Just goes to show, propaganda only works as a one way street. It does not like to be challenged.

From some Extinction Rebellion coordinator notes I found some time ago:

"Don’t have a Q & A. This allows the extreme people who want it to be one way to bring everyone else down. 80% are normal people 20% political absolutists. There to appropriate your energy." While it is not uncommon that organizers prefer people at their venues that are interested in learning what they wish to say and to teach, the problem here is that the other side does not show up.

Perhaps one could also see it from a different angle. The US is a big country, maybe the people invited did not have time. Also we don't know who they invited. If I rent a room in town and invite some, who consider themselves to be big shots, to argue about the climate, it is possible there will be a number of empty chairs. When I watched the video, I also thought what if the climate alarmist turn out to be justified in their alarmism, but not for the reasons they expect? What if those who claim all is as usual turn out to be wrong, but again not for the reasons they expect?

What are the chances of changing the focus for and against global warming? Joe is trying, and I will try to play with his idea in terms of the energy of the large impactor that hit northern Greenland. In the process I will draw on posts in other threads, so for the most part I am not going to link to original papers, because they can be found there. It is going to be long and windy, but at the end I hope to find out how much energy such an object could melt and how much the sea would rise if all the melted ice from the impact left the Greenlandic ice sheet and became a part of the ocean. The conclusion I reach is that the energy of such a large asteroid corresponds to five times the present yearly total global energy consumption. Under the very unrealistic assumption that all the energy of the asteroid was used to melt ice and nothing more, the seas would rise around 2 cm.

Good point. I'd amend that comment then to: "Of course, nowhere in the paper or his awareness is there any reference to the well-known hundredfold spike in the concentration of platinum in Greenland ice cores that are dated to 12,890 YBP that is highly suggestive of a meteorite impact."

Btw, the Greenland crater is disputed largely because some say that it is an "improbably recent impact" and that "an impact of this size should occur only once every few million years".

**Small impact and big impact**
In

Near-Earth objects and close calls I made some calculations that showed that on any day there will be on average from 2000-4000 meteoroids of 2.8 m and more moving between the Earth and out to the distance of the Moon, counting the space all around the Earth and assuming about 6 of those hit the planet each year. If one stays with the 3 that are official then we are down to 1000-2000. Due to fast speeds and both the Earth and the meteoroids moving at thousands of km per hour, even a slow meteoroid moving at only 10,000 m/s (6 miles per second) would stay less than 24 hours in this space, most of them less time and on average probably only a few hours.

The calculated number of flybys per day means that on any starry night we go for a walk and look up to the Moon, then in a sphere around the Earth and out to the average distance of the Moon there should at a few hundred meteoroids flying by with a potential total impact energy of more than 1 kt. Since the space we occupy in the big sphere space reaching out to the Moon only includes 1/200,000 the chances of being hit are fortunately small.

**Working with the energy of a an impactor**
The data from NASA,

mentioned in this post by Pashalis show that a meteoroid with a size of about 2.8 m can be counted on to deliver more than one kt of TNT if it hits, assuming it only travels at 10,800 m/s and most travel faster. This is a starting point to discuss the energy involved when meteoroids or asteroids impact the Earth, including the one that hit Greenland.

To begin with what we know, and for a few visual representations,

here and

here I collected some videos from 2019 alone that together give an idea of what different estimated sizes for fireballs calculated in kt will mean when they are observed, including two that were too small to make the cut for the NASA database.

1 kiloton of TNT (kt) is equivalent to 4.184E+12 joules. (See this

post). 1 kiloton of TNT is the explosive energy contained in a 1000 kg of TNT.

The Hiawatha impact (See this

post) is calculated to have delivered about ~3 × 10^21 J of energy joules. This calculation is based on the size of the crater, an estimate of the size of the impactor, its know composition and estimated speed. Now we can try to find out how many kt the Hiawatha impactor corresponded to.

To make the above numbers more workable, let us write 4.184E+12 joules as 4.184 × 10^12 J To find the total impact energy of the Hiawatha impactor measured in kt we can say that the Hiawatha impactor's energy in kt is 3 × 10^21 J/(4.184 × 10^12 J/kt) =717,017,208 kt ~7 × 10 ^8 kt The size of the

nuclear bombs that the US used to terrorize Japan with was 13-18 kt (Hiroshima) and 18-22 kt (Nagasaki). So this impact in Greenland had an energy that was 35,000,000 larger than the largest of the two bombs. That is too big a number to make any sense, but we can try another.

From

World energy consumption - Wikipedia one finds the Global energy consumption in 2013 to be in the order of 5.67 × 10^20

joules 13,541

Mtoe) In

2018 it was more like 14,000 Mtoe.

