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The third paragraph needs to be rewritten as it's a copyright violation from NASA's press release. AxelBoldt 01:51, 25 Dec 2004 (UTC)

Oops, my bad, NASA is government so it's all in the public domain. AxelBoldt 01:55, 25 Dec 2004 (UTC)

Which made it plagiarism. I added Template:NASA accordingly. -- Cyrius| ✎ 02:12, 25 Dec 2004 (UTC)

Is it enough to stick Template:NASA in there? I think it might need a more specific citation. -- Ld | talk 02:55, 25 Dec 2004 (UTC)

I moved it into a "sources" section just to avoid any question. Remember that plagiarism is a matter of ethics, not copyright. We could legally take the NASA report as-is and post it without a care in the world. -- Cyrius| ✎ 03:53, 25 Dec 2004 (UTC)

Remember, the information is not copyright, just the wording. Is the paragraph better now? Any other problems? It's not usually hard to improve government wording. -- Wetman 23:14, 26 Dec 2004 (UTC)

The wording wasn't copyright either. As a NASA release, it's PD. -- Cyrius| ✎ 22:58, 27 Dec 2004 (UTC)

NASA explicitly renounces copyright in photos and movies here. This page doesn't explicitly include text, but it's a general principle of US copyright law that government agencies, having expended public funds to do their work, are not entitled to copyright in their publicly released work product. I'll find a formal reference on this, and put it in the appropriate place, shortly. -- Baylink 00:14, 28 Dec 2004 (UTC)

The following paragraph of mine was deleted with the comment "removing very strange commentary on the odds":

The original NASA report contained the sentence

"In all likelihood, the possibility of impact will eventually be eliminated as the asteroid continues to be tracked by astronomers around the world."

and this statement found its way into numerous newspaper articles. The statement is false if one interprets "in all likelihood" as anything but "with a probability around 299 in 300". Assuming that the astronomer's prediction capabilities are adequate, the possibility of impact will eventually be eliminated if and only if the impact does not in fact happen. The probability of this event was estimated to be around 299 in 300 at the time of writing of the report.

In addition the sentence "Experts expect that additional data will allow for 2004 MN4 to be re-rated as a level zero (no threat) object." was added, which repeats the above mistake. Maybe my explanation above was too dense, so I try first with an example:

"I predict that, with 40% probability, the Dow Jones will be above 12000 on January 2, 2006. However, I'm almost sure that, as further data comes in, I will revise that probability to 0%."

Those two statements are contradictory; no rational person can believe in them at the same time. If I'm almost sure that the probability will be revised to 0%, then I'm almost sure that the Dow Jones will not be above 12000 on 1/2/2006. On the other hand, if my best guess is that it will happen with 40% probability, then I'm not almost sure, but only 60% sure, that further data will revise the predicted probability to 0%.

It's the same with NASA. If NASA expects the asteroid to hit us with probability 2.4% (latest estimate), then they don't expect "in all likelihood" to eventually eliminate the possibility of impact; they expect this to happen with 97.6% probability.

In fact, NASA noticed their mistake in the Dec 24 update to their report:

Nevertheless, the odds against impact are still high, about 60 to 1, meaning that there is a better than 98% chance that new data in the coming days, weeks, and months will rule out any possibility of impact in 2029.

AxelBoldt 19:27, 25 Dec 2004 (UTC)

The following statement is in error:

If I'm almost sure that the probability will be revised to 0%, then I'm almost sure that the Dow Jones will not be above 12000 on 1/2/2006. On the other hand, if my best guess is that it will happen with 40% probability, then I'm not almost sure, but only 60% sure, that further data will revise the predicted probability to 0%.

You have a prediction computer, and it predicts a spread of possibilities for the stock market's state on 1/2/2006. 40% of these possibilities are above 12000. Thus you say the odds are 40%.

You also know that as 1/2/2006 approaches, the prediction computer will become generally more accurate as the time scales become smaller and data accumulates. Suppose that on 1/1/2006 the computer says that there's a near-0% chance of the stock market closing about 12000.

You were not 60% sure that this was going to happen. In fact, you may have been 99% sure because you know that the prediction computer has always generated high estimates in that time frame.

