<blockquote>
<p>Could an icy/extremely-cold asteroid/comet ever strike and cool the Earth?</p>
</blockquote>
<p>No.</p>
<p>The smallest possible velocity is about 11 km/s velocity, and that would require a near miracle. For that to happen, the object would have to enter the vicinity of the Earth through a very small keyhole near the Sun-Earth L1 or L2 point with a very small velocity (a comet will not have that very small velocity), then be nearly ballistically captured by the Moon, and then fall to Earth. A comet would be more likely to hit the Earth at 40 km/s, or larger.</p>
<p>And that means that it does not matter how cold or how hot the object was prior to impact. The kinetic energy of the impact will dwarf whatever thermal energy the object had prior to the impact.</p>

CC BY-SA 4.0
2020-09-04T21:18:56.733Z
I'm sure this is true (and probably orders of magnitude of true) but some energy numbers to back this up would be nice. By my calculation for your low end velocity asteroid would have 60500Kj/kg. While iron has a specific heat capacity of 440 j/KgC so a "temperature deficit" of about 290°C ends up at 129Kj/kg (so a pathetic contribution)
Richard Tingle
CC BY-SA 4.0
2020-09-04T21:22:59.393Z
(Seems water has one of the highest specific heat capacity at 4182 j/kgC so I should have used that but that doesnt really help much)
Richard Tingle
CC BY-SA 4.0
2020-09-03T09:33:53.653Z
2020-09-03T09:33:53.653Z
David Hammen
<p>No, it's not possible to cool the Earth with an asteroid impact.</p>
<p>The mass of any asteroid that could hit the Earth is far too small to be a heat sink.  The <a href="https://en.wikipedia.org/wiki/Chicxulub_impactor" rel="noreferrer">asteroid that killed the dinosaurs</a> had a mass of around <span class="math-container">$10^{16}$</span> kg.  The Earth has a mass of <span class="math-container">$10^{24}$</span> kg, or a 100 million times more.  And no matter how cold an asteroid is, the kinetic energy released upon impact will be largely converted into heat.</p>

CC BY-SA 4.0
2020-09-04T21:24:04.473Z
"The mass of any asteroid that could hit the Earth is far too small to be a heat sink" I'm not sure what this has to do with anything. The kinetic energy is proportional to mass, so a bigger asteroid would be even worse
Richard Tingle
CC BY-SA 4.0
2020-09-04T22:17:59.697Z
@RichardTingle Yes, that's true.
gerrit
CC BY-SA 4.0
2020-09-04T22:36:13.103Z
@RichardTingle Correct of course, but I think it's useful to point out that regardless of the actual effects, in reality any effect is going to be so tiny as to be essentially 0.  It's like asking "does adding an eyedropper of near freezing water to the ocean change the temperature?" or "does a star 10,000 light years away affect the tides on earth'. Sure, technically the change is non-zero, but as a practical matter it's going to be such a small effect as to be irrelevant vs everything else that is going on.
eps
CC BY-SA 4.0
2020-09-03T09:23:20.173Z
2020-09-03T14:08:04.657Z
2020-09-03T14:08:04.657Z
gerrit
<p>For your purpose (energy in the entire system) it doesn't matter if the asteroid lands gently or not because energy is conserved: If the asteroid slows through some kind of atmospheric braking, then the energy goes into the atmosphere as heat. If it slows through running into the ocean at full tilt then the energy goes into the ocean as heat. If the energy goes into the asteroid but it raises the temperature to higher than the atmospheric temperature, the asteroid will radiate the extra heat away into the atmosphere over time anyway.</p>
<p>Let's say the asteroid starts off as cold as it's possible to get, 0 degrees Kelvin, and the average global temperature is just about 285K. We will leave the material as a question mark for now, but call the specific heat capacity c. The energy that our asteroid can absorb before it becomes warmer than the average atmospheric temperature is (in Joules):</p>
<p><span class="math-container">$$m \cdot \Delta t \cdot c = m\, \mathrm{kg} \cdot 285\, \mathrm{K} \cdot c\,\mathrm{J}\,\mathrm{kg}^{-1}\mathrm{K}^{-1}$$</span></p>
<p>Since the escape velocity of Earth is 11 km/s (11000 m/s) the total energy coming in (in kg m^2/s^2 aka Joules) will be:</p>
<p><span class="math-container">$$\frac{1}{2}mv^2 = \frac{1}{2}  m \mathrm{kg} \cdot 11000^2 \,\mathrm{m^2}\mathrm{s^{-2}}$$</span></p>
<p>We want our asteroid to be capable of absorbing more energy than the energy it has from falling:</p>
<p><span class="math-container">$$m \mathrm{kg} \cdot 285 \mathrm{K} \cdot c\mathrm{J}\,\mathrm{kg}^{-1}\mathrm{K}^{-1} &gt; \frac{1}{2} m \mathrm{kg} \cdot 11000^2 \,\mathrm{m^2}\mathrm{s^{-2}}$$</span></p>
<p>Mass cancels out immediately and after some arithmetic, we end up with (rounding quite a bit) c &gt; 200,000 J/kg K.</p>
<p>Unfortunately, the material with the highest specific heat is hydrogen, with a specific heat of about 14,000 J/kg K, which is nowhere near what we need.</p>
<p>Short answer: No.</p>

