<p>The inverse-fourth-power law you're referring to is valid for light emitted from a source, reflected <em>non-<a href="https://en.wikipedia.org/wiki/Specular_reflection" rel="nofollow noreferrer">specularly</a></em>  — 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 <a href="https://astronomy.stackexchange.com/questions/1443/by-putting-a-mirror-in-space-would-we-be-able-to-see-into-the-past">this answer</a> for a more throrough explanation).</p>

<p>An example of this is a radar. But in our case, it isn't <em>us</em> 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 <a href="http://nssdc.gsfc.nasa.gov/planetary/factsheet/planet_table_ratio.html" rel="nofollow noreferrer">NASA's Planetary Fact Sheet</a>) are:</p>

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

<p>Now the differences between them:</p>

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

<p>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).</p>

<p>Following the same approach for Saturn, I get $m = -0.41$.</p>

<p>$^\dagger$<sub>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.</sub></p>

CC BY-SA 3.0
2016-01-29T12:42:50.127Z
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.
DrZ214
CC BY-SA 3.0
2016-01-29T13:10:30.453Z
@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.
pela
CC BY-SA 3.0
2016-01-29T13:13:15.070Z
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.
DrZ214
CC BY-SA 3.0
2016-01-29T13:23:54.547Z
@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.
pela
CC BY-SA 3.0
2016-01-29T13:37:37.253Z
No problem, I just made the edit but it's pending a peer review. I did not change your math because I saw that 0.88^2 was about to change to 3.56^2 and that was scary! But when you think about it, the distance from Mars to Jupiter should not be 0.88 AU... That is too small, so I think I'm on the right track.
DrZ214
CC BY-SA 3.0
2016-01-29T13:54:46.863Z
@DrZ214: No no, it's not 0.88 AU, it's 0.88 times the distance from Earth to Jupiter. As for your edit, I'm okay with it, I'll just edit it a little…
pela
CC BY-SA 3.0
2016-01-29T15:10:24.473Z
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.
DrZ214
CC BY-SA 3.0
2016-01-29T15:42:53.943Z
@DrZ214: You're welcome, and yes, you're right that the _relative_ inclination could be larger. But although Saturns inclination results in a larger absolute distance from the ecliptic, the relevant number is the difference in Mars' and Earth's inclination, which is 1º51'.
pela
CC BY-SA 3.0
2016-01-29T12:21:08.030Z
2016-02-08T11:10:01.627Z
2017-04-13T12:59:06.910Z
pela
<p>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?</p>

<p>There are equations at <a href="https://en.wikipedia.org/wiki/Apparent_magnitude">the obvious wiki entry</a> 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 <strong>not</strong> 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.</p>

CC BY-SA 3.0
2016-01-29T11:42:13.457Z
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.
Dean
<solar-system><mars><apparent-magnitude>
As viewed from Mars, what are Jupiter's and Saturn's maximum brightness in apparent magnitude?
DrZ214
# As viewed from Mars, what are Jupiter's and Saturn's maximum brightness in apparent magnitude?

(Asked by: username_0 on 2016-01-29)

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](https://en.wikipedia.org/wiki/Apparent_magnitude) 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.

**Tags:** solar-system, mars, apparent-magnitude
> **Comments:**
> **username_1**: 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.  

---

## 1 answers

### Answer by username_2 on 2016-01-29. Score: 9 | Accepted answer

The inverse-fourth-power law you're referring to is valid for light emitted from a source, reflected *non-[specularly](https://en.wikipedia.org/wiki/Specular_reflection)* — 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](https://astronomy.stackexchange.com/questions/1443/by-putting-a-mirror-in-space-would-we-be-able-to-see-into-the-past) 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](http://nssdc.gsfc.nasa.gov/planetary/factsheet/planet_table_ratio.html)) 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.

> **Comments:**
> **username_0**: 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.  
> **username_2**: @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.  
> **username_0**: 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.  
> **username_2**: @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.  
> **username_0**: No problem, I just made the edit but it's pending a peer review. I did not change your math because I saw that 0.88^2 was about to change to 3.56^2 and that was scary! But when you think about it, the distance from Mars to Jupiter should not be 0.88 AU... That is too small, so I think I'm on the right track.  
> **username_2**: @DrZ214: No no, it's not 0.88 AU, it's 0.88 times the distance from Earth to Jupiter. As for your edit, I'm okay with it, I'll just edit it a little…  
> **username_0**: 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.  
> **username_2**: @DrZ214: You're welcome, and yes, you're right that the _relative_ inclination could be larger. But although Saturns inclination results in a larger absolute distance from the ecliptic, the relevant number is the difference in Mars' and Earth's inclination, which is 1º51'.  

---