If I knew something about the prediction computer, then I should use that knowledge to come up with a proper probability prediction in the first place. I should say "my computer says 40%, but I know the program to be defective and the prediction to be false; my best prediction is 1%". So I would never report the 40% number in a press release, and I would certainly never enter bets based on odds of 40%. On the other hand, if I trust the predictions of my computer, then I would report 40% to the media, base my bets on it, and would expect that with 60% probability it will eventually revise the chances to 0%. AxelBoldt 21:56, 25 Dec 2004 (UTC)

There are two numbers that are being discussed here. The "current odds of impact", and the "odds of the current odds of impact going to 0% or 100% with time". You're complaining that the one doesn't equal the other. They don't have to. -- Cyrius| ✎ 21:04, 25 Dec 2004 (UTC)

The two numbers being considered are "current odds of impact" and "current odds that the odds estimate over time will eventually go to 100%". My point is that these two numbers are necessarily the same if the prediction mechanism is not completely out of whack. NASA now acknowledges as much, as you see in the quote above, from the revised report. Note especially the word "meaning" in that quote: the two probabilities are always the same; if you know one, that "means" you know the other. The scientists use all the knowledge they have to come up with the best odds prediction; they don't have any hidden knowledge about their prediction computer that's not reflected in their estimates. AxelBoldt 21:56, 25 Dec 2004 (UTC)

Here's another try. Consider the event E that the asteroid will hit us, and the event F that our odd estimates for impact will eventually reach 100%. I claim that (assuming our prediction capabilities are at all adequate) these two events E and F are equivalent: E happens if and only if F happens. The reason: if E happens, then a few days before the due date, we will be sure of the impact, so at that time our odd estimates will have reached 100%, meaning that F happens. On the other hand, if F happens, then at a certain point in time we will be certain that impact happens. This implies that impact does indeed happen, i.e. that E happens.

So E and F are equivalent events, and equivalent events have the same probability, no matter what. AxelBoldt 22:33, 25 Dec 2004 (UTC)

You're making assumptions about how the issued predictions are calculated. But this is becoming increasingly irrelevant.

My only assumption is that the prediction mechanism be adequate, i.e. that the predicted probabilities zero in on the truth over time. I don't think anybody disputes that; NASA certainly won't. AxelBoldt 02:09, 26 Dec 2004 (UTC)

There's a larger problem here. You're trying to read a press release as if it's an exact mathematical statement. It isn't. To the average person, 299 out of 300 is the same as "in all likelihood". Even if I agreed that those were the odds, that didn't need to be nitpicked for a third of the article's length. -- Cyrius| ✎ 23:46, 25 Dec 2004 (UTC)

I agree; overdetailed nitpicking is best left to the talk page. Regarding the average person: I'm afraid that given the general ignorance about probabilities, initially they will think "Oh, 1 in 300, that's a bit scary", and then they will feel reassured by the next sentence "experts say that the possibility of impact will in all likelihood be eliminated given further observations". They won't realize that the second sentence does not provide any reassurance at all and is nothing but a reformulation of the first, with "in all likelihood" replacing "299 in 300". This replacement is a pretty devious rhethorical trick, so I thought it deserved to be exposed; but I won't press the point and can live with the article as it stands. AxelBoldt 02:09, 26 Dec 2004 (UTC)

Axel is absolutely correct. NASA's press release is an exact mathematical statement, i.e., the probability is, exactly, in their opinion, such-and-such a number. There is no way to say that "it is likely that the odds will eventually become zero" without confusing people. There is one probability and that should be the referenced one. 64.123.58.14 01:35, 26 Dec 2004 (UTC)

Cema: I think the original thesis is incorrect. Probability may be a function of time. In this case, as the celestial body comes closer to Earth and its trajectory can be better estimated, astronomers will be able to reevaluate the probability of an impact. The reevaluation is likely to reduce the probability, according to the experts. This is what the quoted NASA report means.

At the same time, I agree that the rhetorical trick was used and it was intended for the unaware public. Cema 03:40, 27 Dec 2004 (UTC)

I was trying to compute the magnitude of an earthquake that release an energy of 1600 megatons of TNT. A megaton of TNT has 4.2*10²² ergs, so we are looking at 6.72*10²⁵ ergs. I used the formula

Energy(ergs) = Moment(dyne-cm)/20000

from http://www.seismo.unr.edu/ftp/pub/louie/class/100/magnitude.html to find the Moment to be 1.34 * 10³⁰ dyne-cm. Now using the formula

M_W = 2/3 (log Moment(dyne-cm) - 16)

from the same page, I arrive at a magnitude of

M_W = 9.4

Using the other (and almost equivalent) formula

log Energy(ergs) = 11.8 + 1.5 M

and solving for M gives a similar result.