CC BY-SA 4.0
2020-09-04T07:42:55.483Z
+1 Thanks for showing some calculations that are quite easy to understand! I wonder how would the shape of the asteroid come into play. Also, striking on land/ocean doesn't seem to result in any difference according to your answer, which I slightly doubt it. If I shoot an ice bullet to a pool of water, I dont think all the energy would be translated to heat.
y chung
CC BY-SA 4.0
2020-09-03T21:21:29.240Z
2020-09-07T11:54:35.787Z
2020-09-07T11:54:35.787Z
user3067860
<p>I'm gonna sail against the wind and say &quot;yes&quot;, <em>indirectly</em>.</p>
<p>Although you don't need a cold asteroid, a &quot;normal&quot; one would suffice. When striking the Earth, asteroids eject a large amount of dust into the atmosphere, blocking the Sun radiations, thus cooling the planet. The phenomenon is known as <a href="https://en.wikipedia.org/wiki/Impact_winter" rel="noreferrer">impact winter</a> and is similar to volcanic and nuclear winters.</p>

CC BY-SA 4.0
2020-09-03T13:57:54.730Z
That's a very good point that I missed! Do you have any idea would striking the ocean produce a similar effect?
y chung
CC BY-SA 4.0
2020-09-03T19:12:30.923Z
@ychung The one that killed the dinosaurs happened in shallow ocean, but it seems to get a bit more complicated for deep ocean strikes.  Deep ocean means a lot more water vapor is shot up, which can produce a strong greenhouse effect.  But according to the wiki article should still be an overall cooling effect. https://en.wikipedia.org/wiki/Impact_winter "there would still be an enormous amount of debris ejected into the atmosphere".  Los Alamos simulated it: https://www.youtube.com/watch?v=9xUf-tF-OU8
eps
CC BY-SA 4.0
2020-09-04T03:40:34.977Z
I would like to point out that at the moment archeologists do not think that volcanic winters (the only example of such events we have reliable data on) lead to long-term change in climate (see [Lane et al., 2013](https://www.pnas.org/content/110/20/8025)). Surface would be cooled for a time, but the debris in atmosphere would significantly increase its heat capacity, and this energy would lead to relatively quick (in geological terms) restoration of average temperatures.
Danila Smirnov
CC BY-SA 4.0
2020-09-04T07:12:12.470Z
@DanilaSmirnov Absolutely, it's a very short effect. There are some historical examples as well, although with smaller eruptions than Toba. For instance the "year without a summer" after Tambora eruption. Even the (relatively) small Pinatubo eruption cooled the Earth for a few months.
Jean-Marie Prival
CC BY-SA 4.0
2020-09-04T07:46:15.030Z
@Jean-MariePrival my point was it's not really cooling of the Earth, it's just redistributing the heat, with debris-filled atmosphere receiving energy that would usually be received by the surface. Surface gets colder than usual, upper atmosphere gets hotter than usual, but energy of the whole system is mostly unchanged.
Danila Smirnov
CC BY-SA 4.0
2020-09-04T08:48:43.807Z
@DanilaSmirnov When making the distinction between cooling the Earth vs. redistributing the heat we'd have to ask the OP what they mean by "cooling the Earth": whole planet, surface, lower troposphere, whole atmosphere.  The greenhouse effect cools the stratosphere, but we speak of global warming, not global heat redistribution.  And are sure you mean archeologists, not paleoclimatologists? Although archeology can provide clues on historical climate, that's just one piece in the overall puzzle (NB: welcome to Earth Science!).
