One popular theory to explain how Earth got its water is that it was delivered by asteroid/comet/etc. The form this theory usually takes is that many small impacts occured over a long time, each delivering a relatively small amount of water until the planet reached its current water content.

But what if 'all the water' was delivered in a single massive event, a single comet or asteroid, a single impact.

What would this event have "looked" like?

Best answers will include details such as size/speed/etc. of the impactor, angle of impact, effects on the geography caused by the impact (impact crater size, or effects on tectonic motion, etc), whether or not the same impact could account for the formation of the moon, etc.

(optional) Bonus question: Since many of us in this community build worlds on different scales, how would these types of impacts change with the size of the planet? In other words, how different would an impact that covers a smaller planet, like Mercury for example, in about 70% water, or how would it be different for a super-earth with double or triple Earth's mass, to be covered by about 70% water in a single impact?

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    $\begingroup$ @StarfishPrime, I hope we are still doing worldbuilding, not lotteries $\endgroup$
    – L.Dutch
    Sep 24, 2019 at 15:50
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    $\begingroup$ @L.Dutch For all I know, the angle of impact makes no difference, on the other hand it might be crucially important. I only mention it in case it is significant. If it's not, feel free to explain why it's not in your answer, or leave it out of your answer entirely. $\endgroup$
    – Harthag
    Sep 24, 2019 at 15:53
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    $\begingroup$ Note to the answers based on the mass hydrosphere - there is more water in Earth's mantle: Water in Earth's mantle $\endgroup$
    – Alexander
    Sep 24, 2019 at 16:18
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    $\begingroup$ @Dalila The angle of impact could mess with the rotation rate or axial tilt. Venus and Uranus are thought to have gotten their slow spin and high tilt respectively from huge impacts in their past. $\endgroup$ Sep 24, 2019 at 16:42
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    $\begingroup$ Relevant XKCD: what-if.xkcd.com/12 $\endgroup$ Sep 25, 2019 at 5:58

4 Answers 4


So, lets say there's about $1.8*10^{21}$kg of water on the surface of the earth (this excludes hydrates and stuff in the mantle, but the surface stuff seems like the bit most likely to be deposited by impacts after earth's formation).

Given the density of ice, $920kg/m^3$, that much water would form a solid sphere about 776km in radius. That's Quite Big, by the way... the Chixulub impactor that kicked off the Cretaceous-Paleogene extinction event wasn't likely to have been bigger than 81km across. It is bigger than every asteroid (Ceres has a radius under 500km) and as big as some of the larger moons in the solar system... Iapetus is a similar size and mass and is also largely made of ice so it is a good representative for your impactor.

Here's a size comparison of Earth, the Moon and Iapetus, so you can get a handle on what you're asking about.

Size comparison of Earth, the Moon and Iapetus

(By way of a bonus, the massive crater Engelier is just about visible on Iapetus, and makes it look a bit like the death star. It is a mere 500km across, far smaller that anything we'll discuss here.)

It is at least smaller than the Theia impactor believed to have created our moon, which was believed to have been about 6000km across. There are theories suggesting that much of the Earth's water did arrive during the Theia impact. I won't go into the Theia impact here, but instead consider only a single delivery of ice, probably after the moon was formed and the Hadean era ended (otherwise subsequent bombardments might have blown the water away into space).

Lets assume it is hitting crystalline rock, there being no water or sedimentary minerals on a waterless world. You can now throw these handy figures and assumptions into the Earth Impact Effects Program. I picked a conservative impact velocity of 11km/s (it is a bit unlikely for it to be lower than this, and at this speed it is more likely that some of the delivered water will stay) and a 45 degree impact angle (other angles don't make much difference, which isn't entirely surprising). Summary for those of you too lazy to follow the link and fill in the form for yourself:

  • Initial crater 606km deep, 1710km across. Given that Earth's crust is no more than 90km deep, that means the mantle is very definitely exposed. The hole will fill in with ejecta, of which there is quite a lot... it'll end up about 3-4km deep.
  • Final crater diameter: 4540km, once the surrounding land has finished falling into the initial hole. This is vastly bigger than the biggest hypothesised impact structures ever found, MAPCIS.
  • Despite the impact energy being measured in exatonnes, the calculator doesn't suggest that you'll get a really interesting superheated fireball as the impactor vapourises. I'm slightly dubious on this, but as I'm not an expert on banging rocks together and the authors of the application are, I'll defer to them. Certainly, the behaviour of objects undergoing a hypervelocity collision is unintuitive. This increases the chance that some of the water will actually survive the impact and stay put.
  • Debris from the impact (like, lumps of the stuff, not just dust) will fall over 5000km from ground zero.
  • If it hits at the right sort of place (say, at the equator) it could change the day length of the earth by a bit... for a 45 degree impact, the change is of the order of ±15 minutes.

The sedate impact velocity is required to minimise the chances of massive post-impact heating. Hopefully the impact pressures are low enough (relatively speaking) and the energy release spread out over a long enough period of time that what you get is a huge explosion of rock and steam that boils and buries an area larger than North America, rather than a multi-thousand-degree fireball that propels debris out of earth's orbit and generates large quantities of light gases that can escape the atmosphere. If the latter occurred, you'd need to deliver even more water, and then the impact energies would be even higher and more volatiles would be lost... and so on. You can see why many smaller impacts are preferred. Some vapourisation will inevitably occur, but it calculating how much is definitely out of my league.