This value for M seems to be way too high. Does anybody see the mistake? Could it really be a 9.4 magnitude earthquake? AxelBoldt 02:09, 26 Dec 2004 (UTC)

Well, you're assuming total conversion of impact energy to seismic energy, and neglecting the air blast and the kinetic energy of the ejecta. You might want to peruse the paper that comes with the impact effects calculator. -- Cyrius| ✎ 04:11, 26 Dec 2004 (UTC)

This is great! A minor ripple of near-quasi-panic for Armageddon. So get me this straight...this asteroid is seen as the biggest threat from space since those Torino scales were made up.

Biggest threat, yet only a 2.2% chance of hitting Earth. -- tomf688 ( talk) 03:35, Dec 26, 2004 (UTC)

And arguably there's time to deal with it if it was determined to be a potential problem. Pakaran (ark a pan) 04:09, 26 Dec 2004 (UTC)

Well, it has been determined to be a potential problem. Cema 03:43, 27 Dec 2004 (UTC)

Not a planet-killer. City-killer. Maybe small countries. Belgium. -- Cyrius| ✎ 04:23, 26 Dec 2004 (UTC)

I'm just wondering what distance you would have to move away from this impact in order to be "safe". Judging by the comparisons it seems like 100s of miles. - Ld | talk 02:51, 27 Dec 2004 (UTC)

Depends on the accuracy of the impact prediction. If they know the impact location to a mile, putting 200 miles between you and it would likely be enough. On the other hand, if it's got a 500 mile uncertainty... -- Cyrius| ✎ 07:24, 27 Dec 2004 (UTC)

I'd guess it also depends on if you own real estate in impact the area. I advise selling short rather than waiting to see if you've developed an interesting new view of the coast.

The name was twice changed to the subscript notation, the second time with the comment "NASA may not use subscripts, but that is nevertheless the preferred notation". Is there any evidence for this preference, excepting obviously the preference of the person who made the change? I see that NEODyS at the University of Pisa also doesn't use subscripts, nor have I seen any newspaper articles that do so. AxelBoldt 07:17, 26 Dec 2004 (UTC)

Yes. The Minor Planet Center (the organization officially in charge of keeping track of asteroids) say: "When possible, these additional numbers should be indicated using subscript characters." [1]

However, since their own Minor Planet Electronic Circulars are ASCII, they don't use the subscript much themselves. And many if not most other sources, including popular astronomy magazines and scientific journals alike, usually ignore the subscript. In the past I have argued we should do the same, for simplicity, readability, ease of editing and consistency with what most other folks do. -- Curps 07:42, 26 Dec 2004 (UTC)

The problem with that slippery slope is that it then naturally leads to the systematic omission of diacriticals. The MPC lists are all pure ASCII, as is the A/CC Catchall Minor Object Catalog (amongst others); you have to go to the Astronomisches Rechen-Institut or Institute of Applied Astronomy to get the proper names (typically in TeX notation--including subscript notation).

"Consistency with what most other folks do" is the path to mediocrity and the perpetuation of common mistakes, in my opinion. Simplicity and readability are unaffected by the subscript notation. Ease of editing is a false lure; we are not here to ease our job (as writers and editors), but to provide our readers with the best.

Urhixidur 16:27, 2004 Dec 26 (UTC)

It's not the same thing at all. Diacriticals and accents provide critical information (for pronunciation and disambiguation) that would otherwise be missing. But writing the number in an asteroid name as a subscript doesn't provide any additional information compared to not using a subscript.

For chemical formulas, H₂O is universally written with subscripts, because using subscripts makes the individual chemical elements stand out more clearly. For asteroid designations, there's no advantage to having the letters stand out from the numbers. The "MN" has no special significance, since MN, MN1, MN2, etc were not even discovered consecutively at all. Not surprisingly, subscripts are rarely used in practice.