gerrit
CC BY-SA 4.0
2020-09-04T10:04:34.887Z
@gerrit re: warming vs heat redistribution: fair point. I just assumed that cooling effect needs to be able to shift the overall heat balance, which might not be the case with volcanic (and maybe with impact/nuclear) winters, as those seem to be followed by relatively quick return to status quo. But you're right, OP did not say anything such conditions.
Danila Smirnov
CC BY-SA 4.0
2020-09-04T10:05:31.770Z
@gerrit re: archaeologists vs paleoclimatologists: I saw the lead researcher of that paper I linked being referred to as an archaeologist, that's why I used that word. Although I do suppose the work itself is about paleoclimatology.
Danila Smirnov
CC BY-SA 4.0
2020-09-04T16:06:44.890Z
@eps: the Los Alamos video was disappointing (thanks a lot for linking to it - it is just their content that is weak). They show that there would be a tsunami. Nothing about how big, comparison with existing ones, possible effects. Nothing about the effects of the vaporized water.
WoJ
CC BY-SA 4.0
2020-09-04T22:19:46.363Z
@WoJ I think that might be tech insider's editorial choices more than anything else!  Although it's not clear in the video, I believe that it is a summary of work done by Gisler using SAGE simulation code.  These papers go into more detail on tsunami height and speed: https://library.lanl.gov/tsunami/212/gisler.pdf and https://sci.esa.int/documents/33321/35974/1567255443617-ESLAB40-Proc_286769-Gisler.pdf .   I think the lasting effects of vaporized water is beyond the scope of their work (or the simulation code), however.
eps
CC BY-SA 4.0
2020-09-05T06:43:54.367Z
@eps: thank you, the papers are indeed much more precise (at least to someone who is not in the field such as myself). Thanks for the follow-up.
WoJ
CC BY-SA 4.0
2020-09-03T12:01:36.843Z
2020-09-03T12:01:36.843Z
Jean-Marie Prival
<p>One asteroid or comet, no.</p>
<p>Even it it was huge you might get a planetary extinction event like the meteorite that crashed into the Yucatan region resulting in the dinosaurs dying out. That would be due to the amount of dust and soil etc. that would be thrown into the atmosphere.</p>
<p>The other thing, as the asteroid/comet/meteorite entered the Earth's atmosphere is would immediately start to heat up due to friction with the air. Some of the asteroid would be lost in the process.</p>
<p>If the asteroid were to be made of</p>
<blockquote>
<p>materials with high heat capacity</p>
</blockquote>
<p>which might protect it against heating up while entering the atmosphere then the same material, acting as an excellent thermal insulator, would prevent the cold from the core of the asteroid from being released quickly. So no instant cooling.</p>
<p>If the asteroid hit the ocean, water has a high thermal capacity. It is not going to rapidly freeze because a potentially very large super cold rock plopped into the ocean. There would be some localized short term cooling which would regain lost temperature fairly quickly.</p>
<p>If an entire blizzard load of super cold rocks hit the Earth at the same time, the Earth's problem won't be with dealing with any cold it will be in dealing with all the dust raised into the atmosphere and the seismic shocks created by the rocks impacting the ground.</p>

