4
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ 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
    Commented Apr 23, 2018 at 21:53
  • 1
    $\begingroup$ Found while reading for another Q: Wikipedia: Absolute magnitude: Planets $\endgroup$
    – Mike G
    Commented Jun 25, 2019 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
    Commented Jun 25, 2019 at 5:03

2 Answers 2

4
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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
    Commented Apr 23, 2018 at 22:47
  • $\begingroup$ It may be trivial, but I finally got my $\log_{10}(1329)$. $\endgroup$
    – uhoh
    Commented Jan 18, 2022 at 20:11
2
$\begingroup$

A source of this mathematical expression is:

THE ALBEDO DISTRIBUTION OF NEAR EARTH ASTEROIDS, Edward L. Wright et.al.

This mathematical expression appears in the introduction to this article, and is attributed there to:

"Application of photometric models to asteroids", by Bowell, Edward ; Hapke, Bruce ; Domingue, Deborah ; Lumme, Kari ; Peltoniemi, Jouni ; Harris, Alan W. pages 524-556

Best regards.

Updated:

The full demonstration of the expression can be found in Appendix A of:

Binary asteroid population. Angular momentum content. P. Pravec, A.W. Harris

$$d \cdot \sqrt p=K\cdot 10^{-\frac H 5}$$

Where:

$$K=2 \ (AU)_{km} \cdot 10^{\frac{V_s}5}$$

The absolute magnitude of Sun is:

$V_s=-26.762$

$(AU)_{km}=149.6\cdot 10^6 \ km$

$$K=2 \cdot 149.6 \cdot 10^6 \cdot 10^{-\frac{26.762}5}=1329 \ km$$

Finally:

$$\boxed{H=5 \log \frac{1329}{d\cdot \sqrt p}}$$

Best regards again.

$\endgroup$
1
  • $\begingroup$ $\log_{10}(1329)$ and all. Excellent! $\endgroup$
    – uhoh
    Commented Jan 18, 2022 at 20:10

You must log in to answer this question.

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