About "toe" is

The

**tonne of oil equivalent** (

**toe**) is a

unit of energy defined as the amount of energy released by burning one

tonne of

crude oil. It is approximately 42

gigajoules ....

Multiples of the

*toe* are used, in particular the megatoe (Mtoe, one million toe)

If one toe is 42 gigajoules or 42 × 10^9 J, then one Mtoe is a million times more or 42 × 10^15 J If we take 14,000 of those we have 14,000 × 42 × 10^15 J =14 × 42 × 10^18 J = 5.88 × 10^20 J ~ 6 × 10^20 J. If we compare this number with the estimated total impact energy of 3 × 10^21 J as we can do by dividing the larger number with the smaller number then we find that 3 × 10^21 J /6 × 10^20 J = 30 × 10^20 J /6 × 10^20 J = 5 That means the energy of the impactor that hit Greenland had an energy that was 5 times larger than the total human energy consumption in 2018. Moreover according to the report mentioned in the

post the ground under the crater is warmer than should be expected. From research on crater on Mars it is known that residual heat from an impact this size can linger on for 100 ka or 100,000 years. In other words, although this crater due to sufficient water flow will cool down faster, it seems the proposed impact is still making a small contribution to the heat balance of the planet.

The asteroid that hit Greenland hit at the edge of the present day ice sheet, it is not yet clear if there was any ice when it fell. Models show the impactor would make a whole 7 km deep and a crater 20 km wide which then would collapse on itself within about a minute and leave a 31 km wide crater 800 meters deep.

Now let's imagine we could use all the energy of this object to melt ice. The question turns out to be complicated, because what is the temperature in an ice sheet? Which ice sheet? Where on the ice sheet What time of the year? Besides what about secondary effects? When does the ice sheet really move and when does it stay? Maybe the ice would not have to be melted directly to end up in the sea and be melted later?

After presenting af few of the problems with determining the physical properties of the ice, we are going to melt with our asteroid energy.

Here is one picture from

https://geosci.uchicago.edu/~kite/doc/cuffey_ch_9.pdf
Here we see the temperature rises as we go down deeper in the ice sheet. One reason is friction, because as the ice moves across obstacles heat is generated. Heat is also generated, as it is deformed within when one part flows quicker or slower than another or when more pressure is applied from above due to new snow piling up. Another source of heat is coming from the Earth itself. This old paper from 1955

Ice Movement and Temperature Distribution in Glaciers and Ice Sheets* mentions:

The rate of outflow of geothermal heat (

*q*) is roughly uniform over the world at about 38 cal./cm.2/year

6 which gives a gradient of 1° C. per 44 m. at the base of the ice sheet.

In a thick ice sheet the pressure affects the

pressure melting point of water. From

Temperature distribution in glaciers and ice sheets - Environmental Change
The melting temperature of ice decreases with increasing pressure at a rate of 0.072°C per 106 pascals (MPa; lPa = INm-2). As an example, the pressure at the base of a 2000 m thick glacier or ice sheet is about 17.6 MPa, corresponding to a lowering of the melting point to —1.27°C.

The only way the above makes sense is if 106 pascals is in fact 1×10^6 pascal which is 1MPa. At any rate, at the bottom of the ice sheet is may be blow zero and still be like water.

I don't think we will need the next excerpt, but I still found it interesting that they introduce the issue of cycles.

Ice Movement and Temperature Distribution in Glaciers and Ice Sheets*
(b) Catastrophic advances of glaciers

If circumstances outlined in this paper produce basal ice temperatures below melting point, a greater shear stress will be required to produce ice flow. This means that greater surface slopes and ice thickness will be formed. Now if the basal ice temperature in such an ice sheet or glacier is raised by some phenomenon—the most likely one being by an increase in ice movement over the bed—the shear stress due to the greater ice thickness formerly accumulated will be in excess of that required to make the ice flow at this higher temperature. As this is an unstable state the flow may continue to increase until the basal temperature reaches melting point, leading finally to a relatively rapid advance of the glacier front. With basal temperatures now at melting point the flow and ice thickness may remain similar to that of a temperate glacier until conduction and the downward “advection” of cold ice again start to lower the temperature of the ice rock interface with consequent thickening of the glacier.

**Such a chain of events may well be cyclic with a very long period.** The mechanism may be the explanation of such catastrophic glacier advances as those of Bråsvelbreen in Svalbard and of the Kutiàh Glacier

20 in the Karakoram recently.

