Celestia Forums

I'm trying to make a vaguely realistic Spica system, and I'm kinda stuck with figuring out the barycentres (see this thread for details).

But is it even possible to 'nest' barycentres in Celestia yet? ie to have one barycentre orbiting another, which in turn is orbiting another? If so, how would you do this? I need to have all stars being light emitters too, so I don't want to use the 'orbit an invisible object' fudge method.

Check out the definition of Alula Australis in nearstars.stc.

Ah, thanks. That's what I was after.

How does one calculate the orbital periods of the bodies around the barycentres though? It looks like you use the same orbital period for both bodies - how is that calculated?

EDIT: does this sound right to calculate the orbital period?

P in years = SQRT[(separation in AU^3)/(sum of masses in Sols)]

Malenfant wrote:EDIT: does this sound right to calculate the orbital period?

P in years = SQRT[(separation in AU^3)/(sum of masses in Sols)]

I think it should actually be:

P in years = sqrt(separation in AU^3)/(sum of masses in Sols)

--Chris

chris wrote:

Malenfant wrote:EDIT: does this sound right to calculate the orbital period?

P in years = SQRT[(separation in AU^3)/(sum of masses in Sols)]

I think it should actually be:

P in years = sqrt(separation in AU^3)/(sum of masses in Sols)

--Chris

Hm, well http://wwwhip.obspm.fr/~arenou/binary/binary.en.html says that:

Due to the Kepler's third law, the axis, orbital period and masses are related through (a1+a2)^3 = P^2(M1+M2). In this equation, the units are easy to remember, when considering the motion of the sun (1) and earth (2): a1 is negligible, so a1+a2 is about 1 AU, P is one year, and M1 is 1 solar mass, with M2 negligible compared to M1. This gives 1^3=1^2*1.

If you rearrange that, you should get:

((a1+a2)^3)/(M1+M2) = P^2

so P = SQRT[((a1+a2)^3)/(M1+M2)], which is what I said.


Taking Alpha Centauri as an example, from Fridger's binary star file:

a1=10.685
a2=12.873
m1=1.12
m2=0.93

a1+a2 = 23.558
m1+m2 = 2.05

so (a1+a2)^3 = 13074.20386

Divide that by 2.05 = 6377.66042

Square root that = 79.86026 years, which is close to what Fridger has in his data file. (79.85 years).


If however you take the square root of (a1+a2)^3 and then divide that by 2.05 (which looks like what you were suggesting), you get 114.34248/2.05 = 55.77682 years, which is very different.

So I suspect I'm right there.

OK, I think this might actually be working... took a fair bit of calculating to figure this out. The trick seems to be to calculate the orbital periods and distances (using the formula from http://en.wikipedia.org/wiki/Barycenter#Barycenter ) from the bottom up (ie starting with the AB pair and moving up the hierarchy) by hand first. Then when you're defining the STC, you start from the top and work down - ie you start at ABCD-E (which orbits the barycentre defined by the Ra/Dec statement) and work down to the AB one.





Putting a planet around Spica E (the K5 V star), I don't see any darkside illumination by the ABCD stars. Now, the appmag of the K5 V is about -26, and the appmag of the ABCD are -17.06, -14.82, -14.17, and -12.72, all crammed into 11 arcseconds of sky (about 1/3 the diameter of a full moon from Earth). The ABC triple are within about 1 arcsecond of eachother.

I'd guess that the ABC magnitudes would at least be cumulative somehow, wouldn't they? ie. they'd look like a single brighter star? Is there a way to calculate the cumulative magnitude? And should that be enough to light up the darkside (ie to be closer to -26)?

Or would the combined magnitudes probably be something like -17.5 because BCD are so much dimmer than A? Even if it is -17.5, that's still about 100 times brighter than the full moon - surely enough to visibly illuminate the darkside.

chris wrote:

Malenfant wrote:EDIT: does this sound right to calculate the orbital period?

P in years = SQRT[(separation in AU^3)/(sum of masses in Sols)]

I think it should actually be:

P in years = sqrt(separation in AU^3)/(sum of masses in Sols)

--Chris

Oops . . . my version is definitely wrong. The period is proportional to the velocity, which is proportional to the square root of r^3/M. I was ignoring the fact that the kinetic energy of an object is proportional to velocity squared. Asleep at the wheel. Or keyboard.

--Chris