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pumpkinpie
2006-Jun-15, 06:07 PM
A colleague of mine noticed that the Moon's north and south poles are more heavily cratered than the rest of the Moon, and wondered why that might be. Some of us in the office have done some research but haven't been able to come up with any explanation. Whenever craters at the poles are discussed, it's only about the permanently shadowed regions.

Does anybody have an explanation for this?

Kullat Nunu
2006-Jun-15, 06:28 PM
Are they really more cratered than the lunar highlands? It could be only an illusion caused by light coming from low angles which makes craters better visible.

hhEb09'1
2006-Jun-15, 06:39 PM
Yes, how are they measuring "more heavily cratered"?

pumpkinpie
2006-Jun-15, 06:49 PM
She hasn't done any measuring. They just appear that way. Sorry--I should have used "appear" in my original post.
We have an image* of the entire Moon globe projected onto a 6' sphere. We can tilt the image up and down so the poles come to the equator and are more easily viewed.
One of her explanations that it is just an artifact of the light, but she wants to be sure.

*we're not sure where the data come from, but we think from Galileo.

Ara Pacis
2006-Jun-15, 06:56 PM
The poles are more heavily cratered? It must be proof of physical pole reversals! It just happens to be that they reversed so completely to look right side up. Obviously NASA has been hiding this fact from the public... :D

(joke)

hhEb09'1
2006-Jun-15, 07:09 PM
Such a globe would be a patchwork of many images, using different angles of sunlight. Of course, at the moon poles, the sun is low. The views of other areas may be larger, and include higher angle sunlight.

Have you tried to compare it to some online map of craters (as opposed to photos)?

Ara Pacis
2006-Jun-15, 07:12 PM
just go to moon.google.com and have a look-see. zoom in all the way ;)

Ronald Brak
2006-Jun-15, 07:56 PM
Well the damn rabbit is sorta in the middle. (And don't tell me there's a man in the moon. Anyone can see it's a rabbit.)

I think it's just because there doesn't happen to be a lunar sea at the poles. Everywhere pretty much got the same number of impacts, it's just lava erased the old ones in the seas.

pumpkinpie
2006-Jun-15, 08:18 PM
Well the damn rabbit is sorta in the middle. (And don't tell me there's a man in the moon. Anyone can see it's a rabbit.)

I think it's just because there doesn't happen to be a lunar sea at the poles. Everywhere pretty much got the same number of impacts, it's just lava erased the old ones in the seas.

I had thought of that, which brings up another why. Why aren't there as many mare at the poles?

To me, it looks like at the poles, the craters are bigger. Maybe that's a better explanation than more numerous.

hhEb09'1
2006-Jun-15, 08:22 PM
I had thought of that, which brings up another why. Why aren't there as many mare at the poles?Or, why are they all on the side facing us?

Some people would say that the extra mass was tidally locked to us--but its not exactly pointed our way anyway

Ara Pacis
2006-Jun-16, 01:28 AM
Could the impacts on the far side have caused antipodal eruptions of magma on the near side? If there was a sustained or periodic bombardment of the far side with big impactors, and the earth shielded the near side then that could result in what we see... maybe.

Superluminal
2006-Jun-16, 02:39 AM
There is also the theory that Earth's gravity caused more eruptions of lava on the near side. Large asteroids, of the type that formed the larger basins, such as Imbrium, approach the moon at angles near to the lunar equator. Mare would form around the equator leaving the poles to be bombarded by smaller asteroids that have been scattered into more inclined orbits.

hhEb09'1
2006-Jun-16, 08:24 AM
Could the impacts on the far side have caused antipodal eruptions of magma on the near side? If there was a sustained or periodic bombardment of the far side with big impactors, and the earth shielded the near side then that could result in what we see... maybe.I don't think the shield idea really works, in the Earth/moon system. It would be like a tennis ball shielding a marble, over three feet away.

Ara Pacis
2006-Jun-16, 09:37 AM
I'm thinking deflection more than blocking.

hhEb09'1
2006-Jun-16, 09:52 AM
I'm thinking deflection more than blocking.Except it's just as apt to deflect it towards as away, so I still don't think it's a big factor

HenrikOlsen
2006-Jun-16, 12:32 PM
There is also the theory that Earth's gravity caused more eruptions of lava on the near side.
Isn't that the same kind of fuzzy thinking that would have us have only one tide a day?

MAPNUT
2006-Jun-16, 04:06 PM
Do asteroids, large meteoroids or whatever travel in random directions in the solar system? Do they not tend to travel in the orbital plane more often? If they do, craters near the poles should be much less common, and the impact crater recently discovered in the Antarctic should be truly a freakish event. I know meteor showers come from out of the plane, like comets do, but where do the bigger ones come from?

Perhaps a dumb question, but the message at the top of the page is scolding me for not posting in several weeks.

pumpkinpie
2006-Jun-16, 04:20 PM
I finally took a look at the images in question. I would say there's not necessarily *more* craters there. But, as was brought up above, it looks like they stand out because of shadowing. I'm sure if we took counts of craters at the polar regions, vs. a number of other regions of the same size, we would get similar numbers. (excluding maria, of course.)

But thanks for all the discussion!

Kullat Nunu
2006-Jun-16, 05:18 PM
Do asteroids, large meteoroids or whatever travel in random directions in the solar system? Do they not tend to travel in the orbital plane more often? If they do, craters near the poles should be much less common, and the impact crater recently discovered in the Antarctic should be truly a freakish event. I know meteor showers come from out of the plane, like comets do, but where do the bigger ones come from?