The aftermath of the impact will involve a lot of dust lofted into the atmosphere which will also be filled with a great deal of steam. There was significant global cooling after the Chicxulub impact, but that didn't involve pouring a bajillion litres of water into the mantle, so whether the energy stored in all that steam will dissipate and rain out before the dust settles, or whether you'll end up withno significant coolings and a thick, hot water vapour atmosphere for a considerable time afterwards I don't know... again, that sort of guesswork is out of my league.

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    $\begingroup$ That density is way off. Normal water ice is about 920 kg/m^3. There are several exotic forms of water ice with higher and lower densities, but the higher-density ones only exist under extreme pressure and AFAIK none of them come close to 2750 kg/m^3. $\endgroup$
    – G_B
    Sep 24, 2019 at 23:57
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    $\begingroup$ @GeoffreyBrent you are of course absolutely right. No idea how that one didn't trigger my WTF senses, but there you go. On the upside, I now have a nice ewxample of a real-life body to represent the impactor. $\endgroup$ Sep 25, 2019 at 9:14

Estimates for the mass of the Chicxulub impactor that offed the dinosaurs range from 1015 and 4.6 $\times$ 1017 kilograms.

The hydrosphere's mass is currently estimated to be around 1.4 $\times$ 1021 kilograms. That is like 10,000 Chicxulubs in terms of orders of magnitude, if we use the upper bound for Chic's mass. Might be more like 100,000 dino-killing asteroids hitting at once, maybe 1,000,000.

For the record, the mass of the hydrosphere is also just one order of magnitude less than that of the Moon (at about 7 $\times$ 1022 kg).

The good news is since you are starting with a completely dry planet there is nothing to kill.

The bad news is that the impact will have so much energy that most of that water will become gaseous. It will also be awkwardly hot. The atmosphere and the water will be so hot that a lot of that water and a lot of any atmosphere you had will escape to space.

In the end the world's mountains will have been smoothed to different degrees due to strong, hot-steam winds happening during the months or years it will take for the planet to gradually cool down. After things stabilize again You will have an Earth with shallower oceans and a thinner atmosphere. Maybe some life can develop there.

I just remembered something. The megacomet in question is not as massive as the Moon, but it will be coming much faster than the Moon would if it stopped in its tracks. Which reminds me of this other question:

No! Not the moon!

It's about what would happen if the Moon hits us. Consider the effects described in there but with less severity. For example:

  • The impact is also enough to cause ripples on what is left of the crust, tearing it apart. A significant portion of the surface is destroyed; vast chasms open, ejecting lava(...)

  • Most of the water on the surface of the planet becomes gas. A major portion of it escapes into space along with the gods know how much of the atmosphere.

And so on.

  • $\begingroup$ In regards to the water boiling back off in to space, is there any equilibrium point where a larger impactor (so more water to spare) wold leave Earth with the same water content as present after the boil-off is done? Or would the extra energy from the extra mass just boil off even more of it, leaving Earth with even less water, instead of more? $\endgroup$
    – Harthag
    Sep 24, 2019 at 16:05
  • $\begingroup$ @Dalila I have neither the math nor the physics in me to find that out, but I think if there is an equilibrium point, it's for something less massive than the megacomet I imagined when typing my answer. $\endgroup$ Sep 24, 2019 at 16:16
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    $\begingroup$ keep in mind your mega comet is going to eject a lot of rock too, so you need a more massive earth to start with as well. $\endgroup$
    – John
    Sep 24, 2019 at 19:17
  • $\begingroup$ @Dalila: Would you like to try for a lunar gravity assist to steal energy from it first? $\endgroup$
    – Joshua
    Sep 25, 2019 at 3:29
  • $\begingroup$ @Joshua I don't see why not $\endgroup$
    – Harthag
    Sep 25, 2019 at 12:29

If you want to have the total hydrosphere being delivered on a single impact, let's first check the mass

The total mass of Earth's hydrosphere is about $1.4 \cdot 10^{18}$ tonnes

That means $1.4 \cdot 10^{21}$ kg.

As references, Mimas has a mass of $3 \cdot 10^{19}$ kg, while Enceladus has a mass of $1.0 \cdot 10^{20}$ kg.

That's a major impact. As a consequence of the impact, most of the water would be vaporized, and such a vast amount of vapor in the Earth atmosphere would be bad: it would mean a huge greenhouse effect, probably turning the planet into a twin Venus.

Just based on the mass it's impossible to estimate the angle of impact, while the impact velocity would be at least in the order of 10 km/s.


According to the USGS it would look like this:

enter image description here

Source: https://water.usgs.gov/edu/gallery/global-water-volume.html

The larger sphere represents all of Earth's water, while the smaller sphere represents Earth's fresh water. The larger sphere has a diameter of 1384 km, quite large but only 0.13% of the total volume of the Earth.

Ice has a density 92% that of water, so you can imagine an ice ball this size being 8.7% bigger.

  • $\begingroup$ Taking into account the water in the lithosphere (mostly in the hydration of crystals), I think it would have to be bigger than just 8.7%. But that's a good infographic anyway. $\endgroup$ Sep 26, 2019 at 5:22
  • $\begingroup$ @AdrianColomitchi "This sphere includes all of the water in the oceans, ice caps, lakes, rivers, groundwater, atmospheric water, and even the water in you, your dog, and your tomato plant." - I assume that includes the lithosphere. If not, the lithosphere is only 1.7% of the total anyway so it shouldn't change the graphic much. $\endgroup$ Sep 26, 2019 at 15:16

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