Expert: Jayendra Upadhye - 12/2/2006

Question
Hi, its me again. 35kpc its impossible answer because Shapley concluded that Andromeda Galaxy it was in the Milky Way so the distance must be smaller than 10kpc which is the radius of Milky Way...I found in the internet that is about 1600 light years but i can't find the way to solve it...

Answer
Hi,
Sharon, in that case i need to go into the depth of it.
I did not pay attention to detail, as i thought it was some school question with a simple rule of 3s embedded in it. I do not know the actual details of what shapley concluded and how,so i trust it will take more time to get back to you with the answer! [i never remember actual detail and though i remember like that the m31 is 2 million light years away, i do not remember how many lightyears is 1 kpc]
If you have the time, fine, else consider me out of this.
I may take upto 2 days to answer you that's why!
jayen

followup!
Sorry for being late, but one does have to earn a living too!

Ok wikipedia (ref:- http://en.wikipedia.org/wiki/Parsec)
defines a parsec to be 3.261630751 light years.
At that rate, 35000 parsec (35 kilo-parsec) would be 114157.085 or approx 0.114 million light years away!
[meaning observations needed to be more accurate!)

And i was right, as per wikipedia it is 2.5 million light years away.
(ref:- http://en.wikipedia.org/wiki/Andromeda_Galaxy)

Now, my logic regarding rule of 3s is right, (it cant be otherwise), so it may be shapely with his limited instruments, judged it wrong?

Moreover, the s.Amdromedae actually happened in 1885, and shapeley was in 1920s..

Curtis actually noticed that other novae were happening in andromeda and they were 10 times fainter than similar novae in the milky way.

if we recalculate independently using the 9 and 6.2 magnitude data, based on brightness and inverse aquare calculations, we still get similar results, that fall short of the REAL 2.5 million lightyears that is the actual distance to M31.

Now 1st magnitude star is 100 times more bright than 6th magnitude star as (2.512) raised to power (6 - 1)= 5 is 100.
(REF:- http://en.wikipedia.org/wiki/Magnitude_%28astronomy%29)

In that case, 6.2 magnitude star is (2.512) rasied to [(9 - 6.2) = 2.8] = 2.512 raised to 2.8 = 13.184 times more powerful in apparent magnitude .

using the inverse square law for distance, & assuming absolute luminosity to be identical,
(1*k)/(r*r) = (13*k)/R*R) where r = 10kpc and R is distance of the other star; k is constant of proportionality which is arbitrary (it is there on each side and hence does not matter),

then R = sqrt[(13) * 100] = 36.309 Kpc times distant! we got 35 earlier!

You see, both methods are yielding same results.
If in reality, m31 (andromeda on the messier catalog is 2.5 million light years away as the cephied variables show it to be, the earlier photometric observations either "saw wrongly / interpreted wrongly", such as a novae of different intrinsic luminosity that made it brighter, or made wrong observations!
Such things HAVE happened earlier in astronomy.
so no worries there.
Still if you have doubts, do check my calculations out!
I am as fallible as those mighty achilles of astronomy!

Hope that suffcies!
Jayen