Conforming to de facto actual near-universal practice is often the most important thing. According to the Comité International des Poids et Mesures, the official symbol for astronomical unit is "ua" and not "au" [2], but if you try to change it to this in Wikipedia I will probably start an edit war :-)

Perhaps you could e-mail Brian Marsden (see http://www.cfa.harvard.edu/iau/pressinfo/200PHAs.html) and ask him what the purpose of subscripts is, and does it bother him if they are omitted. If so, it would be cool to post the reply.

-- Curps 21:52, 26 Dec 2004 (UTC)

Ok, would it be agreeable to add something like "(often written as 2004 MN4)" in the intro paragraph, in order to tell our readers that the article is about the same thing that NASA talks about?

I guess there's no issue with search engines etc., since our title already uses the more common version of the name.

Also, the above asteroid naming conventions should be explained somewhere if they aren't already, maybe on asteroid, so that the reader can resolve the difference between common name and our name. AxelBoldt 20:15, 26 Dec 2004 (UTC)

They are; see Astronomical naming conventions#Minor_planets. Do you really think anyone will think 2004 MN4 and 2004 MN₄ are different things? I guess it can't hurt to give the ASCII notation.

Urhixidur 19:21, 2004 Dec 27 (UTC)

Yaohua2000 attempted to move this page to the subscript version, which messed things up a lot. I've moved it back. -- Cyrius| ✎ 20:12, 27 Dec 2004 (UTC)

To name it Apophis is an extremely dark joke. It will get funnier - and scarier - if this thing comes to threaten Earth. Subramanian ^talk 01:08, 14 December 2005 (UTC)

Don't make the animation into a thumb, because it will screw it up. Wikipedia cannot properly scale a gif animation. — Cantus… ☎ 18:12, Dec 27, 2004 (UTC)

I don't think the animation belongs in the article. We could put in in the article on asteroid impact prediction, but certainly not here (especially this huge version). I think NASA has a more 2004MN4-specific animation which I would not mind having here, but preferebly they should soon release some sort of a photo. I will proceed to remove the image from this page if noone objects. - Ld | talk 18:18, 27 Dec 2004 (UTC)

I don't like the NASA animation at all, since it simplifies things to the point of unintelligibility. If you have a cone of virtual asteroids, then you cannot predict the future cone at all, unless you also have some information about the velocity vectors. AxelBoldt 21:39, 27 Dec 2004 (UTC)

Hi Yaohua2000,

For the subscripts, we should use ₄ and not a Unicode character. This is because for different asteroids the subscript number could be arbitrarily large, even in the hundreds. So for consistency we should use "sub" for subscripts. -- Curps 20:16, 27 Dec 2004 (UTC)

All subscript and superscript number 0 to 9 can be found in Unicode. So large subscript number can be written in Unicode form without <sub>. — Yaohua2000 20:22, 2004 Dec 27 (UTC)

On my copy of Windows XP, it doesn't come out right: 1999AZ₀₁₂₃₄₅₆₇₈₉

In my font, only the 1 2 3 4 appears, the rest are blank boxes. And they're spaced one character apart, instead of appearing as "1234", they appear as "1 2 3 4". So in practice this doesn't work at all. -- Curps 20:32, 27 Dec 2004 (UTC)

I've also tried to read the characters under Windows XP, unfortunately, only 1234 appear correctly. :(. But the characters seems appear correctly on my Mac.

The source of the page contains a line in its HTML header, that specified the charset as ISO 8859-1, why not UTF-8? Which contains the character 4 in subscript form, and can solve the naming problem. The character subscript four has a unicode of 0x2084. — Yaohua2000 20:17, 2004 Dec 27 (UTC)

Wikipedias in other languages use UTF-8. Only the English Wikipedia doesn't. Apparently there are some technical issues involved in doing the conversion to UTF-8 (too much downtime, etc). -- Curps 20:57, 27 Dec 2004 (UTC)

The article states that the velocity of impact would be 12.59 km/s. How do we know this to 4 significant digits? Wouldn't it vary wildly based on the angle of impact, which is treated as unknown later in the article? -- Chris vLS 20:21, 27 Dec 2004 (UTC)

I suppose this must be the actual speed of the object at the time of impact. As you say, the angle of impact, or whether there will even be one, is not yet known. -- Curps 20:35, 27 Dec 2004 (UTC)

