<p>The absolute magnitude of the Earth is about <span class="math-container">$H = -3.99$</span>. This means, from <span class="math-container">$1$</span> AU away from the Sun and the Earth, the Earth has an apparent magnitude of <span class="math-container">$-3.99$</span>.</p>
<p>Voyager 1 is <span class="math-container">$152.26$</span> AU from the Earth, and <span class="math-container">$151.85$</span> AU from the Sun. Using the formula for apparent magnitude, <span class="math-container">$${\displaystyle m=H+5\log _{10}{\left({\frac {d_{BS}d_{BO}}{d_{0}^{2}}}\right)}-2.5\log _{10}{q(\alpha )}}$$</span></p>
<p>With <span class="math-container">$d_{BO} = 152.26, d_{BS} = 1, d_0=1,$</span> AU, and <span class="math-container">$q(\alpha) = 0.306$</span>, we obtain that Earth's apparent magnitude from Voyager 1 would be about <span class="math-container">$\boxed{8.208}$</span>.</p>

CC BY-SA 4.0
2021-02-05T14:12:47.003Z
Also I am trying to figure out how you calculated $q(\alpha)$. I can't seem to get your number from either the formula that @james-k posted or from the [function on Wikipedia](https://en.wikipedia.org/wiki/Absolute_magnitude#More_advanced_models)
antgiant
CC BY-SA 4.0
2021-02-05T14:51:08.227Z
@antgiant See https://en.wikipedia.org/wiki/Bond_albedo#Phase_integral
WarpPrime
CC BY-SA 4.0
2021-02-05T20:41:37.660Z
That's better, but $q(\alpha)$ = 0.306 with $d_{BO}$ = 1 gets $m = H$ + 1.29 regardless of phase.
Mike G
CC BY-SA 4.0
2021-02-04T21:56:34.127Z
2021-04-08T13:24:50.573Z
2021-04-08T13:24:50.573Z
WarpPrime
<p>The Earth has an absolute magnitude of -4  This means that from 1 AU and at zero phase (&quot;full Earth&quot;) it would have a magnitude of -4.  Of course the only place that satisfies those two requirements is the location of the sun, but that is not the point;</p>
<p>The apparent magnitude depends on the distance from Earth, and the phase angle of the Earth.  This is the angle &quot;sun-Earth-observer&quot;  A zero phase angle is &quot;full Earth&quot; and 180-degree phase angle is &quot;new Earth&quot; (with the unilluminated side of the Earth facing the observer</p>
<p>The apparent magnitude can be estimated as</p>
<p><span class="math-container">$$m = -4 + 5 \log(D) -2.5 \log\left(\frac{1+\cos\beta}{2}\right)$$</span></p>
<p>(from <a href="https://in-the-sky.org/article.php?term=absolute_magnitude" rel="nofollow noreferrer">https://in-the-sky.org/article.php?term=absolute_magnitude</a>)</p>
<p>D is the distance from the Earth to the observer and <span class="math-container">$\beta$</span> is the phase angle. Even for an observer fixed in space, the phase angle will vary as the Earth orbits the sun.</p>
<p>This is an approximation because the Earth is particularly shiny (it has oceans) which means that there is a significant specular reflection from the oceans.  It is also not a uniform colour, so the albedo varies and hence the absolute magnitude varies slightly.</p>

CC BY-SA 4.0
2021-02-05T14:09:27.173Z
Thank you. I am confused however as to why your formula has $log(D)$ for distance and @fasterthanlight has essentially $log(D^2)$. That change creates a pretty large difference between your calculated magnitudes.
antgiant
CC BY-SA 4.0
2021-02-05T14:28:45.143Z
my source is https://in-the-sky.org/article.php?term=absolute_magnitude. The loss of D^2 is because the square has been moved in front of the log to change 2.5 into 5. Magnitude is essentially -2.5 log(Flux) but if flux is proportionanl to 1/D^2 then you get m = -2.5 log(1/D^2) = +5 log(D) (+constant). I think this answer is correct.
James K
CC BY-SA 4.0
2021-02-05T18:14:12.060Z
If I change the last term's coefficient from 2.5 to 5, I get a better fit with JPL HORIZONS estimates of Earth's apparent magnitude as seen from Neptune.
Mike G
CC BY-SA 4.0
2021-02-04T22:13:01.253Z
2021-02-05T14:23:54.230Z
2021-02-05T14:23:54.230Z
James K
<p>I would like to calculate the apparent magnitude of Earth from various distances, say from the Voyager probe at this moment or from various stars in the sky. I know there must be a formula that I can enter in the distance from Earth (or the Sun) and get out the apparent magnitude. However, I am failing to find it as everything I can find assumes that I am on Earth and want to calculate the brightness of other things.</p>

