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At their closest flyby, Earth is ‘new’ and dim from Mars, and Mars is ‘full’ from Earth and brightest as seen from Earth.

When Earth is 39 degrees from the Sun then Earth is brightest as seen from Mars, then at a ‘crescent’, and Mars ‘gibbous’ to Earth.

But which of the two is visually brightest?

Which is brighter, Mars as seen from Earth, or Earth as seen from from Mars, at their respective peak magnitudes?

https://upload.wikimedia.org/wikipedia/commons/1/11/Phases_Venus.jpg Venus is brightest as a crescent

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    $\begingroup$ None of the answers take into account the additional affect of the relative attenuations due to atmosphere. And if you keep going you might need to consider the difference in light pollution. $\endgroup$
    – Dennis Williamson
    May 3 at 15:36
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    $\begingroup$ Upvoting your question because you got two decently upvoted answer which provided opposite conclusions. It will interesting to find why Stellarium and the formula answers differ so much. $\endgroup$
    – Seth Robertson
    May 4 at 16:02
  • $\begingroup$ @Mark Besser, for comparison purposes, are the observers on the planets' ground (perhaps in an optimal locale like Mauna Kea), or should we ignore atmosphere ? I'd expect Earth's atmosphere to dim viewing more than Mars. $\endgroup$
    – chux - Reinstate Monica
    May 5 at 17:02

4 Answers 4

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We can use the expression that is commonly used to estimate the apparent magnitude of a planet or asteroid in the Solar System:

$$\boxed{m=5 \log \frac{1329}{d \cdot \sqrt p}+5 \log (D_s \cdot D)-2.5\log f(F)}$$

Where:

m is the apparent magnitude

log is the decimal logarithm

d is the diameter in km

p is the albedo

Ds is the distance to the Sun in Astronomical Units (AU)

D is the distance Earth-Planet in Astronomical Units (AU)

F is the phase angle

f(F) is the phase function

As a phase function we will use the usual one:

$$f(F)=\dfrac{1+\cos F}2$$

1. Magnitude of Mars. Opposition (maximum)

$d=2\cdot 3389.5$ km

$p=0.15$

$D_s = 1.381497$ AU. Mars Perihelion

$D=1.381497 - 1.016714$ AU. Mars Perihelion and Earth Aphelion

$F=0$ Opposition

$$m \simeq -2.97$$

2. Magnitude of the Earth in quadrature = greatest elongation as seen from Mars

$d=2\cdot 6371$ km

$p=0.367$

$D_s = 1.016714$ AU. Earth Aphelion

$D=\sqrt{1.381497^2 - 1.016714^2 \ }$ AU. Mars Perihelion and Earth Aphelion

$F=\dfrac{\pi}2$ Earth at greatest elongation as seen from Mars

$$m \simeq -3.21$$

Probably the greatest brightness of the Earth as seen from Mars will not be when the Earth is exactly in quadrature, just as the greatest brightness of Venus as seen from the Earth is not exactly at quadrature but when it is somewhat closer to the Earth.

But if already with the Earth in quadrature we obtain $m \simeq -3.21$ that is a greater brightness than $m \simeq -2.97$ we can claim (according to this estimation), that the maximum brightness of the Earth as seen from Mars is slightly higher than the maximum brightness of Mars as seen from the Earth, although the difference between the both is not large.

A similar thread: Formula to calculate the apparent magnitude of Earth from arbitrary distances

Best regards.

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    $\begingroup$ I think we also should account for the fact that Earth is bigger than Mars. If someone has already said that, I guess I just missed it. $\endgroup$
    – Deko Revinio
    May 3 at 16:22
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    $\begingroup$ @Deko, note that the mathematical expression for the apparent magnitude "m" already takes into account the diameter of the body "d". d=2x3389.5 km for Mars and d=2x6371 km for the Earth. $\endgroup$
    – Albert
    May 3 at 16:29
  • $\begingroup$ ah, must have overlooked that $\endgroup$
    – Deko Revinio
    May 3 at 16:33
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    $\begingroup$ +1 there's more about the "unusual" phase function (as opposed to the "usual one") in this answer to Calculating the apparent magnitude of a satellite $\endgroup$
    – uhoh
    May 4 at 0:24
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At it's brightest, Earth is a rather impressive magnitude -2.5 when viewed from Mars1 (the maximum brightness depends on how favourable the elongation is, but it isn't usually brighter than -2)

At the same time, Mars is a mere -0.7 making Earth about five times brighter than Mars.

