16
$\begingroup$

What object is in this photo?

View coordinates: $47.25103 \ \ 38.81697$

Time: $2013$-$11$-$23 ~21$:$00 +4 ~UTC$

Sony $A580, \ 50mm f/1.4 15"$

M31 Andromeda Galaxy

$\endgroup$
4
  • $\begingroup$ It's no night clubs with lasers or strong LED lights. Can this be M31? $\endgroup$
    – George
    Nov 24, 2013 at 19:06
  • $\begingroup$ Camera lay lens up and looked on the zenith. $\endgroup$
    – George
    Nov 24, 2013 at 19:57
  • $\begingroup$ nightskyinfocus.com/2013/04/08/… $\endgroup$
    – George
    Nov 24, 2013 at 20:14
  • $\begingroup$ nova.astrometry.net/user_images/942615 is a fairly helpful tool as well. $\endgroup$
    – user21
    Jan 13, 2016 at 4:01

1 Answer 1

11
$\begingroup$

Your guess was correct. It is the Andromeda Galaxy, M31.

Here is a map of the part of the sky near zenith at the place and time you provided: Sky map for Taganrog, Russia on 11/23/2013 5:00:00 PM UTC. Even the rotation is small. The sky map is rotated approximately 30° counter clockwise relative to the photo. You were approximately facing south when taking it. enter image description here

On your picture there is a part of the Andromeda constellation. I will use only a small bottom part of your picture (below the imaginary horizontal line going through M31) to identify the constellation.

  • The bright star to the left of M31, very close to the edge is μ Andromedae (marked 37 μ And on the sky map). It is the star between Mirach (out of photo) and M31.
  • On the bottom of the right part of the photograph you see three stars on a imaginary vertical line: ψ, κ and ι Andromedae.
  • Between ψ and κ, to the right there is λ Andromedae (marked 16 λ And and connected with μ on the sky map).

The bottom of the three stars in the upper-left corner of your photo is φ Andromedae (connected by the vertical line with 37 μ And on the sky map).

The 1 ο And on the sky map is not on your photo.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .