How to Calculate Apparent Magnitude/Brightness of Planets from Within the same Solar System

I'd like to be able to determine the appearance of the night sky from an Earth-like planet, in a solar system with multiple other planets, moons, and other bodies. Here on Earth, we can directly measure the apparent magnitude of a body simply through observation, but that's obviously impossible for fictional planets.

I understand that the luminosity of the primary star, the albedo and radius of the object being observed, and the range of distances between the observer and the object (which will vary, because orbits) are the primary factors in determining apparent magnitude -- it's the precise mathematical relationships between these factors which is eluding me.

For illustrative purposes, here is an excerpt from the setup of my solar system, with each planet's semi-major axis in AU, bond albedo, and radius in Earth radii listed. I'm including one planet closer to the star than our observer, one planet further out-system but relatively close, and one rather distant planet.

Sun's Luminosity: 2.248 (relative to Sol's 1)

• Planet B: Semi-Major Axis: 0.47 AU, Bond Albedo: 0.93, Radius 4.29: Earth radii
• Planet E (observer's location): SMA: 2.178
• Planet F: SMA: 3.87, BA: 0.21, R: 0.89

If it helps/matters: Planet B is an in-system ice giant, a.k.a. "Hot Neptune," Planet E is a slightly larger Earthlike planet with a comparable atmosphere, and Planet F is a magnesium-silicate terrestrial.

I'd like to be able to figure out what the apparent magnitude of Planets B and F when observed from Planet E, and how that result was reached so that I can replicate the process for the other planets and bodies within the system. I will also happily absorb any and all tangents on planetary apparent magnitude in general. Thank you in advance!

• Can we just assume that the atmosphere's transparent on Planet E and that there are no superbright novas nearby? Mar 30, 2020 at 20:05
• @BLT-Bub Yes, the night sky will appear much as it does on Earth. Assume clear sky weather and a new moon--I can work out sources of light pollution later. Mar 30, 2020 at 20:08
• You only really need one inferior (closer to the star than planet E) and one superior planet (farther away from the star than planet E). Apr 5, 2020 at 22:56
• If so, you may also want to calculate the visibility cycles of the hypothetical planets. Apr 6, 2020 at 20:25
• @A.E.Stephenson, I lost some of my progress writing this answer because Stack Exchange's servers messed up and my last backup didn't contain everything. :( Apr 27, 2020 at 23:13

So i have copied/pasted the answer below. This is not my answer, i am copying/pasting it as it seems an appropriate answer. All credit goes to Milan Minic.

Calculations of the apparent magnitudes m start with calculations of absolute magnitudes H of celestial objects and of their phase integrals q.

Absolute magnitude H gives us the apparent magnitude of the object when it is observed from the Sun and put at some standard distance from the Sun - that is 1 astronomical unit for our Solar system. For a spherical object of diameter D (in kilometers) and albedo p, H is calculated as

H = $$5 \log_{10} {{1329} \over {D \sqrt{p}}}$$

Phase integral q(α) tells us how brightness of an object varies when observed from various angles, α is the angle between the Sun and observer, as seen from the object. Thus 0° means that the object is in the opposition with the Sun (i.e. the observer is exactly between the Sun and the object, like in the case of the full Moon), and 180° means that the object is in the conjunction with the Sun (i.e. the object is exactly between the Sun and the observer, like in the case of the new Moon). If the object is a diffuse reflecting sphere, the phase integral q can be analitically expressed, but for the real celestial objects astronomers have developed empirical formulae for each one of them, to account for the peculiarities of the light reflection from them. For instance, the Moon shows the irregular brightness increase for α = 0° (the so called opposition surge) owing to the forward reflecting properties of the lunar regolith. Or, in the case of Venus,

we can see significant differences from the diffuse reflecting sphere (similar to the Mercury curve, blue line) due to light passing through its atmosphere and peculiar light transmission enhancement at α = 168° because of sulphuric acid droplets.