CC BY-SA 4.0
2020-09-03T09:25:25.277Z
2020-09-03T09:25:25.277Z
Fred
<h1>Thermodynamically No</h1>
<p>Jean-Marie gives a good answer about aerosol effects.  Others observe that impacts add energy to the planet.  I would simply like to look at the energy budget.</p>
<p>The sun dumps on the order of <a href="https://phys.org/news/2011-10-vast-amounts-solar-energy-earth.html" rel="nofollow noreferrer">173 petawatts</a> on the earth, nonstop.  Now, asteroids have a <a href="https://www.lpi.usra.edu/meetings/lpsc2012/pdf/2626.pdf" rel="nofollow noreferrer">heat capacity</a> on the order of <span class="math-container">$10^{18} - 10^{24} J/K$</span>, depending on size.  Note that a petawatt is <span class="math-container">$10^{15} J/s$</span>.  Also note that the sun dumps <span class="math-container">$10^{18} W$</span> on the planet every 5.8 s or so.</p>
<p>Now, let's imagine we find an asteroid floating around interstellar space at a chilly 3 K (or colder or warmer...the exact number doesn't matter).  Let's further suppose we can chuck it at the earth and land it without any significant addition of energy (we use an alien tractor beam to gently lower it to the surface).  This asteroid will &quot;cool&quot; the earth by heating itself up.  So...how much cooling should we expect?</p>
<p>Instead of pretending that the earth is a lone ice box floating in space, and calculating what happens when we drop a large ice cube into it, I think it's more useful to simply ask: &quot;How quickly will the sun melt our ice cube?&quot;</p>
<p>The sun will contribute enough energy to warm the asteroid by 1 K every 6s or so, for a small asteroid.  To raise the asteroid from 3 K to earth mean surface temperature of 288 K will take somewhat less than 1800 s, or about 30 minutes.  That's about how long an ice cube would last in a chilled beverage on a not particularly warm day.</p>
<p>Now, that is roughly equivalent to turning off the sun for 30 minutes.  That would have a noticeable effect on global weather, but I doubt it would remain noticeable even a few days after.  Over the course of a year, this blip would amount to a solar energy reduction of 0.005%.  Hardly enough to detect in the climate record.</p>
<p>Of course, we picked the smallest asteroid for the initial calculation.  If you go towards the large end, you can crank up all these numbers by a factor of a million, more or less.  Ceres, for instance, would suck up about 5000% of the sun's annual energy sent to earth.  That's 50 years of solar energy, and obviously a significant impact.  Given that it's round, we can debate whether Ceres counts as an &quot;asteroid&quot; or not.  <a href="https://en.wikipedia.org/wiki/44_Nysa" rel="nofollow noreferrer">Nysa</a> is more &quot;asteroid-shaped&quot;, but only has about 100x the heat capacity of our initial calculations.  So, a solar dip of 0.5% over the course of a year might be enough to notice, but likely not enough to cause attributable long-term climate effects (given that the Milankovitch cycles produce larger variance).</p>
<p>Note that if you start with a more typical asteroid temperature of about 200 K, you would reduce all the energy budget numbers to about 1/3 of their value (meaning, the smallest ice cube would melt in ~10 minutes).</p>

CC BY-SA 4.0
2020-09-03T20:38:24.090Z
2020-09-03T20:38:24.090Z
Lawnmower Man
<p>Impacts typically occur at a range of 20 to 70 km/s.</p>
<p>Let's look at the 20 km/s one.</p>
<p>1 kg of mass at 20 km/s = 1kg * 20,000 m/s * 20,000m/s /2 = 200 million joules.</p>
<p>To bring ice from absolute zero to melting point requires 1000 * 273 * 0.5 = 136 thousand joules.  Yes, melting it will take energy and so will vapourization, but this is VERY small potatoes compared to the kinetic energy.  (Water ice is the largest part of comet ice.  Ammonia has a somewhat higher specific heat, methane, far lower.)</p>
<p>So as a source of coolth, a comet impact doesn't rate.</p>
<hr />
<p>So let's instead, invite Archanangel Gabriel to come down and gently lay the comet down to melt and chill the area.</p>
<p>Large comet 20 km diameter.  Assume solid ice for the sake of simplicity.  20 km sphere is  4/3 * pi * 10^3  = about 4200 cubic km of ice.</p>
<p>Wow.  Lots of ice cubes for my martini!</p>
<p>Comparison:  Canada has an area of 10 million km2.   If you had 1/10 of a meter of ice (4 inches...) you would have 1/10,000 of a km * 10 million km = 1000 km3.  So your comet has about the four times the chilling effect as spring in Canada. (Ok, 10 cm is arbitrary.)</p>
<p>Or the same as 40 cm of ice on Canada.</p>
<p>Now a square meter with 40 cm of ice would be about 400 kg. (Please, do NOT muddle the waters with the different density of water and ice.  This is a BOTE calc.)</p>
<p>1 kg of ice takes about 500 kJ to melt, assuming it's starting from close to absolute zero.  400 kg takes 200,000 kJ</p>
<p>Solar constant is about 1 kW/m2 at noon.  (Yes this is variable depending on latitude, time of year, cloud cover, and things that otherwise were thought only to affect the flavour of pizza) So 200,000 seconds of noonday sun would melt the ice.  About 55 hours.</p>
<p>Which would be about 200 hours due to night, and low angles.  8 days.  Ish.</p>
<p>However Canada isn't the whole world.  Only about 1/50 of it.  so 8/50 or about 4 hours sunlight over the earth to melt the comet.</p>
<p>Like throwing an ice cube from your drink into a campfire.</p>
<hr />
<p>You're left with nuclear winter possibilities.  Others have addressed that.</p>

