How long would it take for water vapor from an ocean asteroid impact to rain back down?

Question

In my story a consortium of megacorporations spot a new comet 26km wide entering the inner solar system. Earth is in dire need of water because of an already worsening climate change disaster (think now + a hundred years of out of control capitalist greed). Especially the climate change disaster's effect on temperature, capitalist greed against inland aquifers and increasing salinity in coastal aquafers because of sea rise.

In countless places near Earth they are mining for water: the Moon and several near Earth asteroids. But this thing is huge, and timely, and in their greed, these megacorps think they can score a big win.

They try to bring the comet into orbit around the Moon with gravity tractors. In several years time they manage this, but the tractors didn't just bring a ball of ice, but a dirty ball of rock and ice - a rubble pile with thirty or so core rocks ranging from a few hundred meters across to a kilometer across. The gravity of the Earth and Moon finish pulling the thing apart and fourteen of the largest chunks of rock hit the Earth.

Given Earth is kind of a water world, most hit in the oceans, and throw hundreds of gigatons of water vapor into the air. The ones that hit land do the nastiest damage and a couple hit aquifers throwing not just ash and earth into the air, but also gigatons more water.

The mass of the atmosphere is around 5.3x10^15 gigatons, or 530 million gigatons, according to various sources.

The average amount of water in the atmosphere is 0.001% of all the water on Earth, or 1,386,000 cubic km, from the facts listed on the USGS's page on the water cycle. Wiki has a cubic meter of water as about a metric ton. So a cubic kilometer of water would be a gigaton. Given the previous facts, Earth's atmosphere should have about a 1.386 million gigatons of water in it.

I know greenhouse gas-wise, water vapor in the air is complicated. But because of all the dirt and ash in the air, first they'd have to go thru a prolonged nuclear/volcanic winter. After that, during that and for a long time after I expect it will rain like crazy. So the question is how long will it take for all those gigatons of ash and water vapor to come out of the atmosphere? The amount thrown up would only be few millionths of the total water in the atmosphere, but I would have to guess the suddenness of it would have some kind of a short term and long term effect.

Another complicating factor, and one I couldn't find any reference for is if a 1km size impactor would hit hard enough to cause any mantel plume. Various references say various things about how wide and deep a land crater will be for something that size.

Yet a third complicating factor - much of the rubble pile would still be big enough to cause significant air bursts and ground impacts, throwing up even more water and ash.

So given all that context - how long will the ash take to settle out? And how much longer will all that water vapor take to rain back down?

Answer 1

Earth is in dire need of water because of an already worsening climate change disaster (think now + a hundred years of out of control capitalist greed). Especially the climate change disaster's effect on temperature, capitalist greed against inland aquifers and increasing salinity in coastal aquafers because of sea rise.

I appreciate that you are building a world with such problems in mind, both the physics/logistic one and the humanitarian one. However, I think bringing more water from space is not the solution. Allow me to do a frame challenge.

Your comet will have 9.2 × 10¹⁸ kg. That's about 1% of the mass of the ice on the South Pole. You would elevate the oceans by the same amount we would if we just melted 1% of the Antarctic at once.

Consider that the total mass of the oceans is around The Earth's oceans have about 1.4 × 10²¹ kg, which is three orders of magnitude more than the comet. Your comet will flood coastal aquifers even more while causing no perceptible decrease in salinity. You'll just be putting mangroves underwater, where they will rot and release greenhouse gases.

Proper management of the water we already have here is the path to saving ecosystems. Merely shoving more water into it just makes things worse.

Answer 2

Kinetic Energy = (1/2)MV²

Where:

M is the mass of the asteroid just before it strikes the earth V is the velocity of the asteroid just before it strikes the earth

For example, consider an asteroid that is one kilometre in diameter and weighs 1.4 billion tonnes (M = 1.4×10^12 kilograms), and is traveling at 20 kilometres per second (V = 20,000 m/s). The kinetic energy would be equal to (1/2)×1.4×10^12×(20,000)² = 2.8×10^20 Joules.

