<p>A source of this mathematical expression is:</p>
<p><a href="https://authors.library.caltech.edu/70302/1/aj_152_4_79.pdf" rel="nofollow noreferrer"><strong>THE ALBEDO DISTRIBUTION OF NEAR EARTH ASTEROIDS, Edward L. Wright et.al.</strong></a></p>
<p>This mathematical expression appears in the introduction to this article, and is attributed there to:</p>
<p>&quot;<em>Application of photometric models to asteroids</em>&quot;, by Bowell, Edward ; Hapke, Bruce ; Domingue, Deborah ; Lumme, Kari ; Peltoniemi, Jouni ; Harris, Alan W. pages 524-556</p>
<p>Best regards.</p>
<p><strong>Updated</strong>:</p>
<p>The full demonstration of the expression can be found in Appendix A of:</p>
<p><a href="https://www.asu.cas.cz/%7Eppravec/pravecharris07.pdf" rel="nofollow noreferrer"><strong>Binary asteroid population. Angular momentum content. P. Pravec, A.W. Harris</strong></a></p>
<p><span class="math-container">$$d \cdot \sqrt p=K\cdot 10^{-\frac H 5}$$</span></p>
<p>Where:</p>
<p><span class="math-container">$$K=2 \ (AU)_{km} \cdot 10^{\frac{V_s}5}$$</span></p>
<p>The absolute magnitude of Sun is:</p>
<p><span class="math-container">$V_s=-26.762$</span></p>
<p><span class="math-container">$(AU)_{km}=149.6\cdot 10^6 \ km$</span></p>
<p><span class="math-container">$$K=2 \cdot 149.6 \cdot 10^6 \cdot 10^{-\frac{26.762}5}=1329 \ km$$</span></p>
<p>Finally:</p>
<p><span class="math-container">$$\boxed{H=5 \log \frac{1329}{d\cdot \sqrt p}}$$</span></p>
<p>Best regards again.</p>

CC BY-SA 4.0
2022-01-18T20:10:26.767Z
$\log_{10}(1329)$ and all. *Excellent!*
uhoh
CC BY-SA 4.0
2022-01-18T20:07:31.497Z
2022-01-19T07:49:14.067Z
2022-01-19T07:49:14.067Z
Albert
<p>The JPL CNEOS
<a href="https://cneos.jpl.nasa.gov/tools/ast_size_est.html" rel="nofollow noreferrer">asteroid size estimator</a>
and various asteroid albedo papers cite
<a href="https://doi.org/10.1006/icar.1996.5664" rel="nofollow noreferrer">Harris and Harris 1997</a>.
That paper is behind a paywall, but
<a href="http://www.boulder.swri.edu/clark/jssrpb04.pdf" rel="nofollow noreferrer">Stuart and Binzel 2004</a>
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.</p>

CC BY-SA 3.0
2018-04-23T22:47:19.087Z
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](https://meta.stackexchange.com/q/309334/303080)).
uhoh
CC BY-SA 4.0
2022-01-18T20:11:36.780Z
It may be trivial, but I finally got my $\log_{10}(1329)$.
uhoh
CC BY-SA 3.0
2018-04-23T18:11:22.870Z
2018-04-23T18:11:22.870Z
Mike G
<p>In this <a href="https://space.stackexchange.com/a/25121/12102">answer</a> I have rearranged an equation from somewhere, and now I can not relocate the source. My re-aranged form is:</p>

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

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

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

CC BY-SA 4.0
2019-06-25T04:57:01.070Z
Found while reading for another Q: [Wikipedia: Absolute magnitude: Planets](https://en.wikipedia.org/wiki/Absolute_magnitude#Planets)
Mike G
CC BY-SA 4.0
2019-06-25T05:03:14.957Z
@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!
uhoh
<asteroids><absolute-magnitude>
Good source for the relationship between absolute magnitude, diameter, and albedo?
uhoh
# Good source for the relationship between absolute magnitude, diameter, and albedo?

(Asked by: username_0 on 2018-04-23)

In this [answer](https://space.stackexchange.com/a/25121/12102) 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?

**Tags:** asteroids, absolute-magnitude
> **Comments:**
> **username_1**: Found while reading for another Q: [Wikipedia: Absolute magnitude: Planets](https://en.wikipedia.org/wiki/Absolute_magnitude#Planets)  
> **username_0**: @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!  

---

## 2 answers

### Answer by username_2 on 2022-01-18. Score: 2 | Accepted answer

A source of this mathematical expression is:

[**THE ALBEDO DISTRIBUTION OF NEAR EARTH ASTEROIDS, Edward L. Wright et.al.**](https://authors.library.caltech.edu/70302/1/aj_152_4_79.pdf)

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**](https://www.asu.cas.cz/%7Eppravec/pravecharris07.pdf)

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

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.

> **Comments:**
> **username_0**: $\log_{10}(1329)$ and all. *Excellent!*  

---

### Answer by username_1 on 2018-04-23. Score: 4

The JPL CNEOS
[asteroid size estimator](https://cneos.jpl.nasa.gov/tools/ast_size_est.html)
and various asteroid albedo papers cite
[Harris and Harris 1997](https://doi.org/10.1006/icar.1996.5664).
That paper is behind a paywall, but
[Stuart and Binzel 2004](http://www.boulder.swri.edu/clark/jssrpb04.pdf)
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.

> **Comments:**
> **username_0**: 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](https://meta.stackexchange.com/q/309334/303080)).  
> **username_0**: It may be trivial, but I finally got my $\log_{10}(1329)$.  

---

