Talk:99942 Apophis/Archive 1

message: 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)
replies: text: Oops, my bad, NASA is government so it's all in the public domain. AxelBoldt 01:55, 25 Dec 2004 (UTC)
replies: text: Which made it plagiarism. I added :Template:NASA accordingly. -- Cyrius|✎ 02:12, 25 Dec 2004 (UTC)
replies: text: 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)
replies: text: 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)

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

message: 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
:MW = 2/3 (log Moment(dyne-cm) - 16)
from the same page, I arrive at a magnitude of
:MW =  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)
replies: text: 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)
thread_title:  Equivalent earthquake 

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

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

text: 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)
thread_title:  ooooh destruction of Earth 

message: 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)
thread_title: Animation

message: 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)
thread_title:  ISO 8859-1 ? 

message: 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)
replies: text: 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)
replies: text: 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)
thread_title:  Velocity of impact is known? 

message: 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)
replies: text: 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)
replies: text: 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)
replies: text: 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)

text: 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)
thread_title:  Eastern Hemisphere? 

message: ::.
::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)
replies: text: As we now know, it was named Apophis as of July 19 2005. -- Curps 13:17, 20 July 2005 (UTC)
thread_title:  Naming 2004 MN<sub>4</sub> 

message: 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)
replies: text: Geez, it's not! I'll throw it out. --DerHerrMigo 14:43, 2 February 2006 (UTC)
thread_title:  Close approach AND Friday the 13th? 

message: Apophis is no longer at 1 on the Torino scale. The meteor is now at 4. 12-8-2005
:False, look there . Please, note that it's Torino scale and Apophis is an asteroid, not a falling star! Bye! Dread83 23:42, 8 December 2005 (UTC)
thread_title:  Torina Scale