Comments on: A galaxy choked with dust

By: Pete Jackson

Pete Jackson — Fri, 17 Jun 2011 14:51:44 +0000

@19 Nigel: I don’t see any disagreements. Hubble can see and resolve the brighter stars in Centaurus A, as well as stationary fireflies in Tokyo.

By: Nigel Depledge

Nigel Depledge — Fri, 17 Jun 2011 14:34:14 +0000

@ Pete Jackson (17) -
Er … so, are you agreeing with IanS, or not?

By: ozprof

ozprof — Fri, 17 Jun 2011 14:29:29 +0000

Hi BA

NGC 5128 is not an elliptical galaxy. It is classified as an SO.

I found that out the hard way when I lost a bet over is any elliptical galaxy visible to the naked eye. I said “yes” and cited NGC 5128 as my evidence. Then it was shown to me that is was not an elliptical.

By: Pete Jackson

Pete Jackson — Fri, 17 Jun 2011 11:46:55 +0000

@15 Ian: Hubble could resolve the two fireflies if it could see them in the first place. I see fireflies outside my window looking like apparent magnitude m = -4 at a distance of about 10 meters. At 100 meters, m = +1; at 1 km, m = +6; at 10 km, m =+11; at 100 km, m = +16; at 1000 km, m = +21 and finally at 10000 km, m = +26. Problem is that fireflies only stay on for about 10% of the time, so the average magnitude, m, would be about 28.5.

So, against all my intuition, Hubble could both detect and resolve the two fireflies as long as they both stayed within 10 foot boxes for the several hours exposure needed!

By: Nigel Depledge

Nigel Depledge — Fri, 17 Jun 2011 11:42:38 +0000

@ Pete Jackson (12) -
I don’t think it’s a question of whether or not Hubble can detect stars in galaxies like Cen A. It’s a question of whether or not Hubble can resolve those stars into distinct points (as opposed to a smear of light from many thousands of stars relatively close together).

Edit – I just noticed that Ian S (15) beat me to it.

By: IanS

IanS — Fri, 17 Jun 2011 08:10:04 +0000

@12 Pete, that just tells us how bright an object can be detected, to know whether we are seeing individual stars we need to consider the angular resolution of hubble and the distance to cen A

The formula for working out paralax arcseconds can be simplified to:
p=(206265*B)/d where p is the angle in arcseconds B is the seperation between two objects and d is the distance from the us to cen A. Rearranging that formula we get (p*d)/206265=B (someone please confirm this, maths is not really my strong point!).

Hubble’s best angular resolution is 0.04 arcseconds and cen A is 11 million lightyears away so we can give values to p and d thus: (0.04*11000000)/206265=B therefore B= 2.133 lightyears

so Hubble can resolve individual stars in cen A as long as they are seperated by at least 2.1 lightyears. which is damn impessive really

(Nasa have the following to say about hubbles resolving power: “In visible light (at wavelengths near 500 nm) the combination of the Hubble telescope plus its highest resolution cameras achieve an angular resolution of about 0.04 arc seconds. The human eye can resolve objects separated by about 40 arc seconds. In other words the resolution of Hubble exceeds that of the human eye by about a factor of 1000. The science team that developed the new Advanced Camera for Surveys be installed in SM3B points out that the high resolution channel of the ACS should resolve two fireflies separated by about 10 feet at a distance corresponding to the distance between New York and Tokyo.”

Now that is really impressive!)

By: Brian

Brian — Fri, 17 Jun 2011 08:04:25 +0000

Looking at the bottom right corner of the smaller picture there are two bright points that look strangly like eyes. Turn the picture upside down and you can also see an oval shaped head with curtains of hair on both sides. You can even make out a nose. Pareidolia strikes again.

By: Richard Drumm The Astronomy Bum

Richard Drumm The Astronomy Bum — Fri, 17 Jun 2011 02:10:45 +0000

Intergalactic photobomb.
Good one. I like it!

