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What would be the resulting impact crater of a meteorite hitting bodies of water on Earth?

Let's take for example the impactor that resulted in Meteor Crater, Arizona. What would the impact look like if it hit a shallow sea? A deep ocean? Is there any depth of water which would result in no crater at all?

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    $\begingroup$ A cool question. Would be nice to have formulae for crater's true & apparent radius & depth, ejecta volume as functions of meteorite size, impact velocity and angle, depth, & type of seabed. Also interesting whether water density makes a difference. $\endgroup$ – Deer Hunter Oct 18 '15 at 13:19
  • $\begingroup$ For want of a better term, the "cohesion/fusion" of a meteor & what it's made of would be important. A massive chrondrite meteor would create a crater on land but might fall apart as it travelled through deep water whereas as a dense iron-nickel meteor would have a deeper penetration capability in water & maintain more of its kinetic energy as it moved through water. $\endgroup$ – Fred Oct 18 '15 at 13:23
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    $\begingroup$ See onlinelibrary.wiley.com/doi/10.1029/2009RG000308/abstract $\endgroup$ – stali Oct 18 '15 at 22:11
  • $\begingroup$ @Fred actually the strength of the meteorite material has no effect at all. It is totally about the kinetic energy of the object. The event, whether marine or terrestrial, is essentially identical to an explosion of the same energy as the original kinetic energy, centered sightly beneath the surface. $\endgroup$ – Eubie Drew Oct 19 '15 at 1:57
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    $\begingroup$ @stali fascinating article. Full version PDF here: onlinelibrary.wiley.com/doi/10.1029/2009RG000308/epdf I read the conclusion and I was shocked that an important parameter was ignored, namely velocity of the object relative to Earth, which can vary by a factor of 7, therefore the kinetic energy can vary by a factor of 50 for the same object mass! Which completely overwhelms any consideration of object density, which they did address. The initial water cavity is a fairly straightforward exercise, but what happens after, both in the fluid & the seafloor, is incredibly complicated. $\endgroup$ – Eubie Drew Oct 19 '15 at 2:08
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You don't need much of a depth to be able to stop a meteorite. Less than 100 m water depth should be enough to stop such a meteorite as shown in the provided link without any particular crater on the sea bed.

The Physics part of the story is provided here;

The rest is mathematics.

Short. At collision all the kinetic energy is transferred to Pressure, and thereafter this pressure is again transferred to kinetic energy; it is explosion. Very simply said it's Bernoulli principle.

By this reason it doesn't penetrate practically any deeper in water than in Earth, and as the water then smooths up you have no crater. A very simple scaling might be made with density; typical rock / water; 2.3/1 depth.

If you need some visual proof, try this video;

https://www.youtube.com/watch?v=cp5gdUHFGIQ

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  • $\begingroup$ This answer is a bit too vague. It would be cool to develop it with some of the comments in the original post. $\endgroup$ – kwinkunks Oct 29 '15 at 15:40
  • $\begingroup$ @kwinkunks It answers the question. What else you want. Ok, I'll do a basic explanation. $\endgroup$ – Jokela Oct 29 '15 at 16:50
  • $\begingroup$ Sorry for being vague myself. "You don't need much of a depth... Less than 100 m... should be enough... with any particular crater. The rest is mathematics" is, frankly, fluffy geological arm-waving that the link you provided, which is about yoghurt, does little to elucidate or substantiate. What I think would be more convincing is a worked example from a published quantitative model, either theoretical or empirical but either way preferably validated, that shows the relationship between, say, bolide mass, water depth, and crater size. $\endgroup$ – kwinkunks Oct 29 '15 at 17:15
  • $\begingroup$ @kwinkunks My experience comes from physics. But this is simply a collision like any collision. The answer simply points out that there isn't any remarkable difference in penetration to water. Velocity difference is a greater issue. If you like I can delete this answer. $\endgroup$ – Jokela Oct 29 '15 at 17:50
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    $\begingroup$ @JokelaTurbine: Bernoulli's principle can't help you in this case, as this is only valid for laminar flows, without any evaporation. Both will play a major role in determining the penetration depth of a falling rock in water. $\endgroup$ – AtmosphericPrisonEscape Dec 28 '15 at 19:13

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