There's nothing weird about that. Remember that Earth is very small compared to its orbit. An object with only a slight difference in inclination would pass far over Earth's poles. Combined with other orbital parameters, impacting objects can come from any direction as seen from the Earth. Same goes for the Moon, although it rotates very "straight" compared to the Earth.

ngc3314
2006-Jun-16, 05:20 PM
Do asteroids, large meteoroids or whatever travel in random directions in the solar system? Do they not tend to travel in the orbital plane more often? If they do, craters near the poles should be much less common, and the impact crater recently discovered in the Antarctic should be truly a freakish event. I know meteor showers come from out of the plane, like comets do, but where do the bigger ones come from?



Asteroid orbits do concentrate near rthe ecliptic plane (although not nearly as much as major planets - some big asteroids have 40-degrees inclinations). However, what matters for impact directions is not the impactor's inclination, but where it's aimed for (which can be anywhere on a body small compared to the span of the inclined orbit) and, less important, that the relative impact direction on the target body can cover a huge range of angles even for modest relative inclination (example: the NEAR spacecraft did an Earth swingby to pump up its orbital plane to match Eros, and spent a good while almost above the Earth's south pole despite only a few-degree plane change). On top of these, impact craters have almost the same properties whether the impact is vertical or off-vertical until grazing angles of something like 15 degrees above the horizon are involved. So we actually do expect that the cratering density will be the same all over the surface of a single world (except where some other process wipes them out, as on Earth or the lunar maria).

tony873004
2006-Jun-16, 06:32 PM
Saying the same thing in different words, relative to the Sun, the meteroids do not travel in random directions. They heavily favor roughly ecliptic orbits in the prograde direction. Polar orbits around the Sun do not exist. But relative to the Earth or Moon, they can come from all directions.

This is because Earth's orbital velocity, and the orbital velocity of the meteroids is ~30 km/s. So an object striking the Moon's north pole at 10 km/s still has the vast majority of its solar velocity in the prograde ecliptic direction.

Here's a few screen shots illustrating:
a) the trajectory of an object about to collide square with the Moon's north pole as viewed from within the ecliptic plane.

b) The impacting object's solar orbit as viewed from within the ecliptic plane. It's inclination is 17 degrees.

c) The impacting object's solar orbit as viewed from above the ecliptic.

http://orbitsimulator.com/BA/mpolar1.GIF

http://orbitsimulator.com/BA/mpolar2.GIF

http://orbitsimulator.com/BA/mpolar3.GIF

MAPNUT
2006-Jun-16, 07:13 PM
Thanks for all these responses to my questions, which seem to be saying the same thing, but I'm not sure I get it. If meteoroids do travel in the plane, it still seems that the impact density should be less near the poles, of either the earth or the moon, just as the incident sunlight is so much less dense that the poles are cold. Do you mean to say that gravity pulls the objects in from some distance above the poles? That would make sense but I don't think anybody said it.

tony873004
2006-Jun-16, 07:38 PM
It's not so much Earth's gravity, or the Moon's gravity that pulls the asteroids in. It's mostly their trajectory simply intersecting the Earth's or Moon's position at the right time. The gravity will have some effect, causing objects that would otherwise have had a close encounter to impact.

Consider a baseball. No matter what direction you throw it, relative to the Sun you have not significantly altered its velocity vector. It still retains Earth's velocity around the Sun. It is still travelling around the Sun in the ecliptic. So are you. But this doesnt mean that the ball can only hit you in a horizontal direction. If you toss the ball straight up, nothing prevents the ball from dropping straight down relative to you, and landing on your head.

MAPNUT
2006-Jun-16, 09:05 PM
OK, I can see that some objects will have enough of a "north" or "south" component to hit near the poles. But it still seems to me that especially with regard to the moon, where the gravity is much less and there is no atmosphere to slow down an object passing close by, that the density of impacts should be much less, since the large majority of objects travel primarily in the plane. Put it this way, say 20 degrees of latitude around the pole spans a large land area, but the area exposed to the orbital plane is much less. So the density of impacts should be less. No?

ngc3314
2006-Jun-16, 09:57 PM
OK, I can see that some objects will have enough of a "north" or "south" component to hit near the poles. But it still seems to me that especially with regard to the moon, where the gravity is much less and there is no atmosphere to slow down an object passing close by, that the density of impacts should be much less, since the large majority of objects travel primarily in the plane. Put it this way, say 20 degrees of latitude around the pole spans a large land area, but the area exposed to the orbital plane is much less. So the density of impacts should be less. No?

But what matters is the relative motion between target and impactor. Both will share about the same velocity component in the ecliptic plane, so the approach velocity (before target gravity adds its bit) is more isotropic. Potentially impacting objects move within a fat disk-shaped swarm about the ecliptic; the greater the inclination, the greater the "vertical" velocity component when they cross it.

(Hmm - wonder how much effort an actual calculation would be?)

tony873004
2006-Jun-16, 11:00 PM
(Hmm - wonder how much effort an actual calculation would be?)

For a head-on collision with the Moon's north pole (solar inclination in degrees):

inclination =~ -0.0232 * collision speed2 + 2.27571 * collision speed - 2.3809

and

collision speed =~ 0.0047 * inc2 + 0.4 * inc + 1.3486