By: Pete Jackson

Pete Jackson — Fri, 17 Jun 2011 01:17:51 +0000

Hubble will have no problem seeing individual stars in Centaurus A. Use the standard formula

m (apparent magnitude) = M (absolute magnitude) -5 + 5 log d, where d is the distance in parsecs.

d = 11 million light years = 3,374,000 parsecs.

Supergiants have M = -5 or lower (more luminous is more negative)

Formula gives m = 22.6

Hubble can easily see down to about m = 27 (even 29 or 29.5 on long exposures).

By: Messier Tidy Upper

Messier Tidy Upper — Fri, 17 Jun 2011 01:16:04 +0000

See : http://en.wikipedia.org/wiki/Cepheid_variable

http://stars.astro.illinois.edu/sow/deltacep.html

http://www.space.com/9592-50-year-mystery-strange-pulsating-stars-solved.html

for more Cepheids~wise.

By: Messier Tidy Upper

Messier Tidy Upper — Fri, 17 Jun 2011 01:15:20 +0000

@^ Sam H & CB : Very true. There have been bright Cepheid variables detected and followed in other nearby galaxies – that was one major reason in working out that these were other galaxies in the first place. Luminous Blue Variables – stars like Eta Carinae have also been identified in other galaxies.

Stunning, magnificent superluminous image of Centaurus A there. Thanks BA.

By: Sam H

Sam H — Fri, 17 Jun 2011 00:34:32 +0000

@CB #8: That’s what I always thought too – the starlike objects in distant galaxies are simply star clusters viewed from a great distance. However, given the brightness and distance of stars like Deneb and Eta Carinae in our own galaxy, a few of those starlike objects could indeed be massive (and extremely luminous) individual stars.

By: CB

CB — Thu, 16 Jun 2011 22:35:28 +0000

@ Mark T:

No way, it’s tens of thousands of light years across, no way individual stars would be discernible at that scale. I’m not sure what they are, though. If they’re in that galaxy and not foreground objects, then they could be star clusters. No way is one star visible, but 10,000 stars in a relatively small space might be a visible dot in another galaxy, maybe?

By: CB

CB — Thu, 16 Jun 2011 22:13:24 +0000

@ Kappy:

Those are both stars in our galaxy.

By: Kappy

Kappy — Thu, 16 Jun 2011 21:38:01 +0000

The two really large bright points of light with the crosses on top of them (I’m assuming it’s a light artifact), what are they? Are the really big objects in the Cen A Galaxy, or are they foreground stars from the Milky Way that happened to be in the FOV for this image (which I imagine is just a fraction of an arcsecond wide).

EDIT: Looking at wikipedia it appears it’s much bigger than I though (25×20 arcseconds), with an apparent magnitude of ~7, so if I understand that correct, it’s very large in the sky (something approximately moon sized), but probably just slightly too dim to see with the naked eye?

By: Mark T

Mark T — Thu, 16 Jun 2011 20:16:09 +0000

I’m going to ask what might be sort of a dumb question. I had thought that, when we view other galaxies, with few exceptions we’re not really seeing any individual stars (due to the distance), but rather what appear to be individual points of light/stars are actually clusters or clumps of stars.

However, in this image, it really looks like I can make out lots of individual stars. Is that what I am seeing?

By: Mejilan

Mejilan — Thu, 16 Jun 2011 20:15:28 +0000

Mmmmm, s’mores….
I mean…. S’more! I want s’more pics like this!

By: Andy Beaton

Andy Beaton — Thu, 16 Jun 2011 20:08:04 +0000

Man! All the good stuff is always on the wrong side of the planet.

By: Larian LeQuella

Larian LeQuella — Thu, 16 Jun 2011 19:49:33 +0000

And I [i]just[/i] changed my wall paper…

By: abadidea

abadidea — Thu, 16 Jun 2011 19:19:06 +0000

The upper left looks like lightly toasted marshmallow. Mmmm. Peanut butter swirl galaxy.