Chris, the speed (with respect to Earth) does not depend on the angle at all. From the observed data, it is easy to calculate the velocity of both Earth and the asteroid on that day in 2029, with accuracy of 4+ significant digits, and the speed is the absolute value of the difference of these velocities. It's harder to calculate the position with the accuracy of hundreds of kilometers. -- Lumidek 21:06, 27 Dec 2004 (UTC)

any impact is likely to occur in the Eastern Hemisphere (time zones UTC +3 to UTC -10)

• Most of the timezones between +3 and -10 are in the Western Hemisphere, not the Eastern Hemisphere... whats the deal? -- Bletch 20:58, 27 Dec 2004 (UTC)

From reading the reasoning, I would think they meant UTC +3 to UTC +10, but I'm not sure. AxelBoldt 21:39, 27 Dec 2004 (UTC)

Technically I believe the US is considered the eastern hemisphere when you view the UTC as being the dividing line for the day. Regardless, the Americas are often viewed as the "west" so lets fix it. I believe the UTC times are correct and this should be changed to Western Hemisphere as soon as someone can validate it. Additionally at 21:21UTC it would by night for most of the western hemisphere which would seem to make sense if the asteroid is coming from the outside of earth's orbit. Anyone know a source for impact estimation information? I want to get this resolved as soon as possible. -- Benefros 02:23, 28 Dec 2004 (UTC)

The western hemisphere is the one that has the Americas on it. Look at the prime meridian. To its west is the Atlantic Ocean, and to its east is the bulk of Europe and Africa. When you quote latitude, the Americas are always such-and-such west and Asia is such-and-such east. -- Cyrius| ✎ 02:44, 28 Dec 2004 (UTC)

It probably meant UTC -10, except going by way of the date line instead of the Americas. Time zones are not a good way of measuring ranges, especially of that size. There's too much ambiguity. -- Cyrius| ✎ 02:44, 28 Dec 2004 (UTC)

Do you understand the decimal digits of the date (13.88 or 13.89) the same way as me? My result is that the impact would be around 9:07 pm Greenwich time. Did someone get the same conclusion? By the way, one can combine this with the simulation from which direction the asteroid hits, to determine which part of Earth will approximately be affected. -- Lumidek 21:08, 27 Dec 2004 (UTC)

Looks like it's going to hit the Eastern Hemisphere, more or less. Here's how I got there. What follows is probably wildly inaccurate and should not be believed by any reasonable person.

First, we've got the graphic showing the cloud of positions. Helpfully, it's got a direction of motion arrow, and the direction of the sun.

Then we take a jog over to NASA's Solar System Simulator. We tell it the date and time, say we want to look at the Earth from the sun, and we get this image, which shows the eastern Pacific, and a decent view of North and South America.

If you take those two bits of information, you can determine which half of the world is being threatened. Since the rock's relative motion is toward the sun, that means it's going to hit where it's night time. In rough terms, the hemisphere centered on the equator, somewhere in the western Indian Ocean.

This is highly speculative and based on approximations and outright guesses, so I don't want anyone sticking anything about this in the article. Wait until the experts say something about likely impact targets. -- Cyrius| ✎ 23:29, 27 Dec 2004 (UTC)

And now completely irrelevant :) -- Cyrius| ✎ 02:18, 28 Dec 2004 (UTC)

I did some additional work on JPL Horizons System, which uses a ephemeris updated on December 27, 2004. the result shows that, the asteroid will reach the nearest location from the Earth at 20:49 UTC on April 13, 2029, about 38000 kilometers from the center of the Earth. — Yaohua2000 01:58, 2004 Dec 28 (UTC)

38000 km from the center of the Earth? That's within geostationary orbit! The rock might manage to hit something yet. -- Cyrius| ✎ 02:18, 28 Dec 2004 (UTC)

The most recent ephemeris shows that, the most possible minimum distance is 64085 kilometers occurs at 21:50 UTC from the barycenter of the Earth. This distance is much farther than geostationary orbit. — Yaohua2000 06:27, 2004 Dec 28 (UTC)

Though still pretty close at only 0.17 Lunar Distances. [3] shows that only 5 are known to have or will pass closer. Evil Monkey → Talk 06:43, Dec 28, 2004 (UTC)

The Times seems to have broadcasted new impact odds - any chance of an informed update?

Z.

It would be nice if someone could add info on the probable energy (in Megatonnes of TNT?)of the one that caused the extinguishing of the dinos.