CC BY-SA 4.0
2020-09-06T04:30:01.050Z
2020-09-06T04:30:01.050Z
Sherwood Botsford
<p>While comets are cold out in space they dissipate as they get closer to the sun and earth. Upon contact with the atmosphere they burn up. If a comet is large enough bolide they may possess the same climatological concerns as the Cretaceous asteroid impact</p>

CC BY-SA 4.0
2020-09-04T03:32:02.363Z
2020-09-04T03:32:02.363Z
LazyReader
<p>This is obviously an imaginary question, but it seems like something computable (which I don't have the mathematical ability to do so unfortunately). P.S. I do not have in-depth knowledge about comet/asteroid/meteoroid/even any space object at all, so please correct me if I made any mistakes. Also, please answer the question considering the current state of the Earth, not a super-heated Earth.</p>
<p>Some propositions:</p>
<ol>
<li>The asteroid/comet is sizable, cold to the core, and is made up of materials with high heat capacity</li>
<li>The asteroid/comet &quot;strikes&quot; the Earth at low speed</li>
<li>The asteroid/comet strikes the Ocean</li>
</ol>
<p>Related: <a href="https://astronomy.stackexchange.com/questions/8707/how-gently-could-a-comet-asteroid-meteorite-hit-earth">How gently could a comet/asteroid/meteorite “hit” Earth?</a></p>

CC BY-SA 4.0
2020-09-03T09:32:18.897Z
the slowest speed an asteroid will have when it impacts earth is 11 KM/S or 40320 KM/H (based on the information in your link)
trond hansen
CC BY-SA 4.0
2020-09-03T11:36:26.947Z
@Jean-MariePrival Nice!  Can you post that as an answer?
gerrit
<climate-change><meteorite><asteroids>
Could an icy/extremely-cold asteroid/comet ever strike and cool the Earth?
y chung
# Could an icy/extremely-cold asteroid/comet ever strike and cool the Earth?

(Asked by: username_0 on 2020-09-03)

This is obviously an imaginary question, but it seems like something computable (which I don't have the mathematical ability to do so unfortunately). P.S. I do not have in-depth knowledge about comet/asteroid/meteoroid/even any space object at all, so please correct me if I made any mistakes. Also, please answer the question considering the current state of the Earth, not a super-heated Earth.

Some propositions:

1. The asteroid/comet is sizable, cold to the core, and is made up of materials with high heat capacity
2. The asteroid/comet "strikes" the Earth at low speed
3. The asteroid/comet strikes the Ocean

Related: [How gently could a comet/asteroid/meteorite “hit” Earth?](https://astronomy.stackexchange.com/questions/8707/how-gently-could-a-comet-asteroid-meteorite-hit-earth)

**Tags:** climate-change, meteorite, asteroids

> **Comments:**
> **username_1**: the slowest speed an asteroid will have when it impacts earth is 11 KM/S or 40320 KM/H (based on the information in your link)  
> **username_2**: @username_7 Nice!  Can you post that as an answer?  

---

## 8 answers

### Answer by username_3 on 2020-09-03. Score: 22

> Could an icy/extremely-cold asteroid/comet ever strike and cool the Earth?

No.

The smallest possible velocity is about 11 km/s velocity, and that would require a near miracle. For that to happen, the object would have to enter the vicinity of the Earth through a very small keyhole near the Sun-Earth L1 or L2 point with a very small velocity (a comet will not have that very small velocity), then be nearly ballistically captured by the Moon, and then fall to Earth. A comet would be more likely to hit the Earth at 40 km/s, or larger.

And that means that it does not matter how cold or how hot the object was prior to impact. The kinetic energy of the impact will dwarf whatever thermal energy the object had prior to the impact.

> **Comments:**
> **username_4**: I'm sure this is true (and probably orders of magnitude of true) but some energy numbers to back this up would be nice. By my calculation for your low end velocity asteroid would have 60500Kj/kg. While iron has a specific heat capacity of 440 j/KgC so a "temperature deficit" of about 290°C ends up at 129Kj/kg (so a pathetic contribution)  
> **username_4**: (Seems water has one of the highest specific heat capacity at 4182 j/kgC so I should have used that but that doesnt really help much)  

---

### Answer by username_7 on 2020-09-03. Score: 21

I'm gonna sail against the wind and say "yes", *indirectly*.

Although you don't need a cold asteroid, a "normal" one would suffice. When striking the Earth, asteroids eject a large amount of dust into the atmosphere, blocking the Sun radiations, thus cooling the planet. The phenomenon is known as [impact winter](https://en.wikipedia.org/wiki/Impact_winter) and is similar to volcanic and nuclear winters.