Now, when we know H, α and q(α), we can calculate the apparent magnitude m as

m= $$H + 5 \log_{10} {{d_{BS} d_{BO}} \over {d^2_{OS}}} − 2.5\log_{10} q(α)$$

where $$d_{BS}$$, $$d_{BO}$$, $$d_{OS}$$ and α are connected by the cosine theorem:

$$d^2_{OS} = d^2_{BS} + d^2_{BO} − 2d_{BS} d_{BO} \cos(α)$$

Note: The log function in this answer always refers to the logarithm of 10.

There are several steps we need to follow to find the minimum and maximum apparent magnitude of a celestial body as seen from another planet, all of which are shown here.

The following formula gives the apparent magnitude of a celestial body: $$m=M-5\cdot\log(d)+5$$

m = apparent magnitude
M = absolute magnitude
d = distance between the observer and the object in parsecs


We can use the following formula to find absolute magnitude: $$M=m-2.5\cdot\log(L)$$

m = apparent magnitude of the Sun as seen from Earth (approximately -26.74)
L = luminosity of the star relative to the Sun


Now, we can calculate the absolute magnitude of your star: $$M=-26.74-2.5\cdot\log(2.248)\approx-27.98$$

We must now convert the distance of the semi-major axis from AU to parsecs using the factor of 1 parsec ≈ 206,265 AU. We also need to determine the minimum and maximum distances between the planets to determine the minimum and maximum apparent magnitudes.

For Planet B and Planet E:

• Minimum distance: |2.178 - 0.47| AU = 1.708 AU
• Maximum distance: 2.178 + 0.47 AU = 2.648 AU

For Planet F and Planet E:

• Minimum distance: |3.87 - 2.178| AU = 1.692 AU
• Maximum distance: 3.87 + 2.178 AU = 6.048 AU

We can now convert these distances to parsecs.

For Planet B and Planet E:

• Minimum distance: 1.708 AU / 206,265 ≈ 8.28 x 10^(-6) parsecs
• Maximum distance: 2.648 AU / 206,265 ≈ 1.28 x 10^(-5) parsecs

For Planet F and Planet E:

• Minimum distance: 1.692 AU / 206,265 ≈ 8.20 x 10^(-6) parsecs
• Maximum distance: 6.048 AU / 206,265 ≈ 2.93 x 10^(-5) parsecs

Now, we need to find the apparent magnitude of the planets.

We first need to find their absolute magnitudes with this formula: $$m=M+2.58\cdot\log\frac{r^{2}*a}{d^{2}}$$

m = mass of the planet
M = mass of the star
a = bond albedo
d = distance between the planet and the star in AU


For Planet B: $$m=-27.98+2.58\cdot\log\frac{4.29^{2}\cdot0.93}{0.47^{2}}\approx-20.61$$

For Planet F: $$m=-27.98+2.58\cdot\log\frac{0.89^{2}*0.21}{3.87^{2}}\approx-23.34$$

We can now find the apparent magnitude of the planets when seen from Planet E.

Minimum apparent magnitude of Planet B: $$-20.61-5\cdot\log(8.28\cdot10^{-6})+5\approx-6.25$$

Maximum apparent magnitude of Planet B: $$-20.61-5\cdot\log(1.28\cdot10^{-5})+5\approx-5.76$$

Minimum apparent magnitude of Planet F: $$-23.34-5\cdot\log(8.20\cdot10^{-6})+5\approx-8.98$$

Maximum apparent magnitude of Planet F: $$-23.34-5\cdot\log(2.93\cdot10^{-5})+5\approx-7.66$$

It is also important to determine the appearance of the planets when seen from Planet E. Planet B, as an ice giant, would likely appear blue or blue-green as a result of methane in its atmosphere; methane absorbs red light and reflects blue and green light. Planet F, as a magnesium-silicate terrestrial, would likely appear gray or brownish-gray because of magnesium silicates. The exact color of the planets, when seen from Planet E, would depend on the exact composition of the planet's surface and its atmosphere.