<p>Planetary magnitudes vary not only according to the Sun’s luminosity, their own average albedo, and their distance from the Earth, but also from:</p>
<ul>
<li>Variations in their albedo across their surface.</li>
<li>Their phase angle, for planets that we sometimes see as a crescent.</li>
<li>Their inclination, for planets like Saturn and Uranus that have a different albedo at their equator than at their poles.</li>
<li>Neptune keeps getting brighter. No one knows why.</li>
</ul>
<p>All of these effects are detailed in an informative paper <em>Computing Apparent Planetary Magnitudes for The Astronomical Almanac</em> by Mallama and Hilton, revised in 2018:</p>
<p><a href="https://arxiv.org/pdf/1808.01973.pdf" rel="noreferrer">https://arxiv.org/pdf/1808.01973.pdf</a></p>
<p>Their source code for computing planetary magnitudes given all of these effects can be found here:</p>
<p><a href="https://sourceforge.net/projects/planetary-magnitudes/" rel="noreferrer">https://sourceforge.net/projects/planetary-magnitudes/</a></p>

CC BY-SA 4.0
2020-07-30T02:04:02.847Z
Thanks for keeping Skyfield and Stack Exchange up to date!
uhoh
CC BY-SA 4.0
2020-07-30T02:09:08.947Z
just fyi I've just asked [How fast is Neptune getting brighter? When was was this first noticed and reported?](https://astronomy.stackexchange.com/q/37206/7982)
uhoh
CC BY-SA 4.0
2020-07-28T00:34:55.837Z
2020-07-28T00:34:55.837Z
Brandon Rhodes
<p>Lets assume we're dealing with a superior planet, which is to say, the planet is orbiting at a greater distance than the Earth. This effectively ensures that there is no planetary phase to deal with.</p>
<p>Now, the Sun has a luminosity of <span class="math-container">$L_{sun}$</span> such that the Solar flux as seen by the planet is
<span class="math-container">$$
 f_{planet} = \frac{ L_{sun} }{ 4 \pi d_s^2 }.
$$</span></p>
<p>The cross section of the planet is about <span class="math-container">$A = \pi r_p^2$</span>, so for a given albedo <span class="math-container">$a_p$</span> the reflective luminosity of the planet will be
<span class="math-container">\begin{aligned}
 L_{planet} &amp;= a_p f_{planet} A \\ 
            &amp;= a_p \pi r_p^2 \frac{ L_{sun} }{ 4 \pi d_s^2 }.
\end{aligned}</span></p>
<p>Given that we know the absolute magnitude of the Sun, the absolute magnitude of the planet follows as
<span class="math-container">\begin{aligned}
 V_{planet} &amp;= -2.5 \log_{10}\left[ \frac{ L_{planet} }{ L_{sun} } \right] - V_{sun} \\
            &amp;= -2.5 \log_{10}\left[ a_p \frac{ r_p^2 }{ 4 d_s^2 } \right] - V_{sun} 
\end{aligned}</span></p>
<p>The apparent magnitude of the planet as seen from Earth can then be calculated as
<span class="math-container">$$
 m_{planet} = V_{planet} + 5 \log_{10}\left[ d_{e-p} \right] - 5
$$</span>
where the distance between Earth and the planet, <span class="math-container">$d_{e-p}$</span>, should be in parsecs and of course depends on the orbital phase of the two respective planets.</p>
<p>So the one additional parameter required was the Solar luminosity. There are probably a bunch of subtle effects with the albedo and the geometry of the system that are not taken into account here, but it should be a fair approximation.</p>

CC BY-SA 3.0
2016-08-28T13:04:43.230Z
"the distance between Earth and the planet, de−p, should be in parsecs" is incorrect. For magnitudes within the solar system, distances must be in AUs. The formula reflects this in fact, since the absolute magnitude of a solar system object is defined at 1 AU viewing distance.

