointed out the difference in velocity would mean SL-9 would shred even sooner on Jupiter than on Earth, you produced a reference claiming that wouldn't happen.  Now, you're claiming that it won't penetrate because it would have less energy....  The exact same argument I made.  Sorry, but you can't have that one both ways.

Third:  Slowing to 17 km/s and 1/13 the energy also means 1/13 the braking force.  So it will slow much more slowly.

Fourth:  Table 8 strongly suggests an object of the mass you suggest will penetrate to more than 10x the pressures calculated.  (More on that later.)

 

In short... there is no justification for your divide by 13 adjustment.



#514 indydave

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Posted 11 July 2016 - 03:18 PM

Pi>>OK.... went to the Takata paper and reviewed it.  There wasn't even a suggestion of anything that would help your case regarding the 0.1 density, 70km bolide you propose.>>

 

Yes, there was...because it gave penetration depth for a 1.0 object...which could then be compared to the .6 object depth in the K paper.  However, you found very similar data in the chart where you "eyeballed" it.  We can set aside Takata if you wish.

 

>>What calculations are needed to conclude that an object penetrating to an atmospheric pressure of 100 bars on Jupiter would penetrate to the surface of the Earth?  Your whole argument is based on the claim that a proposed bolide with a density of 0.1 and a diameter of 70km will not penetrate Earth's atmosphere. >>

No...if it has a speed of 61km/s it probably would.  If it is 17km/s it would not.  You have just WHIFFED the speed factor because YOU KNOW it kills your argument.  You must show why you believe it WOULD penetrate to the surface and you have done nothing of the kind. 

 

>>The surface pressure of Earth's atmosphere is 1 bar.  If the object survives to that pressure, it has done the equivalent of passing through Earth's atmosphere are reaching the surface.  A pressure of 100 bars is 100x Earth's atmospheric pressure.>>

Um, yeah...I got that a long time ago.  More stalling by you.

 

>>I think if anyone needs to show calcs on that point, it would be you showing how an object can penetrate a planet's atmosphere until the pressure is nearly 1500 psi, but will vaporize before reaching Earth's surface where the pressure is only 14.7 psi.>>

I did that, by doing the calcs and converting for slower speed and lower density.  Do you agree that reducing the density to .1 drops the energy by a factor of 6?  Yes or no?  And then if you drop the speed to 17 then it drops it again by a factor of 13?  Do you agree to this?  You SHOULD.  And then it is simply a question of what effect on penetration there would be when the energy is drastically lower. 

 

>>First:  It's another non-linear relationship so you can't just divide by 13.>>

I don't know what you mean.  Of course it is not linear because energy is proportional to the SQUARE of the speed.  When you reduce speed from 61 to 17 that is 3.6 to 1.  Squaring it is 12.9.  The total energy drops a factor of 77.4 if you include the reduced density.   YOU YOURSELF blundered when you said that the lower density would result in a depth of 1.67 atm, believing that would mean it would hit the surface.  BUT YOU FORGOT TO FIGURE THE SPEED FACTOR.  If it is true that energy at the top of the atm will determine how much the atm will stop an object and how deep it will go, then the depth on Jupiter for the SL-9 object means it would penetrate 1/77th as deep (in bars) on Earth.  If not then you need to explain WHY not. 

>>Second:  IIRC, when I pointed out the difference in velocity would mean SL-9 would shred even sooner on Jupiter than on Earth, you produced a reference claiming that wouldn't happen. >>

I don't want to try to resurrect what we both said then, however I would agree that if you have the same density for 2 objects then the faster one would indeed shred faster (in terms of time, not sure about depth/distance) due to the higher pressure at the front of the object.  BUT a much less dense one would shred faster than a denser one at the SAME SPEED...agreed?  Regardless, the impact effects site says that objects which are .1 traveling at 17km/s would break up at 106km altitude.  That means they would become much smaller and therefore would be decelerated even quicker, thereby depositing their energy higher up.  Do you agree?

 

>>Third:  Slowing to 17 km/s and 1/13 the energy also means 1/13 the braking force.  So it will slow much more slowly.>>

That is absurd.  Of course a faster object would have more impact on the atm, but that does NOT mean it would decelerate quicker.  It would take LONGER to decelerate.  So we would expect a slower object to penetrate LESS deep, not more.  Picture a slow bullet hitting ballistic gel compared to a fast one.  In both cases, the inertia of the gel relmains the same, Regardless of what "braking force" you decide the gel has on each bullet, there is no disputing that the faster bullet would go deeper.  Unless one is being ridiculously obtuse.  Are YOU???



#515 piasan

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Posted 11 July 2016 - 06:24 PM

We are indeed moving into the area of heat which is not what this topic is about so I think we should limit comments about that and try to reach a point of resolution about crater formation on the Moon and on Earth.... you could still question the matter of too much heat or even whether anything could ever be launched at all. But this topic deserves to be put to bed ....

.... I agree this topic needs to be put to bed.  We're more than 500 posts into it and haven't come to agreement on even the most simple things....

