From Earth, Jupiter's maximum brightness is -2.94 and Saturn's is -0.24. But what about from Mars? They should be brighter, but by how much?

There are equations at the obvious wiki entry but I'm not sure I understand them. Plus I'm afraid to attempt this calculation myself because I've heard that this sort of thing does not follow the inverse square law. They're reflecting light (not generating it), and I read somewhere that it should follow an inverse 4th power law because of that. I don't see any 4th powers in the apparent magnitude equations.

  • $\begingroup$ It should follow the inverse square law, magnitude is dependent on distance and luminosity so if your source of light is constant (reflected sunlight should be fairly constant) then the only variable is distance. Also there is no reason to treat reflected sunlight differently to actual sunlight, the basic physics is still the same, they're just photons. $\endgroup$
    – Dean
    Jan 29, 2016 at 11:42
  • $\begingroup$ If you believe the people who wrote Stellarium did the calculations accurately, you could try checking there too (or Celestia or any other program that lets you view from different planets). You might also want to start at en.wikipedia.org/wiki/Extraterrestrial_skies and follow links. $\endgroup$
    – user21
    Jan 29, 2016 at 17:34

1 Answer 1


The inverse-fourth-power law you're referring to is valid for light emitted from a source, reflected non-specularly — i.e. in all directions — from a reflector, and detected by the original emitter. If the reflector is a mirror, the observed flux just follows the normal inverse-square law with the nominator equal to $(2d)^2$ instead of $d^2$, since the light has to go back and forth. But if the reflector scatters light in all directions — i.e. into a $2\pi$ hemisphere — then the detected flux is $\sim r^2/d^4$, where $r$ is the radius of the reflector (see this answer for a more throrough explanation).

An example of this is a radar. But in our case, it isn't us that emit the light, it's the Sun. The amount of light reflected from Jupiter and Saturn depends on their distance to the Sun, and that distance doesn't change if you move to Mars. The relevant distances (that I got from NASA's Planetary Fact Sheet) are:

  • Earth semi-major axis $d_\mathrm{E} = 1.00\,\mathrm{AU}$
  • Mars aphelion$^\dagger$ $d_\mathrm{M} = 1.64\,\mathrm{AU}$
  • Jupiter semi-major axis $d_\mathrm{J} = 5.20\,\mathrm{AU}$
  • Saturn semi-major axis $d_\mathrm{S} = 9.58\,\mathrm{AU}$

Now the differences between them:

  • Earth to Mars $d_\mathrm{M-E} = 0.64\,\mathrm{AU}$
  • Earth to Jupiter $d_\mathrm{J-E} = 4.20 \,\mathrm{AU}$
  • Earth to Saturn $d_\mathrm{S-E} = 8.58\,\mathrm{AU}$
  • Mars to Jupiter $d_\mathrm{J-M} = 3.56 \,\mathrm{AU}$
  • Mars to Saturn $d_\mathrm{S-M} = 7.94 \,\mathrm{AU}$

Hence, from Mars the distance to Jupiter is $d_\mathrm{M-J} = 0.85 d_\mathrm{J-E}$, and the received flux is thus $1/0.85^2 = 1.4$ times that on Earth. The change in apparent magnitude is then $$ \Delta m = -2.5 \log\left( \frac{0.85^2}{1}\right) = -0.36, $$ i.e. Jupiter would be $m = -2.94 - 0.36 = -3.30$ as viewed from Mars (assuming the values you provide are correct; I didn't check this).

Following the same approach for Saturn, I get $m = -0.41$.

$^\dagger$The reason I used Mars' aphelion instead of Mars' semi-major axis is because Mars has a rather eccentric orbit ($e\sim0.09$). Jupiter and Saturn are somewhat closer to circular orbits, though ($e\sim0.05$). This of course is still an approximation. If you wish to take into account the eccentricities of all orbits, you also need to know the angle between their semi-major axes. This also doesn't take into account the inclination of the orbit; however, these are only 1º-2º. And of course this minimum distance will not occur every Martian year.

  • $\begingroup$ No offense but that link references the average distance between planets. I'm pretty what we want is the minimum distance between planets. The Jupiter value you got, for example, can't be right because Jupiter's max brightness from Mars must be brighter than its max brightness from Earth, because Mars gets closer to Jupiter than we do. $\endgroup$
    – DrZ214
    Jan 29, 2016 at 12:42
  • $\begingroup$ @DrZ214: Aw man, of course you're right. The formula is right, except I flipped fraction between fluxes; of course $\Delta m$ should be negative. I'll edit. Concerning the distances, the orbits of these planets have rather small eccentricities. You're right that these are average distances. To get the precise minimum distances you need to know exactly how their axes are aligned. $\endgroup$
    – pela
    Jan 29, 2016 at 13:10
  • $\begingroup$ Hmm, well most planets have very little eccentricity. But Mars' orbit is fairly eccentric so I don't wanna negate it. In fact I think the eccentricity of Mars is second only to Mercury. I'm okay with approximating Jupiter and Saturn as circular orbits, though. I'll edit your answer to include a table of the distances, then let you take it from there. $\endgroup$
    – DrZ214
    Jan 29, 2016 at 13:13
  • $\begingroup$ @DrZ214: Okay, I just saw you last comment after my edit. If you're concerned with the eccentricities, the problem becomes a bit more comvoluted, as I wrote in the footnote. $\endgroup$
    – pela
    Jan 29, 2016 at 13:23
  • 1
    $\begingroup$ Oh I see now. 0.85 is the proportional adjustment multiplied to d_J-E, which wasn't written as an inline number. Nice bit of cooperative editing there, and thanks for the math. But be careful in the footnote assumptions. It doesn't matter for this problem, but if orbital incs are up to 2º from ecliptic, then 2 other planets could be 4º inc relative to each other. Also, bigger orbits (going out to Saturn here) will have a bigger z-difference, to say nothing of aphelion or node alignment or lack thereof which you mentioned. So who knows! But I'm satisfied with the approx. here. Thanks again. $\endgroup$
    – DrZ214
    Jan 29, 2016 at 15:10

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