Charles, , your not actually from this planet are you? I'm guessing Kapax? close?

Aww, man, you ruined it. Yes, the GRB is stronger along the polar axis… however, something able to be seen with the naked eye from half the universe away is going to go right through whatever dust may be obscuring it, and there's no way to tell at what inclination the GRB is to the host galaxy… conservative sense tells me to assume that its along the axis of the host galaxy, but that isn't necessarily true, is it.

No it won't ionize the whole nucleus of the host galaxy, but! a few dozen parsecs is a cubic crapton of gas… and all that gas in the assumed accretion disk of the core will be blasted back.

Peter Knapp,

It is safe to assume that GRB's of this type (if one could even say this is a 'type' of GRB) don't happen anywhere anymore… they are bygones of an ancient era of time when supermassive black holes in the center of galaxies were still eating material. However, whatever causes them (SMBH swallowing a star, or neutron stars colliding or something) obviously gives off alot of energy, and likely blasts back any accretion disk, and in my opinion signals the end of an active galactic nucleus phase, as all the food for the massive black hole is ionized and blasted into the disk of the galaxy, sparking new star formation and ending the black hole's access to material.

So the answer is yes, the Milky Way probably had one at one time, or even several, and the Sun and Earth are likely spawn of one such, but these days the Milky Way is quiet… so thankfully we don't expect to see except from really far away.

p. tanwani,

Aw shucks, it weren't nothin.

No reason to ruin a moment of contemplation!

sorry to burst your bubble of destruction a bit, but the effects of a GRB on it's host galaxy are not very large. First of all, GRBs are strongly collimated. They are invisible form the side (you only see the following supernove), and the most powerful blast effects only hit 2 x a few square degrees of sky.

Also, strong ionization takes place only within a few dozen parsec of the GRB.

If such an event occured in the core of our galaxy, we would not see it, because there's dozens of magnitudes of extinction by dust. But the gamma blast would fry all electronics on Earth, at the very least…

From my Astronomy text, I can figure the absolute magnitude of an object if I know its distance and apparent magnitude. The equation is this:

m = M + 5 log(d/10)

where m is apparent magnitude, M is absolute magnitude (its magnitude at 10 pc) and d is distance in parsecs.

If the object is 7.5 billion light years away, that is 2.3 billion parsecs, or 2.3×10^9.

This give an M value (absolute magnitude) of -36.8

The apparent magnitude of the Sun as seen from Earth is -26.5 or thereabouts.

If I take the same equation, solving for d to find out how far this object would appear in the sky as bright as the sun, I get an equation of

d = { [ (m - M) / 5 ] ^10 } x 10

d = 13761.7 pc,

OR

44,863.15 Light Years.

That is the distance that the object appears as bright in the sky as the sun on a warm summer day.

Given that any habitable planet will already have a sun in the sky, this event would be catastrophic for any system at that distance or closer, and probably even half again as far. A four minute burst at that distance could be enought ot boil half the oceans and kill everyone outside at the time.

So lets see… volume of a sphere is 4/3 x pi x r^3
That comes out to 3.782318 x 10^14 cubic light years of space.

Or for better visualization, a sphere with a diameter of 89726.3 Light years…

The Milky Way's disk is thought to extend about 40 kpc in diameter.thats 40,000 pc, or 130,400 light years. Thats about 34% of the disk of the galaxy.

Interstellar dust extinguishes light by about one magnitude for every thousand parsecs… so if there was a planet on one edge of the galaxy and this object was on the other edge, the GRB would appear to be a 3.2 magnitude, shining right through its galaxy to the other side with a brightness in an alien sky equivalent to the faintest stars visible in an urban neighborhood with naked eye.

In all likelihood, the grb would be in the center of its host galaxy. This means it would

The distance from Earth to the center of our galaxy is 8 kpc, 8000 pc, 26,080 light years. If it happened in the center of our galaxy, it would shine in our sky with an apparent magnitude of -9.52, roughly an eighth of the brightness of the full moon.

The galaxy would be completely dead in the central third, and there would be damage control issues all over its host galaxy. Likely life would have to start over for millions of worlds, if it was even possible after such an extreme event.

The TORTOREM video camera attached to the REM robotic telescope in Chile caught the GRB with a time resolution of 0.13 Seconds!!!!

Check out the videos here:

Full time resolution:

http://vo.astronet.ru/~karpov/grb080319b.avi

10 Frame binning:

http://vo.astronet.ru/~karpov/grb080319b_lowres.avi

The weird movement at ~ 30 seconds is the REM telescope slewing to the GRB, after that, it's in the middle of the FOV.

By the way, in the Pi of the Sky movie, the first frames, where theafterglow is at naked-eye visibility, are not even included in this video!!! Same as TORTOREM, PotS was bserving GRB 080319A, and the flash was in the left upper quadrant.

Don Alexander

I wonder though…how large a volume of space would this explosion have sterilized? eeek!

Adios.

Vancouver, B.C.

No matter how many times I run through it in my head, it still blows me away that an event half way across the visible universe can appear as bright as a 5th mag. star, however briefly that may be.