Celestia Forums

krypton wrote:I should have been more specific.
"Full"=+ or - 12 hours of full phase. (Assume a 24 hour day)

Still ambiguous. +-12 hours for the outer moon (spanning 21 degrees) or +- 12 hours for the inner moon (spanning 60 degrees)?

How would one go about calculating the conjunctions of moons? What I''m interested in is finding out how often two moons would appear in conjunction in their full phase. So, assuming the moons have nearly identical orbital planes,how often would the inhabitants of a planet see as shown below? Assum...

Yes the sun moves across the Milky Way, but all the stars move and we never observed this phenomena around nearby stars ... what does that lead scientists to believe that the sun is different ? The heliosphere is hard to observe because it is a tenuous and dim objects. We do not observe the winds o...

The quoted uncertainty of brightness is +- 20%. At the lower end, 62e receives exactly as much heat as Earth. Upper end is 40 % more - Venus is 80 % more. F-s 41% terrestrial heat is close match to Mars. Viewed from f, the maximum elongation of e is slightly smaller than Venus?s. It also is bigger. ...

It is quite irritating how exoplanet.eu and many other sources give the period with 4 significant figures - but then semimajor axis with 1. Given that the mass of B has 3 significant figures but the last is uncertain, I got 2 figures for the axis and it was 0,042 AU. The distance between Bb and Bc a...

How would you estimate the maximum elongation and visual magnitude of Bb from a planet, say Bc, in habitable zone of B? Maximum elongation is the easier part. Do I get the Bb semimajor axis right at 0,042 AU? Then, the habitable zone is 0,7 AU... maximum elongation would be around 3 and a half degre...

Kooky. That'd be much more illuminated than full moon on a clear night on earth, right? So there really would be no true 'night' until the companion was below the horizon. Approximately. Depends on what you are willing to call "true night". 1/500 sunlight is what A gives at 28 AU (near ap...

How do you derive this image? What is the angular size of the gas planet on the horizon? Haven't you seen this picture before? This is from the movie Avatar, gas giant Polyphemus with it's inhabited moon Pandora. Polyphemus orbiting the "Goldilocks zone" around Alpha Centauri A. http://ja...

How do you derive this image?

What is the angular size of the gas planet on the horizon?

To the contrary - this demonstrates the error in your calculations. Seriously? You resuscitate a decade-old thread for this? The alternative would have been post a completely new thread on Alpha Centauri, without reference to the inspiring thread. My "calculation" involved nothing more co...

Well, if we assume an Earth-like atmosphere, then light is going to scatter in exactly the same way, and so blue light will be spread over the whole vault of the sky, but at 1/500th of its intensity in Earth's daytime sky. To be visible during the day on Earth, a star or planet needs to have a magn...

Helium forms small molecules... so you would get a lot of Rayleigh Scattering. The power of Rayleigh Scattering is inversely proportional to the fourth power of the wavelength... Actually, no. Helium has very little polarizability. Which means Rayleigh scattering is very weak. About 70 times weaker...

The illumination of twilight at horizontal surface has been quite difficult to search for, but I did find it. At: http://stjarnhimlen.se/comp/radfaq.html#10 Salient points: Sun overhead (angle +90 degrees above horizon) - 129 klux This is total - direct Sun disc plus blue sky combined. As the Sun fa...