# Star density in the Omega Cluster

## Posers and Puzzles

###### Fast and Curious
54d

So I posted this question in Science with no response, maybe you all could look at it hear, the Omega has some 10 million stars inside and a diameter of some 150 light years, a lot of stars crammed into a relatively small volume, I calculated the stars would be on average about 1/5th of a light year apart, close to one tenth the distance to OUR closest star, Alpha Centauri, about 4 ish ly away.
Is my number anywhere near correct?

###### chemist
54d

@sonhouse

I didn't realize it was a question.
So you calculated a sphere with 75 Light year radius giving roughly 1.7 million ly as volume for the sphere, having 10 million stars we find about 5 per cubic light year. So the estimate is okay.

###### Fast and Curious
53d

@Ponderable
Yeah, that's what I got. Can you imagine what the night sky would look like if Earth were somewhere in the middle of THAT mess?🙂

###### chemist
53d

@sonhouse
I can imagine a world, where there is no night. If life is possible is a totally different question.

###### Fast and Curious
53d

@Ponderable
Well I would think if there was a star with an Earthlike planet buried somewhere in the middle of Omega, since the closest star would still be 1/5th of a light year away, it would not be much brighter than say Serius which shines brightly in our sky, so lets say there were four such and then more at 2/5ths of a light year, and more at 3/5ths of a light year, it would be an interesting calculation to see whether the increase in luminosity on average would go up the further you got away from the home planet out to the radius of 75 light years or so.
Maybe there would be no real night like Earth but more like moonlight 100% of the time, what do you think? How could we calculate that?

###### Fast and Curious
53d

@sonhouse
It is obviously an integral problem but I was a bit short on calculus, I got to where in calc 101, 'this is an integral and THIS is differential🙂

###### Fast and Curious
53d

@sonhouse
Two variables, good old inverse square law VS increasing stars with distance.

###### chemist
53d

@sonhouse said
@sonhouse
It is obviously an integral problem but I was a bit short on calculus, I got to where in calc 101, 'this is an integral and THIS is differential🙂
The problem is a bit tricky, the model would be to put earth around sun and then add the stars. Of curse there are different stars with different size and thus luminosity. Plus we can't just take the next neighbours, we would have to add as many as are signifacant light source (and we have the distribution problame again).

So give me the parameters I make a model, but it will be a weak model anyway.

###### Fast and Curious
53d

@Ponderable
Yep, there would be a Sol to make water liquid and we would want a planet with a nice fat magnetic field to ward off solar flairs and such but just making an assumption the stars are all exactly the same luminous level as Sol and that makes it easier to model I would think. And of course our Earth 2 would rotate in 24 hours and tilted 20 odd degrees to produce winter/summer.....But the main thrust would be the radiance seen by a being on Earth 2.

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