Faulkner has already addressed the launch process and finds it doubtful Brown's process can penetrate the atmosphere as cleanly as claimed.  And that's for a one time blowout.... the atmospheric penetration, and backpressure, would need to be dealt with each time Brown's "flutter" cycle repeats.

All the sudden you want to change the subject from craters!

Really?  You say I could "still question ... whether anything could ever be launched at all."  I respond that "Faulkner has already addressed the launch process...."  In what way does my response that the launch process has already been addressed mean I want to change the subject?

 

Shall we talk about your repeated attempts to redirect this topic to heat .... almost from your very first post on this thread.  In fact, there are three posts of yours on this very page discussing heat.



#516 indydave

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Posted 11 July 2016 - 09:13 PM

This topic was set up as tangential to your topic about heat and as soon as we complete this one we should continue with the obvious question of where the energy that is deposited into the atmosphere...not on the surface of Earth, making craters... goes. I have not addressed the heat question here other than to reply to what you said previously in one of your recent posts. However there is an obvious shift that should be made as soon as you will concede that no craters on Earth would be made by the objects we are discussing. You should stop being so stubborn. An object that is slower and less dense than SL - 9 would not cause impact craters on Earth. The energy is 1/77th as much.

#517 indydave

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Posted 11 July 2016 - 09:19 PM

To try to simplify things, let's go back to where you said that if the object is .1 density then it would still penetrate to 1.67 atmospheres. You said that without any consideration at all of the reduction in speed, so why don't you tell us what effect the slower speed would have on that penetration. The energy from slower speed would be 1/13. So how deep do you think this .1 object would go if it was slower? Let's hear it, Pi. Don't dodge!

#518 indydave

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Posted 12 July 2016 - 02:00 PM

Whether Pi likes it or not, I'm going to mention a major point that helps also to remove heat when an impactor enters the atmosphere. The energy is not simply deposited into the atmosphere and remains there as what has been implied so far. Instead the process involves the development of a plume which travels UPWARD and pulls much of the heated atmosphere up into space where it takes a while before it falls back downward. It could involve about 15 to 20 minutes or so and during that time the very bright plume cools off and is no longer bright enough to show up as hot when viewed in telescopes. That means that all of the heated atmosphere molecules have a chance to be cooled down dramatically before they even are permanently back inside the atmosphere of the planet. This was observed with Jupiter. This is not just my conjecture. If I can ever get Pi to concede the point of this thread which is the question about cratering, then we can move on to look at what effects there would be of heating in the upper atmosphere...and the upward development of a plume is part of that consideration. Pi needs to swallow his pride and get on with tapping out on this argument. There has been enough of his crapping on the board!

#519 indydave

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Posted 13 July 2016 - 01:15 PM

From Takata's paper...re. a 2km impactor:

 

At t = 82 sec, the expanding fireball accelerates upward and the expansion velocity reaches —10 km/sec at the shock front. Because of the expansion of the highly pressurized gas into a near vacuum, the plume easily reaches an altitude of —350 km (above the 1 atm level...high above the cloud tops--INDY) (-1-gbar pressure level), i.e., >15 scale heights above the 1-bar level where the plume originates. Now a mushroom cloud-like plume develops more symmetric radial expansion. The expanding plume radius is approximately 250 km (250x the radius of the impactor--INDY) at this stage. The shock temperature of plume head is 1000K. The temperature is still 10000K inside of the fireball, since shock-heated gas with high internal energy is continuously funneled upward from deeper within the cometary trajectory to the base of the fireball—plume. The pressure at the shock front is still 100 times the ambient pressure at the same altitude, so the plume will continue to rise, until all the kinetic energy and internal heat is transformed into gravitational potential energy (i.e. gravity halts its upward rise--INDY) or is radiated to space.

 

This suggests to me that all the energy of the impact (in the atm...not on the surface) is shot back upward along the path of the impactor and into space where it radiates away ALL THE HEAT, and only the energy of the residue falling back onto the cloud tops is left to heat the atm.  And that would be probably no more than 1/10000th or so (my estimate)...since the energy of falling from 350km to (say) 160km is vastly less than a speed of 61km/s or even 17km/s.  

 

This means that there would be no cratering at all AND there also would be virtually no heating of the atm...and what heating there may be would be above the cloud layer so the ark occupants would be protected. 



#520 indydave

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Posted 13 July 2016 - 02:42 PM

I wrote:>>This suggests to me that all the energy of the impact (in the atm...not on the surface) is shot back upward along the path of the impactor and into space where it radiates away ALL THE HEAT, and only the energy of the residue falling back onto the cloud tops is left to heat the atm. >>

 

As I reflect on the quote, it is possible that "until all the kinetic energy and internal heat is transformed into gravitational potential energy" is referring to all the energy IN THE PLUME, rather than all the energy of the impactor.  However, since the plume is made up of 40x more atm mass than impactor mass, then it is likely that the heated atm molecules would be most of what contributes heat to the plume.  I suppose it would depend on how fast the heat could propagate inside the atm before the atm is sucked up into the plume.  It is quite possible that all the heated gases are sent up into space TO COOL DOWN, before they fall back with much much less energy.






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