<earth><apparent-magnitude><magnitude>
Formula to calculate the apparent magnitude of Earth from arbitrary distances
antgiant
# Formula to calculate the apparent magnitude of Earth from arbitrary distances

(Asked by: username_0 on 2021-02-04)

I would like to calculate the apparent magnitude of Earth from various distances, say from the Voyager probe at this moment or from various stars in the sky. I know there must be a formula that I can enter in the distance from Earth (or the Sun) and get out the apparent magnitude. However, I am failing to find it as everything I can find assumes that I am on Earth and want to calculate the brightness of other things.

**Tags:** earth, apparent-magnitude, magnitude

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## 2 answers

### Answer by username_1 on 2021-02-04. Score: 4

The absolute magnitude of the Earth is about $H = -3.99$. This means, from $1$ AU away from the Sun and the Earth, the Earth has an apparent magnitude of $-3.99$.

Voyager 1 is $152.26$ AU from the Earth, and $151.85$ AU from the Sun. Using the formula for apparent magnitude, $${\displaystyle m=H+5\log \_{10}{\left({\frac {d\_{BS}d\_{BO}}{d\_{0}^{2}}}\right)}-2.5\log \_{10}{q(\alpha )}}$$

With $d\_{BO} = 152.26, d\_{BS} = 1, d\_0=1,$ AU, and $q(\alpha) = 0.306$, we obtain that Earth's apparent magnitude from Voyager 1 would be about $\boxed{8.208}$.

> **Comments:**
> **username_0**: Also I am trying to figure out how you calculated $q(\alpha)$. I can't seem to get your number from either the formula that @james-k posted or from the [function on Wikipedia](https://en.wikipedia.org/wiki/Absolute_magnitude#More_advanced_models)  
> **username_1**: @antgiant See https://en.wikipedia.org/wiki/Bond_albedo#Phase_integral  
> **username_2**: That's better, but $q(\alpha)$ = 0.306 with $d_{BO}$ = 1 gets $m = H$ + 1.29 regardless of phase.  

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### Answer by username_3 on 2021-02-04. Score: 3

The Earth has an absolute magnitude of -4 This means that from 1 AU and at zero phase ("full Earth") it would have a magnitude of -4. Of course the only place that satisfies those two requirements is the location of the sun, but that is not the point;

The apparent magnitude depends on the distance from Earth, and the phase angle of the Earth. This is the angle "sun-Earth-observer" A zero phase angle is "full Earth" and 180-degree phase angle is "new Earth" (with the unilluminated side of the Earth facing the observer

The apparent magnitude can be estimated as

$$m = -4 + 5 \log(D) -2.5 \log\left(\frac{1+\cos\beta}{2}\right)$$

(from <https://in-the-sky.org/article.php?term=absolute_magnitude>)

D is the distance from the Earth to the observer and $\beta$ is the phase angle. Even for an observer fixed in space, the phase angle will vary as the Earth orbits the sun.

This is an approximation because the Earth is particularly shiny (it has oceans) which means that there is a significant specular reflection from the oceans. It is also not a uniform colour, so the albedo varies and hence the absolute magnitude varies slightly.

> **Comments:**
> **username_0**: Thank you. I am confused however as to why your formula has $log(D)$ for distance and @fasterthanlight has essentially $log(D^2)$. That change creates a pretty large difference between your calculated magnitudes.  
> **username_3**: my source is https://in-the-sky.org/article.php?term=absolute_magnitude. The loss of D^2 is because the square has been moved in front of the log to change 2.5 into 5. Magnitude is essentially -2.5 log(Flux) but if flux is proportionanl to 1/D^2 then you get m = -2.5 log(1/D^2) = +5 log(D) (+constant). I think this answer is correct.  
> **username_2**: If I change the last term's coefficient from 2.5 to 5, I get a better fit with JPL HORIZONS estimates of Earth's apparent magnitude as seen from Neptune.  

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