.

Both hard to remember and hard to say when reading it, this asteroid is still waiting for a name. Since his near-by flight will be on friday the 13th, are there any chances one of the following to be chosen :

• Selkirk
• Robinson
• de Foe
• Daniel
• vendredi
• Crusoe
• Frígedag

82.224.88.52 15:48, 27 May 2005 (UTC)

As we now know, it was named Apophis as of July 19 2005. -- Curps 13:17, 20 July 2005 (UTC)

Not a very good sign. -- Ixfd64 2005 July 8 01:36 (UTC)

They've calculated the trajectory to umpty-zillion decimal places. While the Friday the 13th thing was interesting when the numbers showed it could impact, there's no cause for alarm. -- Cyrius| ✎ 06:36, 12 July 2005 (UTC)

Is it really appropriate to include this in the article? Ardric47 03:49, 2 February 2006 (UTC)

Geez, it's not! I'll throw it out. -- DerHerrMigo 14:43, 2 February 2006 (UTC)

• May I ask why a mention of this is inappropriate? It has been calculated that the day is a Friday, and should be at least noted. -- Riffsyphon1024 00:25, 5 July 2006 (UTC)

Images from Deep Impact show that impacts on Minor Plants happen, if anyone really needed proof.-- Jirate 14:54, July 17, 2005 (UTC)

User:Irate insists on including the following:

It should be noted that these estimates rely on a complete knowledge of the inner solar system and all the interactions that 2004 MN4 has over the elapsed period. The knowledge of the inner solar system is far from complete and any small change in the path due to an encounter with an unknown object will produce a bigger variation in the final postion.

I think it should not be included as it is overemphasising an extremely remote possibility and gives further credit to the popular misconception that the solar system is full of asteroids that collide and change paths all the time as seen in many movies and games and whatnot. It's wrong. Space is mostly empty (even in the main asteroid belt) and that an asteroid should collide with a smaller object and measurably get its calculated trajectory for the next 100 years changed is extremely, extremely unlikely. And that's why it is not taken into consideration when dealing with impact solution and risk-modeling. I encourage Irate to try finding any scientific reference or document that makes the note that he wants to include here.

That asteroids have craters and traces of being hit in their 4.5 billion year history is of course true, but, first, most of these were formed in the early history of the solar system when collisions were more frequent, and second, most of these impacts were very minor and didn't change the asteroid path in any measurable degree for the following years. If a 1 m rock hits a 320 m asteroid, there will be a crater formed, but the asteroid will not get its immediate trajectory noticeably changed (we're talking about a 1:30.000.000 mass ratio). But the most important thing here is the fact that the collisions Irate wants to address are so rare and improbable that his note is just misleading to the reader. So I'm reverting him again now. Shanes 02:38, 18 July 2005 (UTC)

I'm reverting back until learn the difference between M and MV.-- Jirate 10:59, July 20, 2005 (UTC)

I concur. The only possible instance of such a collision having been observed in the last two hundred years is 3 Juno: « According to James L. Hilton (1999), Juno's orbit was observed to change (slightly) in 1839, "very likely" due to an "unmodeled encounter" — that is to say, some other object passed very close to it or actually hit it. ».

Subtle effects such as the Yarkovsky effect or the YORP effect are more likely to be important sources of orbit evolution than random collisions.

Urhixidur 04:06, 2005 July 18 (UTC)

So that's one encounter atleast in the last 200 years. If you consider the knowledge of the inners solar system and the relative inability to make highly accurate observations and solve the orbital functions, then one is a highish count. The number of minor planets being observed and tracked has increased rapidly in the last 10 years, this is simply not enough observing time. I think you also forgetting the amplification effects of the earth encounters in the this case, a change in velocity of a fraction of milimeter per second will produce enough movment over a year to signifcantly allet the position. I think you failing to take into account the very long term consequences.-- Jirate 11:12, July 20, 2005 (UTC)

The problem is where the comment is being put --it does not belong in this article, but rather (once cleaned up) in the Asteroid article.

Urhixidur 12:25, 2005 July 20 (UTC)

The evidence for change of trajectory for Juno is very speculative and based on measurements taken in the early 19th century, when positions were not measured as accurately as today. This is one astronomer's opinion, not a generally accepted theory (references to the contrary?).