14 hit earth (1 in land 13 in sea)

so it 13 x 2.8×10^20 = 3.64 x 10^21 Joules or 3.64 x 10^19 KiloJoules

ΔQ=cmΔt specific heat capacity of water(which equals 4.18710^3 J/(kg℃)) Average sea temperature is 30°C It is given by the equation Q=mL

Q= energy input m=mass of the matter to vaporize L=the specific latent heat

According to Wikipedia the value of the latent heat of vaporization of water is: L=2264.705 kj/kg

3.64 x 10^19 = (2264.705 x M) + (M x 4187 x (100-30))

3.64 x 10^19 = (2,264.705 x M) + (M x 4,187 x (100-30)) 3.64 x 10^19 = 295,354.705 M

M = 123,241,646,006,621 Kilogram of water

M = 123 Gigatons of water will be convert up into the atmosphere

1 mile² thunderstorm cloud = 72,000 tons

15 mile² thunderstorm cloud = 1,080,000 tons

15 mile² thunderstorm cloud last 30 mins

so our 123 Gigatons will continue to form into 123 millions of 15 mile² thunderstorm cloud above initial impact point and spreading around which take 30 mins to empty their bucket back down to earth

(123,000,000 x 30 )/( 60 x 24 x 365 ) =7020 years

but that is single one at a time which is not how reality work ,the thunderstorm would formed and spread faster

earth surface area is 510,100,000 km² or 196,950,711 mile²

13 million of 15 mile² thunderstorm cloud can fit in that

so.. if the asteroid hit and spread evenly (13 of 1 km diameter asteroid hit around the earth)

it will be downpour for (123 x 10^6 / 13 x 10^6) x 30 mins

5 hours..

but if compact into certain area say your capitalism corp pull it down to sea nearby USA (easy for them to go catch those rain by predicted that the wind will bring those thunderstorm cloud into land)and the said 13 asteroids spreading impact zone about the same size as USA outside the shore (with 1 miscalculated landing on New york instead)

9.834 million km² or 3.8 million mile²

200,000 of 15 mile² thunderstorm cloud can fit in that

it will be 123,000,000 / 200,000 x 30 mins so vaguely about 13-20 days

so if you want a good period of time for your story conflict like "capitalism corp covering up their mess" need about 3-4 month or so then just triple asteroid mass (or the impact zone must be more compact / or throw in more meteor or increased it mass)

but if you want something like a year of raining oh well you'll need quite bigger asteroids to do the job

(btw the dust from only 1 asteroid hitting land would dispersed faster giving it'll be raining a lot so those dust probably last around 30% of the time period)

Source

15 mile² thunderstorm cloud > https://wxguys.ssec.wisc.edu/2011/09/12/how-much-condensed-liquid-water-is-in-a-cubic-mile-of-fog/

asteroid energy > https://www.real-world-physics-problems.com/asteroid-impact.html

Sea temperature > https://www.seatemperature.org/

calculate water to stream > https://socratic.org/questions/calculate-the-amount-of-heat-required-to-convert-45-0-g-of-water-at-43-0-c-to-st#:~:text=Explanation%3A,the%20water%20to%20100%20degrees.&text=c%3D4187%20Joules%20per%20kilogram,specific%20heat%20capacity%20of%20water.

how long thunder storm last >https://www.google.com/search?newwindow=1&rlz=1C1OKWM_enTH866TH867&hl=en&biw=899&bih=923&sxsrf=ALeKk00Q2Jx2nBOHGsRrabrk60l9YppPQA%3A1594779746751&ei=YmgOX8C0LYfn9QOrq5KgAQ&q=how+long+thunderstorm+last&oq=how+long+thunderstr&gs_lcp=CgZwc3ktYWIQAxgAMgYIABAWEB4yBggAEBYQHjIICAAQFhAKEB4yBggAEBYQHjIGCAAQFhAeMggIABAWEAoQHjIICAAQFhAKEB4yBggAEBYQHjoECCMQJzoFCAAQkQI6BAgAEEM6CAgAELEDEIMBOgsILhCxAxDHARCjAjoCCC46BQguEJECOgUILhCxAzoCCABQi-0CWNmWA2CZnQNoAnAAeACAAXCIAdIMkgEEMjAuMZgBAKABAaoBB2d3cy13aXo&sclient=psy-ab