The calculation of Vplanet is not affected, as long as rp and ds are expressed using the same unit of distance.
Urhixidur
CC BY-SA 3.0
2017-05-23T04:40:05.883Z
Is this albedo the Bond albedo or the Geometric albedo?
Kate Bertelsen
CC BY-SA 4.0
2024-08-09T15:14:48.037Z
@KateBertelsen See my answer below, this answer has a 4 missing.
Firestar-Reimu
CC BY-SA 3.0
2014-07-29T13:29:30.550Z
Thanks for answering! One question, though: can solar luminosity not be computed from the sun's absolute visual magnitude (or the other way around) because the total luminosity does not have to all be in visual light? Or is there some other reason?
feralin
CC BY-SA 3.0
2014-07-29T19:16:04.147Z
@feralin I guess you can, but you'd still need a second parameter to normalize the conversion, be it either the relative magnitude of the sun or an absolute magnitude of a calibration star. Using the relative magnitude of the sun would make sense here and probably gives a more direct final expression, but I think using luminosity gives a better sense of what's physically going on.
Michael B.
CC BY-SA 3.0
2014-07-29T19:23:15.100Z
@feralin As a second note; if you want to model an inferior planet you can simply make the reflective surface - A - depend on (synodic) orbital phase.
Michael B.
CC BY-SA 3.0
2014-07-29T20:11:59.270Z
sorry, but can you explain what "relative magnitude" is? I'm not getting good results from google...
feralin
CC BY-SA 3.0
2014-07-29T20:29:08.893Z
@feralin sorry, I meant to say apparent magnitude
Michael B.
CC BY-SA 4.0
2014-07-29T08:16:14.570Z
2020-07-31T14:58:37.703Z
2020-07-31T14:58:37.703Z
Michael B.
<p>This answer is from <em>Fundamental Astronomy 6ed</em>, Section 7.8</p>
<p>The first several steps are the same with @Brandon Rhodes</p>
<p>We got: (<span class="math-container">$A$</span> is the <strong>Bond albedo</strong>)</p>
<p><span class="math-container">$$
F = AC\Phi(\alpha) \frac{1}{4\pi\Delta^2} \pi R^2 \frac{L_\odot}{4\pi r^2}
$$</span></p>
<p>The difference is this <strong>normalisation constant</strong> <span class="math-container">$C$</span>:</p>
<p>Since all the radiation reflected from the planet
is found somewhere on the surface of the sphere,
we must have: (<span class="math-container">$\mathrm{d}S = \Delta^2\mathrm{d}\Omega$</span>)</p>
<p><span class="math-container">$$
\frac{C}{4\pi\Delta^2} \int_S \Phi(\alpha)\,\mathrm{d}S = 1
$$</span></p>
<p><span class="math-container">$$
C = \frac{2}{\int_0^\pi \Phi(\alpha)\sin\alpha\,\mathrm{d}\alpha}
$$</span></p>
<p>We denote the <strong>phase integral</strong>:</p>
<p><span class="math-container">$$
q = 2\int_0^\pi \Phi(\alpha)\sin\alpha\,\mathrm{d}\alpha = \frac{4}{C}
$$</span></p>
<p>and <strong>geometric albedo</strong>:</p>
<p><span class="math-container">$$
p = \frac{AC}{4}
$$</span></p>
<p>For a <strong>Lambertian surface</strong> (defined as <span class="math-container">$A = 1$</span>):</p>
<p><span class="math-container">$$
\Phi(\alpha) = \begin{cases}&amp; \cos\alpha,\quad 0\leq\alpha\leq\pi/2 \\ &amp; 0,\quad \text{otherwise}\end{cases}
$$</span></p>
<p>which leads to:</p>
<p><span class="math-container">$$
C = 4,\quad A = 1,\quad p = q = 1
$$</span></p>
<p>If we replace the planet with a Lambertian surface
of the same size, we get:</p>
<p><span class="math-container">$$
F_L = 4 \Phi(\alpha) \frac{1}{4\pi\Delta^2} \pi R^2 \frac{L_\odot}{4\pi r^2}
$$</span></p>
<p>the <strong>geometric albedo</strong> (which we see on wikipedia etc.) is the ratio of the flux densities at phase angle <span class="math-container">$\alpha = 0$</span> reflected by
a planet and a Lambertian surface of the same
cross section.</p>
<p><span class="math-container">$$
p = \frac{F}{F_L}
$$</span></p>
<p>and finally we have:</p>
<p><span class="math-container">$$
F = p\Phi(\alpha) \frac{R^2}{\Delta^2} \frac{L_\odot}{4\pi r^2}
$$</span></p>
<p>which is different of a <span class="math-container">$4$</span> factor from the first equation, that is because the normalisation constant <span class="math-container">$C = 4$</span>.</p>
<hr />
<p>I ran into this problem as I am working on the albedo of the Moon. I saw the last equation, but I do not know why the <span class="math-container">$4$</span> is missing.</p>
<p>But if I calculate using the first equation, I came to the Moon's albedo is <span class="math-container">$\sim 0.45$</span>, about four times the correct data (<span class="math-container">$\sim 0.13$</span>).</p>
<p>I use the apparent magnitude of the full moon (-12.74mag) and the sun (-26.83mag), the solar luminosity (3.828e+26W), the moon's radius (1738km), the moon's distance to earth (384400km) and the sun (1.5e+8km).</p>

CC BY-SA 4.0
2024-08-09T15:13:55.930Z
2024-08-09T15:13:55.930Z
Firestar-Reimu
<p>In a hypothetical solar system there exist:</p>

