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.


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 $\boxed{8.208}$.

  • $\begingroup$ 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 $\endgroup$ – antgiant Feb 5 at 14:12
  • $\begingroup$ @antgiant See en.wikipedia.org/wiki/Bond_albedo#Phase_integral $\endgroup$ – fasterthanlight Feb 5 at 14:51
  • $\begingroup$ That's better, but $q(\alpha)$ = 0.306 with $d_{BO}$ = 1 gets $m = H$ + 1.29 regardless of phase. $\endgroup$ – Mike G Feb 5 at 20:41

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.

  • $\begingroup$ 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. $\endgroup$ – antgiant Feb 5 at 14:09
  • $\begingroup$ my source is 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. $\endgroup$ – James K Feb 5 at 14:28
  • $\begingroup$ 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. $\endgroup$ – Mike G Feb 5 at 18:14

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