-----------------------------------------------------------------------------------
Nitpicker, thank you for that answer, which appeals to my mathematical side. So then the formula for apparent magnitude, given distance and absolute magnitude, would be:Nitpicker wrote:According to what I have just read about Normal Type Ia supernovae, they all have an absolute magnitude of around -19.3, meaning that this one will have an apparent magnitude of about +8.5. So, visible in binoculars, but not to the unaided eye.BDanielMayfield wrote:Since it is expected to brighten for another two weeks, and since a supernova can breifly out-shine an entire galaxy, how bright can we expect this to become? Will this become a naked eye object?
8.5 = -19.3 - 5 (1- log10[3,700,000pc])
Edit: this formula is not particularly accurate for objects beyond the Milky Way.
M(app) = M(abs)-5*(1 – log10[D]), with distance in parsecs, and not taking any reddening factors into account.
Reddening must be quite strong in the exploding cigar though as shown by the infrared image Ann posted. One estimate I read mentioned 10.1 as a predicted maximum for SN 2014J.
Please excuse my ignorance of something that I ought to know, but at a fairly dark site what might the limits of a pair of 10x50 binoculars be?
Dirty Harry: "A man’s got to know his limitations.”