<ul>
<li>a sun of radius $r_s$ and absolute visual magnitude $V$</li>
<li>the Earth, with radius $r_e$ and distance from the sun of $1\,AU$</li>
<li>another planet, wit radius $r_p$, albedo $a$, distance from Earth $d_e$ (in units of $AU$), and distance from the sun $d_s$ ($AU$)</li>
</ul>

<p>Assume that the second planet is magically lit up entirely by the sun (so it does not have phases when viewed from Earth).</p>

<p>Now how can I compute the apparent visual magnitude of the second planet with respect to Earth?</p>

<p>If there are other parameters required, please tell me.</p>

CC BY-SA 3.0
2014-07-23T02:31:26.940Z
Do you also assume that the second planet is $d$ + 1 $\rm{AU}$ from the star?
Aaron
CC BY-SA 3.0
2014-07-23T14:07:36.673Z
@Aaron no, sorry. I'll edit the question.
feralin
CC BY-SA 3.0
2014-07-24T20:42:04.087Z
http://en.wikipedia.org/wiki/Absolute_magnitude#Solar_System_bodies_.28H.29
Florin Andrei
CC BY-SA 3.0
2014-07-24T21:03:32.433Z
@FlorinAndrei looking at that link, how can I compute the absolute magnitude $H$ of a planet?
feralin
<solar-system><planet>
Compute Planet's Apparent Visual Magnitude
feralin
# Compute Planet's Apparent Visual Magnitude

(Asked by: username_0 on 2014-07-23)

In a hypothetical solar system there exist:

* a sun of radius $r\_s$ and absolute visual magnitude $V$
* the Earth, with radius $r\_e$ and distance from the sun of $1\,AU$
* another planet, wit radius $r\_p$, albedo $a$, distance from Earth $d\_e$ (in units of $AU$), and distance from the sun $d\_s$ ($AU$)

Assume that the second planet is magically lit up entirely by the sun (so it does not have phases when viewed from Earth).

Now how can I compute the apparent visual magnitude of the second planet with respect to Earth?

If there are other parameters required, please tell me.

**Tags:** solar-system, planet
> **Comments:**
> **username_1**: Do you also assume that the second planet is $d$ + 1 $\rm{AU}$ from the star?  
> **username_0**: @Aaron no, sorry. I'll edit the question.  
> **username_2**: http://en.wikipedia.org/wiki/Absolute_magnitude#Solar_System_bodies_.28H.29  
> **username_0**: @FlorinAndrei looking at that link, how can I compute the absolute magnitude $H$ of a planet?  

---

## 3 answers

### Answer by username_3 on 2020-07-28. Score: 7

Planetary magnitudes vary not only according to the Sun’s luminosity, their own average albedo, and their distance from the Earth, but also from:

* Variations in their albedo across their surface.
* Their phase angle, for planets that we sometimes see as a crescent.
* Their inclination, for planets like Saturn and Uranus that have a different albedo at their equator than at their poles.
* Neptune keeps getting brighter. No one knows why.

All of these effects are detailed in an informative paper *Computing Apparent Planetary Magnitudes for The Astronomical Almanac* by Mallama and Hilton, revised in 2018:

<https://arxiv.org/pdf/1808.01973.pdf>

Their source code for computing planetary magnitudes given all of these effects can be found here:

<https://sourceforge.net/projects/planetary-magnitudes/>

> **Comments:**
> **username_4**: Thanks for keeping Skyfield and Stack Exchange up to date!  
> **username_4**: just fyi I've just asked [How fast is Neptune getting brighter? When was was this first noticed and reported?](https://astronomy.stackexchange.com/q/37206/7982)  

---

### Answer by username_5 on 2014-07-29. Score: 7

Lets assume we're dealing with a superior planet, which is to say, the planet is orbiting at a greater distance than the Earth. This effectively ensures that there is no planetary phase to deal with.

Now, the Sun has a luminosity of $L\_{sun}$ such that the Solar flux as seen by the planet is
$$
f\_{planet} = \frac{ L\_{sun} }{ 4 \pi d\_s^2 }.
$$

The cross section of the planet is about $A = \pi r\_p^2$, so for a given albedo $a\_p$ the reflective luminosity of the planet will be
\begin{aligned}
L\_{planet} &= a\_p f\_{planet} A \\
&= a\_p \pi r\_p^2 \frac{ L\_{sun} }{ 4 \pi d\_s^2 }.
\end{aligned}

Given that we know the absolute magnitude of the Sun, the absolute magnitude of the planet follows as
\begin{aligned}
V\_{planet} &= -2.5 \log\_{10}\left[ \frac{ L\_{planet} }{ L\_{sun} } \right] - V\_{sun} \\
&= -2.5 \log\_{10}\left[ a\_p \frac{ r\_p^2 }{ 4 d\_s^2 } \right] - V\_{sun}
\end{aligned}

The apparent magnitude of the planet as seen from Earth can then be calculated as
$$
m\_{planet} = V\_{planet} + 5 \log\_{10}\left[ d\_{e-p} \right] - 5
$$
where the distance between Earth and the planet, $d\_{e-p}$, should be in parsecs and of course depends on the orbital phase of the two respective planets.