> **Comments:**
> **username_0**: That's a very good point that I missed! Do you have any idea would striking the ocean produce a similar effect?  
> **username_5**: @username_0 The one that killed the dinosaurs happened in shallow ocean, but it seems to get a bit more complicated for deep ocean strikes.  Deep ocean means a lot more water vapor is shot up, which can produce a strong greenhouse effect.  But according to the wiki article should still be an overall cooling effect. https://en.wikipedia.org/wiki/Impact_winter "there would still be an enormous amount of debris ejected into the atmosphere".  Los Alamos simulated it: https://www.youtube.com/watch?v=9xUf-tF-OU8  
> **username_8**: I would like to point out that at the moment archeologists do not think that volcanic winters (the only example of such events we have reliable data on) lead to long-term change in climate (see [Lane et al., 2013](https://www.pnas.org/content/110/20/8025)). Surface would be cooled for a time, but the debris in atmosphere would significantly increase its heat capacity, and this energy would lead to relatively quick (in geological terms) restoration of average temperatures.  
> **username_7**: @username_8 Absolutely, it's a very short effect. There are some historical examples as well, although with smaller eruptions than Toba. For instance the "year without a summer" after Tambora eruption. Even the (relatively) small Pinatubo eruption cooled the Earth for a few months.  
> **username_8**: @username_7 my point was it's not really cooling of the Earth, it's just redistributing the heat, with debris-filled atmosphere receiving energy that would usually be received by the surface. Surface gets colder than usual, upper atmosphere gets hotter than usual, but energy of the whole system is mostly unchanged.  
> **username_2**: @username_8 When making the distinction between cooling the Earth vs. redistributing the heat we'd have to ask the OP what they mean by "cooling the Earth": whole planet, surface, lower troposphere, whole atmosphere.  The greenhouse effect cools the stratosphere, but we speak of global warming, not global heat redistribution.  And are sure you mean archeologists, not paleoclimatologists? Although archeology can provide clues on historical climate, that's just one piece in the overall puzzle (NB: welcome to Earth Science!).  
> **username_8**: @username_2 re: warming vs heat redistribution: fair point. I just assumed that cooling effect needs to be able to shift the overall heat balance, which might not be the case with volcanic (and maybe with impact/nuclear) winters, as those seem to be followed by relatively quick return to status quo. But you're right, OP did not say anything such conditions.  
> **username_8**: @username_2 re: archaeologists vs paleoclimatologists: I saw the lead researcher of that paper I linked being referred to as an archaeologist, that's why I used that word. Although I do suppose the work itself is about paleoclimatology.  
> **username_9**: @username_5: the Los Alamos video was disappointing (thanks a lot for linking to it - it is just their content that is weak). They show that there would be a tsunami. Nothing about how big, comparison with existing ones, possible effects. Nothing about the effects of the vaporized water.  
> **username_5**: @username_9 I think that might be tech insider's editorial choices more than anything else!  Although it's not clear in the video, I believe that it is a summary of work done by Gisler using SAGE simulation code.  These papers go into more detail on tsunami height and speed: https://library.lanl.gov/tsunami/212/gisler.pdf and https://sci.esa.int/documents/33321/35974/1567255443617-ESLAB40-Proc_286769-Gisler.pdf .   I think the lasting effects of vaporized water is beyond the scope of their work (or the simulation code), however.  
> **username_9**: @username_5: thank you, the papers are indeed much more precise (at least to someone who is not in the field such as myself). Thanks for the follow-up.  

---

### Answer by username_2 on 2020-09-03. Score: 13

No, it's not possible to cool the Earth with an asteroid impact.

The mass of any asteroid that could hit the Earth is far too small to be a heat sink. The [asteroid that killed the dinosaurs](https://en.wikipedia.org/wiki/Chicxulub_impactor) had a mass of around $10^{16}$ kg. The Earth has a mass of $10^{24}$ kg, or a 100 million times more. And no matter how cold an asteroid is, the kinetic energy released upon impact will be largely converted into heat.

> **Comments:**
> **username_4**: "The mass of any asteroid that could hit the Earth is far too small to be a heat sink" I'm not sure what this has to do with anything. The kinetic energy is proportional to mass, so a bigger asteroid would be even worse  
> **username_2**: @username_4 Yes, that's true.  
> **username_5**: @username_4 Correct of course, but I think it's useful to point out that regardless of the actual effects, in reality any effect is going to be so tiny as to be essentially 0.  It's like asking "does adding an eyedropper of near freezing water to the ocean change the temperature?" or "does a star 10,000 light years away affect the tides on earth'. Sure, technically the change is non-zero, but as a practical matter it's going to be such a small effect as to be irrelevant vs everything else that is going on.  

