<p> and 0 must appear because the vast majority of telescopes cannot point to the Sun and an asteroid at the same time. They can’t even start taking photons at the same time- sky’s too bright.</p>
<p>All telescope measurements are to some level arbitrary- if not due to background (sky) noise, then to optical flaws and instrument noise. Any target  worth reporting in peer literature, then, must be calibrated, or it’s little more than some person’s sense of “kinda bright” vs. “really bright”.</p>
<p>Lacking the ability to point at the Sun as a ‘test pattern’, we point at calibration stars instead… but which cal stars? Vega is the chief nighttime reference, but it’s down some of the time (including <em>all the time</em> if you’re South enough), at low altitude/high airmass some of the remaining time (our atmosphere has a filtering effect, dimming and reddening it), and just too bright for some instruments.</p>
<p>We then turn (ha) to other, well-studied stars as light standards, but this transfers <em>and multiplies</em> the issue: there is now star-to-star difference, aside from Vega-to-Sun. This is the cause of the  and 0 terms. The star-to-star differences have been measured by now to arbitrarily good degree, and can be looked up and plugged in.</p>
<p>The specific case of 0 takes out the phase issue. A star is a star, from most angles. Only a few stars are significant enough to have an equator-to-pole difference, causing them to appear brighter or dimmer when viewed pole-on vs. bulge-on. Asteroids aren’t stars. Asteroids show phases, in the sense of the moon, and ‘phases’ for individual boulders, cobbles, <em>and particles</em> on their surfaces. Google “opposition effect”, or specifically, “shadow hiding”. There is also some effect from translucent minerals, multiple bounces, etc. These factors sum or null as the viewing angle changes, so 0 is an attempt to account for that.</p>
<p>The whole concept of asteroid absolute magnitude is one of those things that’s straightforward enough in broad brushstrokes, but knottier and knottier as you demand more decimal places. You’re confused because you can grasp the brushstrokes, then you encountered Pravec, Harris, Muinonen, Belskaya, Johnson, Tonry, et al. who demand those decimal places.</p>

CC BY-SA 4.0
2022-08-15T12:52:32.417Z
Thank you for the answer. Things are much clearer now. I had not thought a lot earlier about the zero-point flux of the measuring instrument/system and the appearance of V_sun was a conundrum to me. I could now appreciate even more the effort required to find good enough cal stars.
mysterium
CC BY-SA 4.0
2022-08-12T14:41:53.707Z
2022-08-12T14:41:53.707Z
caInstrument
<p>I've done some research and found that one could derive the absolute magnitude <span class="math-container">$H$</span> of an asteroid in the following way. <span class="math-container">$H$</span> is the brightness of the asteroid, observed on Earth, if the asteroid were at a distance of 1 AU from the Sun and 1 AU from the Earth, while the phase angle (angle subtended at the asteroid by the vectors/lines joining Sun-asteroid and Earth-asteroid) would be <span class="math-container">$0°$</span>, which is a hypothetical scenario in itself, as this orbital configuration is never achieved in the reality (at least in the currently known 3D world).</p>
<p>My understanding is based on <a href="https://astronomy.stackexchange.com/a/48218">this answer</a>, which cites <a href="https://www.asu.cas.cz/%7Eppravec/pravecharris07.pdf" rel="nofollow noreferrer">Pravec &amp; Harris 2007</a>, where they define <span class="math-container">$H$</span> in the appendix A as follows (note that obj is referred to the asteroid).</p>
<blockquote>
<p>The absolute magnitude, H , of a Solar System object is defined as
the apparent magnitude of the object illuminated by the solar light
flux at 1 AU and observed from the distance of 1 AU and at zero phase
angle. From that, we get</p>
<p><span class="math-container">$$H  = V_{sun} -2.5   \log  \frac{F_{obj}(0,1 AU)}{F_{0}} $$</span></p>
<p>where, <br />
<span class="math-container">$V_{sun}$</span> - apparent magnitude of Sun, which is −26.762 ± 0.017 as they report citing Campins et al. (1985). <br />
<span class="math-container">$F_{obj}(0, 1AU)$</span> - the light flux from the object at zero phase angle.<br />
<span class="math-container">$F_{0}$</span> - incident light flux.</p>
</blockquote>
<p>Then they replace <span class="math-container">$\frac{F_{obj}(0,1 AU)}{F_{0}}$</span> by an expression they derive for the geometric albedo and put forth the relation between the absolute magnitude, geometric albedo and the diameter of an asteroid. I'm able to follow the derivation they have for the geometric albedo (equations A.1, A.2 and A.3 in the aforementioned paper).</p>
<p>What I don't understand is why <span class="math-container">$V_{sun}$</span> appear in this expression. Nor do I understand the role of <span class="math-container">$F_{0}$</span> in the equation. My line of logic is that the luminosity of the Sun needs to get scaled to the 1 AU using the inverse square law and it should be already embedded in the expression <span class="math-container">$F_{obj}(0, 1AU)$</span>, without having to put it separately as <span class="math-container">$V_{sun}$</span>. But obviously I'm missing something here. Any help will be much appreciated.</p>
<p>On the contrary I find it much easier to grasp the idea of stellar absolute magnitude <span class="math-container">$M$</span>, where the luminosity of the star gets scaled to the 10 parsec distance using the inverse square law.</p>
<p>Thank you !</p>

