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:

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