Starship Asterisk*

Counting according to Douglas Adams, is the only thing for us humans, to proove that we are independent of computers. I stead of counting sheep, lets count globular star clusters around our Galaxy:

1. Apod 1996-02-21: 160
2. Apod 1998-11-07: 200
3. Apod 2000-10-15: 200
4. Apod 2001-04-22: 200
5. Apod 2002-04-16: 200
6. Apod 2005-09-05: 200
7. Apod 2006-05-20: 150
8. Apod 2007-04-19: 150
9. Apod 2008-05-01: 200

An odd 200. Note further that this subject is favourite during spring and autumn.

What is more intriguing to me is: how do they combine two photographs, regarding to their different sizes? Plain old resampling using some kind of filter isn't so accurate.The sizes of the images must be precisely the same, not to mention the orientation. And around te edges the image gets distorted. The distortion of two telescopes are never the same, so the images must have been processed in order to remove the distortion.

henk21cm wrote:globular star clusters around our Galaxy: 160/200/200/200/200/200/150/150/200.
An odd 200.

158 known globular clusters [1] with perhaps 10–20 more undiscovered [2].

henk21cm wrote:How do they combine two photographs, regarding to their different sizes?

If I can do it to under half-a-pixel precision with Photoshop (I used an overlay to identify SN 1987A in another thread), then I'm sure the bright minds at NASA, JPL-Caltech and Univ. Minnesota can do that even better. I expect them to have done so many times before with specialized software.

Would a planet in a relatively dense part of these clusters have no nightime? (the night sky being as bright as the day sky?) Beautiful creations, these clusters!

Sputnick wrote:Would a planet in a relatively dense part of these clusters have no nightime? (the night sky being as bright as the day sky?)

I don't think so. Two simple estimates.

1) uniform distribution of stars. The diameter of the globular cluster is 150 ly. The volume is (4*π*R^3)/3, so 14 million ly^3. There are 10 million stars, so 0.7 star per qubic ly. That is comparable to our Sol.

2) distribution inversely proportional to the distance to the center. The distribution of stars, ρ is therefore ρ = α/r We do not yet know the value of α , however the number of stars, N, is know and it is equal to:

N = ∫ dr ∫ d φ ∫ d θ ρ cos(θ ) r^2

while integrating over r=0,R, φ =0,2π , θ = -π/2, π/2

Evaluating these integrals leads to:

N= 2παR^2, so α = 1E7/(2π*150^2) ≅ 70 /ly^3

When we locate our planet at 1 ly from the center, we count 70 stars per qubic ly. For simplicity sake we assume 64 stars per ly^3. That is easy, so the distance between the stars is then 0.25 ly. When we see our sun at a distance of 0.25 ly, i.e. 16 kAU, his brightness will be 256 million times less than at earth. Since a magnitude is a factor 2.5, the star will be 21 magnitudes less bright. Our sun at earth has a magnitude of -27, so the brightness of the closest star is -6, roughly the same as Venus, when she is brightest.

There are 4 of these bright Venusses. Then there are 4 lesser bright stars, at 1.4 times the distance, so twice less bright. And so on. As a result i do not see daylight during night. Not even as much as during full moon.

Note that this sort of approximation implicitly uses extinction by dusty clouds, since i stopped calculating at the next nearest neighbours. Otherwise it might lead to something similar to Olbers Paradox.

So, Henck, did you do that math in your head or did you have a little help, eh, snuck a little calculator into the classroom eh.

Thanks Henck .. especially for putting it into language I can understand, 'four Venuses' - 'less bright than a full moon'. Good stuff.

Henck - I think if you pointed your disc at http://antwrp.gsfc.nasa.gov/apod/ap080422.html
you might hear singing .. and I'm serious about this.