But at a favourable opposition, Mars can reach -2.7 or even slightly brighter. So Mars at its brightest from Earth is slightly brighter than Earth at its brightest from Mars.

1These magnitudes are as simulated by Stellarium.

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    $\begingroup$ These magnitudes are as simulated by Stellarium. $\endgroup$
    – James K
    May 3 at 5:16
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    $\begingroup$ It was there all along and I didn't see it? I'm currently taking a muscle relaxant for a strained muscle and so I'm probably seeing things as if I were on Mars. Sorry about that! $\endgroup$
    – uhoh
    May 3 at 8:36
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Without much math, the answer is relatively simple when we look at two facts:

Earth is bigger than Mars. Size of the cross-section (or rather illuminated area) matters for brightness.

The path of the light from the Sun via Earth to Mars is always shorter than the path of the light from Mars to the Earth. Thus the inverse square law for the brightness of a source will dictate that less light arrives on Earth from Mars than vice versa.

A small factor is also the Albedo (thus ability to reflect light): Earth has twice the Albedo than Mars, so everything else equal, it would appear twice as bright (thanks Pm2Ring for the link).

All this taken together, Earth will appear brighter on Mars than Mars appears on Earth.

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    $\begingroup$ The albedo of Earth is roughly double that of Mars. astronomy.swin.edu.au/cosmos/a/Albedo $\endgroup$
    – PM 2Ring
    May 2 at 18:43
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    $\begingroup$ "The path of the light from the Sun via Earth to Mars is always shorter than the path of the light from Mars to the Earth." You seem to have missed the fact that we are looking at two different configurations, one where earth is brightest as seen from Mars, and one where Mars is brightest as seen from earth (they do not happen at the same time) . The Sun-Mars-Earth path when all 3 are in a straight line is not necessarily longer than the Sun-Earth-Mars path when they are in an out-of-phase configuration. We need at least a little math to determine which is longer. $\endgroup$
    – Nuclear Hoagie
    May 2 at 19:02
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    $\begingroup$ You've also ignored the difference in illuminated area. Mars has 1/4 the cross-section of earth, but is fully illuminated when it appears brightest. Earth is much bigger, but is a crescent when it appears brightest, so only about 1/4 of the visible surface is illuminated. With 4x the area but only 1/4 of it illuminated, it's pretty much a wash. $\endgroup$
    – Nuclear Hoagie
    May 2 at 19:11
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    $\begingroup$ @uhoh for an estimate without calculation, a linear influence is small when comparing the impact with respect to the distance with a power of two. Both planets obviously are not pieces of coal either. James simulation with stellarium, and nucleas's comments, shows that these order of magnitude (no pun intended), can still go somewhat wrong. This I shall edit it accordingly this evening in a few hours $\endgroup$
    – planetmaker
    May 3 at 7:22
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    $\begingroup$ @NuclearHoagie: A better way of expressing that would be to observe that when Mars and Earth are closest to each other in their orbits, the illuminated side of Mars would face Earth, but the illuminated side of Earth would face away from Mars. $\endgroup$
    – supercat
    May 3 at 20:21
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Mars at its brightest as seen from Earth can be brighter than Earth at its brightest as seen from Mars. But it's complicated because of the relatively large eccentricity of the orbit of Mars, so the relevant distances vary in a somewhat irregular pattern. Also, there's an extra complication due to the colours of the two planets. People don't necessarily agree on the relative brightnesses of red vs blue lights of the same power.

JPL Horizons can compute the apparent magnitude of Solar System bodies.

  1. Visual magnitude & surface brightness

Approximate airless visual magnitude & surface brightness, where surface brightness is the average airless visual magnitude of a square-arcsecond of the illuminated portion of the apparent disk.

Here are approximate magnitude plots covering a little over one synodic period of Mars (~780 days), with a 7 day time step. The calculations are done for an observer at the body centre and ignore atmospheric refraction and extinction. (Remember, higher magnitudes are dimmer). Over this time span, Mars is the winner, with a magnitude -2.66 at its brightest, whereas Earth is only -2.44 at its brightest. However, most of the time, Earth is significantly brighter than Mars.

Mars mag, full

Earth mag, full

The following plots zoom into the time spans when each planet is at its brightest, with a 1 day time step.

Mars mag, zoom

Earth mag, zoom

Here is my magnitude plotting script if you want to explore other time spans (or the magnitudes of other Solar System bodies). Please see the Horizons manual for info on specifying bodies and time spans.

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