A HUGE fireball has appeared over Chelyabinsk, Russia, this morning. And with HUGE I mean: HUGE. Apparent brightness rivalling the sun, and very strong sonic booms leading to glass damage and people being wounded by flying glass. This must have been a seizable object entering the atmosphere.
Phil Plait, the "Bad Astronomer" has very good coverage including some amazing videos here, so I will refer to him for imagery and the general story (apart from two I include below: one showing the meteor, the other one the arrival of the shock wave).
Below, I will briefly explain why this fireball cannot have been a fragment of 2012 DA14, the ~50 meter wide asteroid that will pass very close to earth coming evening (Feb 15, 2013).
First of all (and Phil Plait points this out as well), the fireball in Russia came from the wrong direction. Several of the videos show it appearing in the east near the rising sun, coming from a N-NE direction. That is the wrong direction: fragments of 2012 DA14 are on a south-north trajectory.
What is even more important: fragments of 2012 DA14 could never enter the atmosphere as far north as latitude 55 N (Chelyabinsk). Fragments in orbits similar to that of the asteroid, have a theoretical geocentric radiant at declination -81 degrees, i.e. almost at the southern celestial pole. They hence approach earth from due south. This means that the northern hemisphere is out of reach of these fragments: the northern hemisphere represents (as seen from these approaching fragments) the "far side" of the earth.
[video added 18/02/2013]
[added 18/02/2013] In the above video I explain this more visually, with the help of an orange. In reality, it is slightly more complicated than I present it in the video, as objects grazing the earth's limb are actually slightly attracted by earths gravity and can end up a little bit over the line between "front" and "far" side of the earth. Ending up at latitude 55 N is nevertheless out of the question.
The funny thing is that the latitude of Chelyabinsk and the approach direction of 2012 DA14 (and fragments in a swarm around it) are well established facts, even if the trajectory of the Russian fireball is less so at the moment. So it is quite nice that from the encounter geometry with the 2012 DA14 orbit and the latitude of the Russian meteor alone, we can actualy already exclude a connection between the two with a quite strong certainty.
Fragments in 2012 DA14-like orbits and the Russian fireball itself are also too fast to be temporarily captured in earth-orbit, so that is no explanation either.
This fireball was not man-made space junk either. Besides coming from an unlikely direction, it is too fast and much too bright for that.
These are amazing times: the reentry of a Russian rocket stage seen from NW Europe on the evening of the 13th, then this hughe meteoric fireball over Russia this morning, and a close pass of asteroid 2012 DA14 tonight. Wow!
1 comment:
Thanks for the informative post. What I was wondering about is whether the calculated trajectory for the Russian meteor was including the effects of Earth's gravity. I'm inclined to think it should not be included. To give the best view of what the true orbit of the meteor was, you should give it as if the Earth was not there to interrupt it and change its trajectory with its gravity.
On the other hand I was able to show using a Hohmann orbit calculator that using the delta-V you get from a close in Earth encounter, up to 11.2 km/s, can change a 2012 DA14 type orbit into the one claimed for the Russian meteor, with one big caveat. The Hohmann calculator assumes you get the delta-V boost all at once. So it's not precisely correct to apply it to a case when you get it from a gravitating body where it's applied over time.
Here's the calculator:
Simple Hohmann Transfer Orbit Calculator.
http://web.archive.org/web/20100128043604/http://home.att.net/~ntdoug/smplhmn.html
Select the orbit around the Sun and input a little less than 1 AU, say, 135,000,000 km, for the beginning orbit radius, which is that of 2012 DA14, and 2.5 AU, about 375,000,000 for the ending orbit radius, the aphelion for the Russian meteor. You'll see the delta-v needed to change the smaller orbit into an elliptical one with the larger, ending radius as its aphelion is well within that which can be delivered by the Earth's gravity. (Hohmann orbit changes use two delta-v burns. The first changes the original smaller, circular orbit into a larger elliptical one. The second circularizes this into one with that larger radius. For this application we only need it to have the elliptical shape of the Russian meteor's orbit so that first delta-v of the calculator is the only one I'm using.)
Bob Clark
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