I am new to astronomy and I was given the task of finding out if there are objects which can't be resolved by a telescope of resolution 5 arcseconds.

I was given the values of RA and Dec of these objects in degrees.

   RA              DEC
 201.1999388    27.49294258
 201.2048319    27.49212959
 201.1984702    27.4978016
 201.2025157    27.51610266
 201.1997827    27.51743447 ...etc

How can I find these in the simplest way? Is there any simple formula which can be used to achieve this which is computationally efficient as well?

  • $\begingroup$ Not clear what you mean. Do you mean you have a list of object positions and need to find out which pairs are unresolved or which objects are confused with at least one other object. $\endgroup$ – Rob Jeffries Oct 8 '17 at 13:20
  • $\begingroup$ @RobJeffries Yes, how to find which pairs that don't lie within 5 arcsec of each other. So by that definition find the pairs which might lie within that 5 arcsec resolution which might be confused with each other. $\endgroup$ – Rahul Aedula Oct 8 '17 at 14:36
  • $\begingroup$ Google "angular distance sphere"; you can actually use a shortcut if you know the points are close, but "premature optimization is the root of all evil", so try the precise formula first. $\endgroup$ – barrycarter Oct 8 '17 at 17:02
  • $\begingroup$ Is this an astronomy issue? If you know how to calculate the angular separation (simple trig formula), then the rest is an algorithmic problem, which doesn;t really belong here. $\endgroup$ – Rob Jeffries Oct 8 '17 at 18:06

This would be your best resource. Just note that the variables are expressed in geographical latitude and longitude, and your values are Right Ascension and Declination. Longitude and RA are the same, but Declination is 0 at the equator and is positive toward the north pole, whereas latitude is 0 at the north pole and is positive toward the south pole.

Don't forget to mind your units (it'll probably be in degrees since that's what your RA and Dec are). If you're doing this with a computer, you'll probably want to convert to radians first since most computer trig functions are in radians. Whatever program you use, just be sure to check what units it prefers.

Finally, be sure to convert your separation to arcseconds and you should be all set.

  • $\begingroup$ Thank you, this was really helpful. But is there any way to shorten my range of objects I can apply this formula to?. My dataset is very mixed up and to find the objects under that arc length would require me to search through the entire dataset. Which is huge. Is there any small approximation I can perform where I can narrow the range of objects I can search using this formula? Maybe if I sort the dataset based on RA and DEC can I shorten the range? $\endgroup$ – Rahul Aedula Oct 9 '17 at 7:12
  • $\begingroup$ en.wikipedia.org/wiki/CURE_data_clustering_algorithm or googling "efficient clustering algorithm" (no quotes) may be helpful. $\endgroup$ – barrycarter Oct 9 '17 at 13:54
  • $\begingroup$ "whereas latitude is 0 at the north pole and is positive toward the south pole". I have never heard of this convention before and do not see this on the referenced page. The formula on Wikipedia is correct for the latitude defined at 0 at the equator, +90 at the north pole, and -90 at the south pole. Am I missing something? $\endgroup$ – JohnHoltz Oct 9 '17 at 15:57

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