So the one additional parameter required was the Solar luminosity. There are probably a bunch of subtle effects with the albedo and the geometry of the system that are not taken into account here, but it should be a fair approximation.

> **Comments:**
> **username_0**: Thanks for answering! One question, though: can solar luminosity not be computed from the sun's absolute visual magnitude (or the other way around) because the total luminosity does not have to all be in visual light? Or is there some other reason?  
> **username_5**: @feralin I guess you can, but you'd still need a second parameter to normalize the conversion, be it either the relative magnitude of the sun or an absolute magnitude of a calibration star. Using the relative magnitude of the sun would make sense here and probably gives a more direct final expression, but I think using luminosity gives a better sense of what's physically going on.  
> **username_5**: @feralin As a second note; if you want to model an inferior planet you can simply make the reflective surface - A - depend on (synodic) orbital phase.  
> **username_0**: sorry, but can you explain what "relative magnitude" is? I'm not getting good results from google...  
> **username_5**: @feralin sorry, I meant to say apparent magnitude  
> **username_6**: "the distance between Earth and the planet, de−p, should be in parsecs" is incorrect. For magnitudes within the solar system, distances must be in AUs. The formula reflects this in fact, since the absolute magnitude of a solar system object is defined at 1 AU viewing distance.  The calculation of Vplanet is not affected, as long as rp and ds are expressed using the same unit of distance.  
> **username_7**: Is this albedo the Bond albedo or the Geometric albedo?  
> **username_8**: @KateBertelsen See my answer below, this answer has a 4 missing.  

---

### Answer by username_8 on 2024-08-09. Score: 0

This answer is from *Fundamental Astronomy 6ed*, Section 7.8

The first several steps are the same with @Brandon Rhodes

We got: ($A$ is the **Bond albedo**)

$$
F = AC\Phi(\alpha) \frac{1}{4\pi\Delta^2} \pi R^2 \frac{L\_\odot}{4\pi r^2}
$$

The difference is this **normalisation constant** $C$:

Since all the radiation reflected from the planet
is found somewhere on the surface of the sphere,
we must have: ($\mathrm{d}S = \Delta^2\mathrm{d}\Omega$)

$$
\frac{C}{4\pi\Delta^2} \int\_S \Phi(\alpha)\,\mathrm{d}S = 1
$$

$$
C = \frac{2}{\int\_0^\pi \Phi(\alpha)\sin\alpha\,\mathrm{d}\alpha}
$$

We denote the **phase integral**:

$$
q = 2\int\_0^\pi \Phi(\alpha)\sin\alpha\,\mathrm{d}\alpha = \frac{4}{C}
$$

and **geometric albedo**:

$$
p = \frac{AC}{4}
$$

For a **Lambertian surface** (defined as $A = 1$):

$$
\Phi(\alpha) = \begin{cases}& \cos\alpha,\quad 0\leq\alpha\leq\pi/2 \\ & 0,\quad \text{otherwise}\end{cases}
$$

which leads to:

$$
C = 4,\quad A = 1,\quad p = q = 1
$$

If we replace the planet with a Lambertian surface
of the same size, we get:

$$
F\_L = 4 \Phi(\alpha) \frac{1}{4\pi\Delta^2} \pi R^2 \frac{L\_\odot}{4\pi r^2}
$$

the **geometric albedo** (which we see on wikipedia etc.) is the ratio of the flux densities at phase angle $\alpha = 0$ reflected by
a planet and a Lambertian surface of the same
cross section.

$$
p = \frac{F}{F\_L}
$$

and finally we have:

$$
F = p\Phi(\alpha) \frac{R^2}{\Delta^2} \frac{L\_\odot}{4\pi r^2}
$$

which is different of a $4$ factor from the first equation, that is because the normalisation constant $C = 4$.

---

I ran into this problem as I am working on the albedo of the Moon. I saw the last equation, but I do not know why the $4$ is missing.

But if I calculate using the first equation, I came to the Moon's albedo is $\sim 0.45$, about four times the correct data ($\sim 0.13$).

I use the apparent magnitude of the full moon (-12.74mag) and the sun (-26.83mag), the solar luminosity (3.828e+26W), the moon's radius (1738km), the moon's distance to earth (384400km) and the sun (1.5e+8km).


---