Now I want to get more practical. If I look at the diagramme above showing the temperatures in different ice sheets across the globe, I decide the energy from the asteroide is going to hit the center of the ice sheet in

Greenland, GISP2 for instance, where the temperature appears to vary between minus 30 and minus 10. I average the profile and say it is minus 25 degrees Celsius, and I decide the energy is only going to be used to melt ice, not to hit the ground, make a crater which it no doubt is going to do, or to boil some of the water, so it evaporates and heats up the atmosphere, instead of melting more ice, or blow some of the ice far far away which is also going to happen, only in this experiment it does not happen. We only want to melt ice. To do that we first have to heat it from -25 Celsius to a melting point of 0 Celsius. At 0 degrees Celsius, ice will melt, but we have to add energy to break the crystal structure and transform the water in solid state to water in a fluid state.

From

Ice - Thermal Properties
Looking at the above table I will average the specific heat values between the ice which is -25 C and then the temperature I want to reach which is zero degrees Celsius. A reasonable approximation, that is easy to use is 2.0 kJ/kg K This means I need to add 2.0 kJ to heat one kg or liter of water by one degree. For more on specific heat capacity

see this Wiki.

On

Latent Heat of Melting for some common Materials one finds that

The latent heat of fusion of a substance is the amount of heat required to convert a unit mass of the solid into liquid without change in temperature

They have a table and claim the transition from ice to water at 0 degrees Celsius or 273 degrees Kelvin requires 334 kJ/kg

Now I'm ready to first heat up 1 kg of ice from minus 25 C to 1 kg of ice at zero degrees and then to make sure I continue to add so much heat that it actually melts. Since 1 Celsius is the same as 1 Kelvin I, we can say we need to raise the temperature of the ice 25 K

For that I need 25 K×2.0 kJ/kgK that is 50 kJ/kg Now I have one kg of ice at zero degrees, and I need to add heat to get one kg of water at zero degrees. That is simple, as the table gave us that we need to add 334 kJ/kg this means that for one kg of ice at minus 25 C we need 50 kJ/kg plus 334 kJ/kg =384 kJ/kg.

What we now need to do is to divide the total energy of the asteroid 3 × 10^21 J by 384 kJ/kg This will give us the number of kg ice we can melt.

1 kJ means 1 kilojoule or a 1000 joule. 384 kJ/kg is the same as 384 × 10^3 J/kg or 3.84 × 10^5 J/kg We enter the numbers in a calculator and find that 3 × 10^21 J / 3.84 × 10^5 J/kg = 3E21/3.84E5= 7,812,500,000,000,000 kg =7.8125E15 We know the names kilo, Mega, Giga, Tera, Peta and this is 7.8 Peta kg of ice that can be melted. How much is that. One cubic meter has a thousand kg so our number is 7,812,500,000,000 m^3. One cubic kilometer (km3) has 1,000,000,000 cubic meters, so we have 7812,5 km3, but this is water not ice. To find out the volume of ice we have melted with our astronomical amount of energy we need to convert the volume of water into volume of ice. The density of ice -25 degrees Celsius is according to the table from the engineering toolbox 919.6 kg/m^3 We take the number 7,812,500,000,000,000 kg and divide by 919.6 kg/m^3 to get 8,495,541,539,799,9 m3 and with one cubic kilometer still being 1,000,000,000 m^3 the above will give about 8495 km^3. To find the area we divided this volume by the thickness of the ice which was 3 km and this gives us 2832 km^2. If we assume this is a square plot then the side of this area is the square root of 2832 km^2 which is 53.2 km. If we rather like to think of the area as round, we use the formular for a the area of a circle as pi×r^2. If we divide 2832 km^2 by pi we get 901.45 km^2 and taking the square root of that will give a radius of around 30 km.

The whole ice sheet on Greenland has a volume of

Greenland ice sheet - Wikipedia 2,850,000 cubic kilometres. Comparing our figure of a possible melt of 8495 km^3 of ice with 2,850,000 km^3 of ice this would give us 8,495/2,850,000 which is 0.00298 or around 3 ‰. The same Wiki says that if the whole ice sheet on Greenland melted, then the sea would rise 7.2 meters. Because the coast line often is flat on the sides, the first cubic kilometers of water would make more difference than the last, but if we forgive ourselves for the possible error then 3 ‰ of 7.2 meters is 3 ‰ of 7200 millimeter (mm) which is 3 × 7.2 mm which is 21.6 mm which is less than an inch or about 2 cm. One problem with this water is that it would only be at zero degrees. It would be colder than most of the ocean water. It would also be lighter as it would it fresh water. This means it would be more on the surface in the beginning, and it may take some time to heat up. As a result there is no guarantee such an event would lead to a warmer weather, rather on the contrary.