In this answer I have rearranged an equation from somewhere, and now I can not relocate the source. My re-aranged form is:

$$M_{Abs} = 5 \left(\log_{10}(1329) -\frac{1}{2}\log_{10}(\text{albedo}) -\log_{10}(D_{km})\right)$$

This is the relationship between the absolute magnitude of an object and its diameter and albedo, assuming it is spherical.

Question: What is a good source for any form of this equation to which I can link in my answer?

  • $\begingroup$ This shouldn't be hard to derive. Twice as large = 4 times as bright, double the albedo = double the brightness, convert both to changes in magnitude, and apply to original absolute magnitude. $\endgroup$
    – user21
    Apr 23 '18 at 21:53
  • 1
    $\begingroup$ Found while reading for another Q: Wikipedia: Absolute magnitude: Planets $\endgroup$
    – Mike G
    Jun 25 '19 at 4:57
  • $\begingroup$ @MikeG Yep, thanks! While I don't see the exact expression there verbatim but I see enough pieces of it in similar form that it is certainly there in spirit. Thanks! $\endgroup$
    – uhoh
    Jun 25 '19 at 5:03

The JPL CNEOS asteroid size estimator and various asteroid albedo papers cite Harris and Harris 1997. That paper is behind a paywall, but Stuart and Binzel 2004 attribute this formula to it: $$ H = C - 5 \log_{10} D - 2.5 \log_{10} p_V$$ where $H$ is absolute magnitude, $p_V$ is albedo, and $C$ = 15.618.

  • $\begingroup$ That's numerically equivalent and so it will do, thank you! I'm still going to try to find the source that has the exact expression log${}_{10}$(1329) in my own quixocity because it bothers me that I've lost a reference (I suspect due to a variant of this). $\endgroup$
    – uhoh
    Apr 23 '18 at 22:47

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