<solar-system><asteroids><apparent-magnitude><magnitude><absolute-magnitude>
absolute magnitude of an asteroid
mysterium
# absolute magnitude of an asteroid

(Asked by: username_0 on 2022-08-10)

I've done some research and found that one could derive the absolute magnitude $H$ of an asteroid in the following way. $H$ is the brightness of the asteroid, observed on Earth, if the asteroid were at a distance of 1 AU from the Sun and 1 AU from the Earth, while the phase angle (angle subtended at the asteroid by the vectors/lines joining Sun-asteroid and Earth-asteroid) would be $0°$, which is a hypothetical scenario in itself, as this orbital configuration is never achieved in the reality (at least in the currently known 3D world).

My understanding is based on [this answer](https://astronomy.stackexchange.com/a/48218), which cites [Pravec & Harris 2007](https://www.asu.cas.cz/%7Eppravec/pravecharris07.pdf), where they define $H$ in the appendix A as follows (note that obj is referred to the asteroid).

> The absolute magnitude, H , of a Solar System object is defined as
> the apparent magnitude of the object illuminated by the solar light
> flux at 1 AU and observed from the distance of 1 AU and at zero phase
> angle. From that, we get
>
> $$H = V\_{sun} -2.5 \log \frac{F\_{obj}(0,1 AU)}{F\_{0}} $$
>
> where,   
> $V\_{sun}$ - apparent magnitude of Sun, which is −26.762 ± 0.017 as they report citing Campins et al. (1985).   
> $F\_{obj}(0, 1AU)$ - the light flux from the object at zero phase angle.  
> $F\_{0}$ - incident light flux.

Then they replace $\frac{F\_{obj}(0,1 AU)}{F\_{0}}$ by an expression they derive for the geometric albedo and put forth the relation between the absolute magnitude, geometric albedo and the diameter of an asteroid. I'm able to follow the derivation they have for the geometric albedo (equations A.1, A.2 and A.3 in the aforementioned paper).

What I don't understand is why $V\_{sun}$ appear in this expression. Nor do I understand the role of $F\_{0}$ in the equation. My line of logic is that the luminosity of the Sun needs to get scaled to the 1 AU using the inverse square law and it should be already embedded in the expression $F\_{obj}(0, 1AU)$, without having to put it separately as $V\_{sun}$. But obviously I'm missing something here. Any help will be much appreciated.

On the contrary I find it much easier to grasp the idea of stellar absolute magnitude $M$, where the luminosity of the star gets scaled to the 10 parsec distance using the inverse square law.

Thank you !

**Tags:** solar-system, asteroids, apparent-magnitude, magnitude, absolute-magnitude

---

## 1 answers

### Answer by username_1 on 2022-08-12. Score: 2 | Accepted answer

and 0 must appear because the vast majority of telescopes cannot point to the Sun and an asteroid at the same time. They can’t even start taking photons at the same time- sky’s too bright.

All telescope measurements are to some level arbitrary- if not due to background (sky) noise, then to optical flaws and instrument noise. Any target worth reporting in peer literature, then, must be calibrated, or it’s little more than some person’s sense of “kinda bright” vs. “really bright”.

Lacking the ability to point at the Sun as a ‘test pattern’, we point at calibration stars instead… but which cal stars? Vega is the chief nighttime reference, but it’s down some of the time (including *all the time* if you’re South enough), at low altitude/high airmass some of the remaining time (our atmosphere has a filtering effect, dimming and reddening it), and just too bright for some instruments.

We then turn (ha) to other, well-studied stars as light standards, but this transfers *and multiplies* the issue: there is now star-to-star difference, aside from Vega-to-Sun. This is the cause of the and 0 terms. The star-to-star differences have been measured by now to arbitrarily good degree, and can be looked up and plugged in.

The specific case of 0 takes out the phase issue. A star is a star, from most angles. Only a few stars are significant enough to have an equator-to-pole difference, causing them to appear brighter or dimmer when viewed pole-on vs. bulge-on. Asteroids aren’t stars. Asteroids show phases, in the sense of the moon, and ‘phases’ for individual boulders, cobbles, *and particles* on their surfaces. Google “opposition effect”, or specifically, “shadow hiding”. There is also some effect from translucent minerals, multiple bounces, etc. These factors sum or null as the viewing angle changes, so 0 is an attempt to account for that.

The whole concept of asteroid absolute magnitude is one of those things that’s straightforward enough in broad brushstrokes, but knottier and knottier as you demand more decimal places. You’re confused because you can grasp the brushstrokes, then you encountered Pravec, Harris, Muinonen, Belskaya, Johnson, Tonry, et al. who demand those decimal places.

> **Comments:**
> **username_0**: Thank you for the answer. Things are much clearer now. I had not thought a lot earlier about the zero-point flux of the measuring instrument/system and the appearance of V_sun was a conundrum to me. I could now appreciate even more the effort required to find good enough cal stars.  

---