---

### Answer by username_6 on 2020-09-03. Score: 12

For your purpose (energy in the entire system) it doesn't matter if the asteroid lands gently or not because energy is conserved: If the asteroid slows through some kind of atmospheric braking, then the energy goes into the atmosphere as heat. If it slows through running into the ocean at full tilt then the energy goes into the ocean as heat. If the energy goes into the asteroid but it raises the temperature to higher than the atmospheric temperature, the asteroid will radiate the extra heat away into the atmosphere over time anyway.

Let's say the asteroid starts off as cold as it's possible to get, 0 degrees Kelvin, and the average global temperature is just about 285K. We will leave the material as a question mark for now, but call the specific heat capacity c. The energy that our asteroid can absorb before it becomes warmer than the average atmospheric temperature is (in Joules):

$$m \cdot \Delta t \cdot c = m\, \mathrm{kg} \cdot 285\, \mathrm{K} \cdot c\,\mathrm{J}\,\mathrm{kg}^{-1}\mathrm{K}^{-1}$$

Since the escape velocity of Earth is 11 km/s (11000 m/s) the total energy coming in (in kg m^2/s^2 aka Joules) will be:

$$\frac{1}{2}mv^2 = \frac{1}{2} m \mathrm{kg} \cdot 11000^2 \,\mathrm{m^2}\mathrm{s^{-2}}$$

We want our asteroid to be capable of absorbing more energy than the energy it has from falling:

$$m \mathrm{kg} \cdot 285 \mathrm{K} \cdot c\mathrm{J}\,\mathrm{kg}^{-1}\mathrm{K}^{-1} > \frac{1}{2} m \mathrm{kg} \cdot 11000^2 \,\mathrm{m^2}\mathrm{s^{-2}}$$

Mass cancels out immediately and after some arithmetic, we end up with (rounding quite a bit) c > 200,000 J/kg K.

Unfortunately, the material with the highest specific heat is hydrogen, with a specific heat of about 14,000 J/kg K, which is nowhere near what we need.

Short answer: No.

> **Comments:**
> **username_0**: +1 Thanks for showing some calculations that are quite easy to understand! I wonder how would the shape of the asteroid come into play. Also, striking on land/ocean doesn't seem to result in any difference according to your answer, which I slightly doubt it. If I shoot an ice bullet to a pool of water, I dont think all the energy would be translated to heat.  

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### Answer by username_10 on 2020-09-03. Score: 5

One asteroid or comet, no.

Even it it was huge you might get a planetary extinction event like the meteorite that crashed into the Yucatan region resulting in the dinosaurs dying out. That would be due to the amount of dust and soil etc. that would be thrown into the atmosphere.

The other thing, as the asteroid/comet/meteorite entered the Earth's atmosphere is would immediately start to heat up due to friction with the air. Some of the asteroid would be lost in the process.

If the asteroid were to be made of

> materials with high heat capacity

which might protect it against heating up while entering the atmosphere then the same material, acting as an excellent thermal insulator, would prevent the cold from the core of the asteroid from being released quickly. So no instant cooling.

If the asteroid hit the ocean, water has a high thermal capacity. It is not going to rapidly freeze because a potentially very large super cold rock plopped into the ocean. There would be some localized short term cooling which would regain lost temperature fairly quickly.

If an entire blizzard load of super cold rocks hit the Earth at the same time, the Earth's problem won't be with dealing with any cold it will be in dealing with all the dust raised into the atmosphere and the seismic shocks created by the rocks impacting the ground.


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### Answer by username_11 on 2020-09-03. Score: 4

# Thermodynamically No

username_7 gives a good answer about aerosol effects. Others observe that impacts add energy to the planet. I would simply like to look at the energy budget.

The sun dumps on the order of [173 petawatts](https://phys.org/news/2011-10-vast-amounts-solar-energy-earth.html) on the earth, nonstop. Now, asteroids have a [heat capacity](https://www.lpi.usra.edu/meetings/lpsc2012/pdf/2626.pdf) on the order of $10^{18} - 10^{24} J/K$, depending on size. Note that a petawatt is $10^{15} J/s$. Also note that the sun dumps $10^{18} W$ on the planet every 5.8 s or so.