On the other hand, I have no idea what Irate is talking about when he talks about "relative inability to make highly accurate observations" today. Today we can indeed make such observations (including radar observations) and we know there will not be an impact in 2029. We also know that the 2029 encounter will alter the objects orbit in such a way that we can't yet predict with sufficient accuracy and may lead to future impacts (as the article mentions). -- Curps 13:15, 20 July 2005 (UTC)

For most of the last 200 years we have not been able to make very acurate predictions, but one interaction was been seen 150 years ago. The monitoring of asteriods is a very patchy buisness and the majority of minor planets are unknown. Add to that the various craters we see on everything in the solar system and the 40,000 tonnes of stuff which collides with earth every year, it is safe to conclude that things hit each other in the solar system on a fairly regular basis.-- Jirate 15:21, July 21, 2005 (UTC)

Once again, the "interaction" supposedly involving Juno 150 years ago is very speculative and frankly doubtful.

Nearly all the craters we see (on Mercury, Mars, the Moon, etc) are a few billion years old, created during an early period of intense collisions and impacts. This very process, however, cleared out much of the flying debris and now such impacts are very much rarer. See our article on the lunar crater Copernicus, which is a "young" crater because it is "only" 800 million years old. So if you are using the existence of such craters as evidence of impact rates today, this is very much misleading. It's like counting dinosaur fossils to estimate the population of dinosaurs on Earth today.

If a big target like the Moon (diameter 3500 km) has only had such a small number of impacts in the last billion years, what is the likelihood of impact for a much smaller target like Apophis, whose diameter is around 320 meters (or 0.32 km). With a diameter 10000 times smaller, its cross-sectional target area for impact is 10¹⁰ (10 billion) times smaller. -- Curps 15:56, 21 July 2005 (UTC)

The evidence for the age of the moon craters is not very good. On Mars a crater has been found by Opportunity, is very recent. We know less about the far side of the moon. The evidence is that all objects no mater what size get hit on a regular basis. There are many unexplained lunar events every year which are sometime attributed to cold out gasing but could be anything. Your chossing to dismiss all the impact evidence in the solar system and things like tanguska. I think you should be demonstrating the impossibility of impacts. Your view is the out dated steady state non catastrophic view of the universe. The target area may 10 billion times smaller but that means that it's volume is 1*10^15 smaller and there it's mass also that much smaller. Therefore any interaction will produce a 1*10^15 greater change in velocity. Basically the results of any collision will by 1*10^15 greater so while rare they will rduce greater perterbation in the orbit.-- Jirate 18:07, July 21, 2005 (UTC)

There are recent craters on Earth too, if by "recent" you mean 50,000 years ago. There is a strong scientific consensus for the ages of lunar craters. If you are setting forth your own original research or advocating a "young Earth" creationist viewpoint, this is not really in keeping with the mainstream and majority opinions that Wikipedia science articles nearly always reflect.

Of course impacts still occur in the solar system. We could probably expect a Tunguska-type object to hit Earth at least every thousand years or so, perhaps every few hundred years. But for two Tunguska objects to hit each other is much, much rarer, and for a specific Tunguska object to get hit is even rarer. And all of this, you seem to be expecting to occur within a 24-year time frame, a blink of an eye in the lifetime of the solar system.

Here's an analogy: consider Army A with machine guns storming a fort, and Army B inside the fort with machine guns defending it. The odds are extremely good that the fort will get hit by many bullets during the battle. The odds are quite low that a bullet fired by Army A will collide in mid-air with a bullet from Army B. The odds are even lower that one specific bullet by Seargent Apophis from Army A will collide in mid-air with any bullet fired by Army B. The ratio of the size of a fort to the size of a bullet is roughly the same order of magnitude as the size of the Moon to the size of Apophis. -- Curps 19:58, 21 July 2005 (UTC)

There are several major problems with you analogy. Firstly if you wanted to calculate the odds knowing the number of bullets would be required, we don't have a bullet count, having such a limit number of directions would also be nice. You seem to have limited it to collision with bullets from the other side. It could be anyside's and anything shrapnel, debris. Dating of lunar craters only acounts for the big ones. We know a rock hit the moon in 800 AD the Chinese recorded it and the moon still resonates. It you that seems to be arging for a clockwork universe, the odds may be small, but they are still odds, this rocks path makes it's orbit particularly sensitive due to the chaotic nature of it's earth approaches.-- Jirate 20:11, July 21, 2005 (UTC)

An impact on the Moon is like a bullet hitting the fort. We're talking about bullets colliding with one another midair.

