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I have two planets that share an orbit around a pair of stars at a distance of 2.05 Earth-AU from the Barycenter of those two stars. (I can provide more details about the two stars, if needed)

Using this site, I can roughly estimate that the bond albedo for either planet is 65%. (For reference: Venus, brightest "star" in the sky, has an albedo of 75%, while the Moon has an albedo of 12%). Given that one of the planets is at the L5 (lagrange point 5) of the other, how bright would these planets appear in each others' skies?

Additional question: One of these planets has rings mostly made up of silicate rock. Would that change how visible it is to the other planet? Would these rings be visible from the other planet's surface?

I understand that any answers given would be guesswork, but I'm hoping for some ideas that will help add to the cultures I'm building.

Edit: Additional information

  • Primary star has a mass of 1.3 sols and a luminosity of 2.197 sols
  • Secondary star has a mass of 0.75 sols and a luminosity of 0.422 sols
  • Primary planet has physical characteristics identical to Earth (aside from orbit of 2.05 au), except for the silicate rings
  • Secondary planet has a radius of 5735 km (90% of Earth's) and a mass of 4.776 x 10^24 kg (80% of Earth's)

While I'm very grateful for all the advice about the orbit stability, I am much more concerned about the apparent brightness of the two planets to each other. Thank you so much to those who are taking the time to try to this out!

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  • $\begingroup$ Somehow I don't expect your orbits to be fully stable. Someone with better math than me can do the computations, but I expect the orbits would look more like an interleaved pair of Lorenz attractor functions sharing common epicenters. The barycenter is a useful point for estimating, but over the timescales of the lifespan of a solar system, it may not be the most useful point. $\endgroup$
    – pojo-guy
    Jun 13, 2018 at 19:39
  • $\begingroup$ @pojo-guy, I know that multiple planets (especially similar in size) on the same orbit are unlikely, but most references seem to suggest that if it does happen, Legrange points 4 and 5 are the most stable spots to place the second planet. Mostly, though, I'm hoping someone can give me an idea of how bright the planets will look to each other (ie, twice as bright as Venus, or half as bright as the moon, etc) $\endgroup$
    – z2a
    Jun 13, 2018 at 20:16
  • $\begingroup$ I cannot imagine any way for even a pseudo-stable orbit to be formed from these four bodies. Sharing an orbit is a non-starter IMO. But to give you an answer people will need to know the luminosities of the stars and the distances (orbital parameters) of all the bodies. $\endgroup$ Jun 13, 2018 at 20:54
  • $\begingroup$ You didn't give the brightness of stars and plant's diameter, so it's not possible to do the exact calculation. My rough guess is that it will be fainter than Venus, but brighter than Sirius. $\endgroup$
    – Alexander
    Jun 13, 2018 at 21:33
  • $\begingroup$ @pojo-guy Orbits at L4 and L5 points ("Trojan" orbits) are indeed only stable if the mass at the Lagrange point is much smaller - on the order of 1:100. However, they could co-orbit, like Pluto and Charon. (The question then becomes "why aren't they the same planet" which is rather complex.) $\endgroup$
    – Cadence
    Jun 13, 2018 at 22:57

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I'll give this one a shot. I'm not doing anything with orbital mechanics or stability; that's way beyond me.

Right, first off: the rings. Yes, they'd affect the brightness of the planet, depending on their composition. In general, they'd make it brighter. I found a paper on Saturn that, based on a cursory reading, said that. It needs to be rather fine particles, but if your planet has life, that's kind of a given, lest you have a lot of impacts.

Because of the rings, your planetary bond albedo will be a little higher. As a completely random guess, let's say $5\%$ higher at $70\%$. Totally random. That means that $70\%$ of incoming light will be reflected back out.

Next, need to find out what percent of light hits the planet. I'll assume an Earth-sized planet to make it easier on myself. Earth receives $1.75\times10^{17}W$ of energy from the sun, at about $1370W/m^2$. With the inverse square law, a planet at $2AU$ receives $\frac{1}{4}$ of this, or $4.375\times10^{16}W$. However, due to the two suns, I'll double this. That's probably not how it works in real life, but I've not an astrophysicist. So, the planet gets $8.75\times10^{16}W$ of energy.

With a $70\%$ albedo, this means that reflected light is $6.125\times10^{16}W$ from the planet. Now, here is where I simplify the question a lot. Of course, only half of the planet (probably) gets the light but I'm ignoring this to make this calculable for myself. By definition, the L5 point means that the second planet is as far from the first as they both are from the central suns. Again, simplifying the suns into a single body (probably wrong). This means that that the energy reaching the second planet from the first is $\frac{6.125\times10^{16}}{4\pi(2AU)^2}$. This is very small.

Now, brightness from Earth is called the apparent magnitude. This, with the values from above, is...(this is about an hour later than when I wrote the last sentence) is about $-0.79$. This is about as bright as the star Canopus, or a bit more than half as bright as Sirius. You could see it around dawn/dusk and at night. Brighter than Arcturus but less than a new moon.

I made a lot of assumptions but I'm fairly sure this is correct. Considering the $2au$ distance, this will be a lot dimmer than Venus or Mars. Please upvote; I just spent an about 70 minutes trawling through papers, university lectures, and other resources to get this. Hope it helps.

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  • $\begingroup$ I'll be honest, I had to look up some stuff to fully understand this answer, but this is exactly what I was looking for! Thank you for including how the planetary rings affect things, but thank you especially for explaining in laymen's terms how bright it would look! $\endgroup$
    – z2a
    Jun 14, 2018 at 14:15
  • $\begingroup$ You're welcome! It's probably not exactly right but it'd be quite close. You'd definitely see it at night $\endgroup$
    – Serenical
    Jun 14, 2018 at 23:52

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