Starship Asterisk*

A recent Messenger picture of Mercury has found an area of 'tear-drop' shaped craters:
http://messenger.jhuapl.edu/gallery/sci ... .nomap.jpg
The accompanying text is less than explanatory; " The elongated shape results from formation on the sloping face of the larger crater's wall. "
Eh, you what?

So can we puzzle this out? NOT low angle impacts, as those are circular too. Not secondary craters as those are also circular, except when they fall in a tight group or line and blur together - these are discrete. A layer of the surface that favours slumping in one direction seems unlikely, and anyway, there are many other truely circular craters in the same area.

What could cause 'tear-drop' craters?

JOhn

Here's my guess John: I noticed that the area where these teardropped shaped crators are found is darker, indicating to me that it might be on a sloping, shaded area. If an impacter hits on an inclined surface, couldn't teardrop-like craters be formed, with the pointed end of the crater pointed up slope?

Bruce

Just noticed more of your question,

JohnD wrote:The accompanying text is less than explanatory; " The elongated shape results from formation on the sloping face of the crater's wall."

This slope must be significant, or the same rounding effect that happens with low angle impacts would have happened here too.

Let me start out by saying that I really don't know.

But I do wonder if the heat generated by the impact might be enough to melt sufficient rock momentarily, such that when on an incline, the crater slumps downhill a little before solidifying.

JohnD wrote:A recent Messenger picture of Mercury has found an area of 'tear-drop' shaped craters:
http://messenger.jhuapl.edu/gallery/sci ... .nomap.jpg
The accompanying text is less than explanatory; " The elongated shape results from formation on the sloping face of the larger crater's wall. "
Eh, you what?

So can we puzzle this out? NOT low angle impacts, as those are circular too. Not secondary craters as those are also circular, except when they fall in a tight group or line and blur together - these are discrete. A layer of the surface that favours slumping in one direction seems unlikely, and anyway, there are many other truely circular craters in the same area.

What could cause 'tear-drop' craters?

Actually, very low angle impacts are not round. But that's not what's going on here. Impacts are round with respect to the gravity normal. The dynamics are different for impacts on sloped surfaces, since you'll have asymmetrical forces and directed down slope slumping.

Nitpicker wrote:But I do wonder if the heat generated by the impact might be enough to melt sufficient rock momentarily, such that when on an incline, the crater slumps downhill a little before solidifying.

Plus it's Mercury, where the ground can be a lot closer to melting even before the impact.

BDanielMayfield wrote:

Nitpicker wrote:But I do wonder if the heat generated by the impact might be enough to melt sufficient rock momentarily, such that when on an incline, the crater slumps downhill a little before solidifying.

Plus it's Mercury, where the ground can be a lot closer to melting even before the impact.

Depends if the impact was during the day or night. Very long days and nights on Mercury.

Nitpicker wrote:

BDanielMayfield wrote:

Nitpicker wrote:But I do wonder if the heat generated by the impact might be enough to melt sufficient rock momentarily, such that when on an incline, the crater slumps downhill a little before solidifying.

Plus it's Mercury, where the ground can be a lot closer to melting even before the impact.

Depends if the impact was during the day or night. Very long days and nights on Mercury.

Right, which is why I wrote "can be" instead of will be closer to melting.

Ah! So the word "sloping" referred to the original ground surface!
And as Chris and BDan suggest, the uphill end of the crater slumped more.

A "low angle impact" is less than 12 degrees, I believe. What ground slope is required to produce a tear-drop?

John

JohnD wrote:Ah! So the word "sloping" referred to the original ground surface!
And as Chris and BDan suggest, the uphill end of the crater slumped more.

A "low angle impact" is less than 12 degrees, I believe. What ground slope is required to produce a tear-drop?

No idea. My guess would be that any angle at all would lead to some degree of distortion. There's probably one or more critical angles related to the nature of the impacted surface.

FWIW, this isn't going to be related to melting in any way. There isn't any significant melt in a crater, except for what's vaporized away. What happens beneath the impact is a fracturing process. On a sloped stone surface, we could see a variety of structures in the crater. Take a look at the meteor crater in Arizona. It's almost square. That's because of asymmetrical weathering, which follows natural faulting in the rock. But if that area had a significant slope, we might expect the newly formed crater to have immediately slumped along the same natural faults, resulting in an off-round structure.

But crater rims aren't solid rock. Presumably they're more like sand, without any complex underlying structure. So I'm thinking that small craters on large crater rims will tend to show very similar structure, determined only by the slope. And that does seem to be the case with the Messenger image.

I think we can illuminate this!

A staple of civil engineering is slope stability. See: http://en.wikipedia.org/wiki/Slope_stability
And that depends on the angle of repose of the material that makes up the crater wall.
If, as Chris suggests a new crater wall is more like sand than rock, then it will have a repose angle of about 34 degrees (figure for dry sand).
The calculations for slope stability are a bit complex - see this paper which attempts to simplify it by offering charts, sort of ready reckoners!
http://www-personal.umich.edu/~rlmich/i ... 202002.pdf
But lacking any background in this, I find the charts inpenetrable - we don't need an astronomer, we ned a civil engineer!

John

The angle of repose is also gravity dependent. I presume steeper with a lower g. I've worked on a few big civil engineering projects, and the responsibility for determination of (compacted and uncompacted) slope stability falls to the geotechnical engineer. These engineers rely on surprisingly small amounts of data and lots of "rules of thumb". They tend to be the most risk-averse people involved in a civil project, and so they tend not to "push the envelope" very much. This is probably a good thing, because it is somewhat impossible to be precise in this field.

(And thanks for dispelling my wonderings about melting rock, Chris.)