Now, let's imagine we find an asteroid floating around interstellar space at a chilly 3 K (or colder or warmer...the exact number doesn't matter). Let's further suppose we can chuck it at the earth and land it without any significant addition of energy (we use an alien tractor beam to gently lower it to the surface). This asteroid will "cool" the earth by heating itself up. So...how much cooling should we expect?

Instead of pretending that the earth is a lone ice box floating in space, and calculating what happens when we drop a large ice cube into it, I think it's more useful to simply ask: "How quickly will the sun melt our ice cube?"

The sun will contribute enough energy to warm the asteroid by 1 K every 6s or so, for a small asteroid. To raise the asteroid from 3 K to earth mean surface temperature of 288 K will take somewhat less than 1800 s, or about 30 minutes. That's about how long an ice cube would last in a chilled beverage on a not particularly warm day.

Now, that is roughly equivalent to turning off the sun for 30 minutes. That would have a noticeable effect on global weather, but I doubt it would remain noticeable even a few days after. Over the course of a year, this blip would amount to a solar energy reduction of 0.005%. Hardly enough to detect in the climate record.

Of course, we picked the smallest asteroid for the initial calculation. If you go towards the large end, you can crank up all these numbers by a factor of a million, more or less. Ceres, for instance, would suck up about 5000% of the sun's annual energy sent to earth. That's 50 years of solar energy, and obviously a significant impact. Given that it's round, we can debate whether Ceres counts as an "asteroid" or not. [Nysa](https://en.wikipedia.org/wiki/44_Nysa) is more "asteroid-shaped", but only has about 100x the heat capacity of our initial calculations. So, a solar dip of 0.5% over the course of a year might be enough to notice, but likely not enough to cause attributable long-term climate effects (given that the Milankovitch cycles produce larger variance).

Note that if you start with a more typical asteroid temperature of about 200 K, you would reduce all the energy budget numbers to about 1/3 of their value (meaning, the smallest ice cube would melt in ~10 minutes).


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### Answer by username_12 on 2020-09-06. Score: 1

Impacts typically occur at a range of 20 to 70 km/s.

Let's look at the 20 km/s one.

1 kg of mass at 20 km/s = 1kg \* 20,000 m/s \* 20,000m/s /2 = 200 million joules.

To bring ice from absolute zero to melting point requires 1000 \* 273 \* 0.5 = 136 thousand joules. Yes, melting it will take energy and so will vapourization, but this is VERY small potatoes compared to the kinetic energy. (Water ice is the largest part of comet ice. Ammonia has a somewhat higher specific heat, methane, far lower.)

So as a source of coolth, a comet impact doesn't rate.

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So let's instead, invite Archanangel Gabriel to come down and gently lay the comet down to melt and chill the area.

Large comet 20 km diameter. Assume solid ice for the sake of simplicity. 20 km sphere is 4/3 \* pi \* 10^3 = about 4200 cubic km of ice.

Wow. Lots of ice cubes for my martini!

Comparison: Canada has an area of 10 million km2. If you had 1/10 of a meter of ice (4 inches...) you would have 1/10,000 of a km \* 10 million km = 1000 km3. So your comet has about the four times the chilling effect as spring in Canada. (Ok, 10 cm is arbitrary.)

Or the same as 40 cm of ice on Canada.

Now a square meter with 40 cm of ice would be about 400 kg. (Please, do NOT muddle the waters with the different density of water and ice. This is a BOTE calc.)

1 kg of ice takes about 500 kJ to melt, assuming it's starting from close to absolute zero. 400 kg takes 200,000 kJ

Solar constant is about 1 kW/m2 at noon. (Yes this is variable depending on latitude, time of year, cloud cover, and things that otherwise were thought only to affect the flavour of pizza) So 200,000 seconds of noonday sun would melt the ice. About 55 hours.

Which would be about 200 hours due to night, and low angles. 8 days. Ish.

However Canada isn't the whole world. Only about 1/50 of it. so 8/50 or about 4 hours sunlight over the earth to melt the comet.

Like throwing an ice cube from your drink into a campfire.

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You're left with nuclear winter possibilities. Others have addressed that.


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### Answer by username_13 on 2020-09-04. Score: 0

While comets are cold out in space they dissipate as they get closer to the sun and earth. Upon contact with the atmosphere they burn up. If a comet is large enough bolide they may possess the same climatological concerns as the Cretaceous asteroid impact


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