Just to clarify what I wrote earlier: the above machine-gun battle is assumed to last only a day. Naturally, if it lasts many thousands of years, you'd eventually get bullets hitting each other. Similarly, various spacecraft have visited a number of asteroids so far, and they do indeed have cratered surfaces. Again, just like for the Moon, Mars and Mercury, most of those impacts occurred in the early days of the solar system and over a very long time scale. No one is disputing the possibility of impacts on a sufficiently long time scale, it's just that for your proposed paragraph to be relevant, such a bullet-colliding-midair-with-bullet impact would have to occur for this specific asteroid (Apophis) within the next few decades.

Look, here's an article for you pointing out the possibility that a meteoroid collided with the asteroid Eros: http://www.newscientistspace.com/article.ns?id=dn7699 . The problem is, this probably occurred a few hundred million years ago, not last century.

Impacts occur. They're just much, much rarer than you seem to realize. We don't know how many bullets are out there, but we've been following a lot of asteroids for quite a few decades now, and where are the bullets-colliding-midair events that you speak of? If you're saying this is going to happen with one very specific asteroid in the next few decades, why haven't we seen it in hundreds of other asteroids so far? -- Curps 20:24, 21 July 2005 (UTC)

I'm not saying there will be an impact. I'm just qualifying the accuracy of the predictions and what they rely upon. note, I haven't given odds, you've read it as a prediction for some reason.-- Jirate 21:42, July 21, 2005 (UTC)

Well, what I'm saying is that the odds of such an impact (another sufficiently large object hitting Apophis within the next few decades) are so infinitesimally small that they don't quantitatively change the currently computed odds of Apophis hitting Earth during that time period (which are calculated under the assumption that Apophis's orbit will be affected by nothing but gravity; the Yarkovsky effect is minimal over such a short time period and there are no outgassing non-gravitational forces as for comets). And thus the proposed paragraph is simply wrong, because it suggests that we need to mathematically take the possibility of such an impact hitting Apophis into consideration.

In other words, if the odds of something happening are, say, 0.0016 to four decimal places of accuracy, and taking X into account only alters the odds by, say, ±0.0000001, then X doesn't meaningfully alter the original odds and discussion of X is superfluous in arguing over whether the event will occur. It's like introducing a paragraph on the possibility of an magnitude-8 earthquake hitting London into our article on Wimbledon because it might affect who wins next year's tournament. Over the next few thousand years, we might expect one such earthquake (who knows?), but probably not in the next eleven months. -- Curps 22:03, 21 July 2005 (UTC)

It suggests no such thing. It spells out what unknowns the prediction relies on. The predictions also cover the period after the encounter, where even small movment can be magnified. Again it is not the size of the object but it's momentum MV.-- Jirate 22:17, July 21, 2005 (UTC)

But the probability of this particular unknown is simply much, much, much too small to be relevant or to warrant mention, over any reasonable time scale that we care about (the next few hundred or thousand years or so). If you buy a 300m × 300m piece of real estate here on Earth, what are the odds of a meteorite landing on it? Not landing a mile away or anywhere else on Earth, mind you, but landing directly on your particular piece of land? Yes the Earth's atmosphere would burn up most such meteorites, but only the ones that would be too small anyway to alter Apophis's trajectory. Thus, the odds of what you are suggesting are essentially equivalent to the odds of a large meteorite landing within 150m of your house between now and 2029. It's certainly not utterly impossible, but no one would bother to take this possibility into account for the purpose of calculating real estate values. -- Curps 04:23, 22 July 2005 (UTC)

Who is the we? The odds wiing the UK national lottery are smaller than the chances of being killed by a metorite. Do you think the National lottery requires a mention? I'm not calculating real estate values. I'm trying to spell out what the predictions rely upon. You seem to be reading it in a very strange way. -- Jirate 12:35, July 22, 2005 (UTC)

Apophis is no longer at 1 on the Torino scale. The meteor is now at 4. 12-8-2005

False, look there [4]. Please, note that it's Torino scale and Apophis is an asteroid, not a falling star! Bye! Dread83 23:42, 8 December 2005 (UTC) (italian user)