Celestia Forums

Posted: **08.12.2003, 05:34**

Let's say I have a binary system consisting of two stars exactly like the sun (same radius, temperature, mass, luminosity, age, etc), and the separation between them is 1 AU. I think this means that in practice they are always located at opposite ends of a circular orbit around a centre of mass that is 0.5 AU from each star, right?

But that assumes that their orbit around the centre of mass is circular, doesn't it? Can this orbit be eccentric if both stars have the same mass? If so, does the centre of mass move to one of the foci of the ellipse - so the distance between each star and the centre of mass varies over the orbit, even though the masses of the stars are the same?

Also, how long would it take each star to complete an orbit? And how do you figure out the orbital period of a planet that orbits the centre of mass outside the pair - do you assume the central mass is the same as the combined mass of the stars?

I've always had trouble visualising this sort of thing... I keep getting confused with reference frames

Posted: **08.12.2003, 09:34**

I don't know the math to figure out your questions, but plugging this stuff into a simulator lets me answer most of your questions.

Yes, they are located at opposite ends of a circular orbit, .5 AU from their center of mass.
Here's 2 solar mass stars. The yellow star is locked to the center of the screen:

Unlocking the yellow star from the center of the screen reveals their center of mass:

Yes, the orbits can be eccentric if the stars have the same mass, with the center of mass at a foci, and their distance from the center of mass will vary.
Here's 2 solar mass stars in eccentric orbits. The yellow star is locked to the center of the screen:

And unlocked:

The orbital period for 2 sun-mass stars 1 AU apart in circular orbits is about 8.5 months. I don't know how to compute this. I just looked at the date on the simulator and watched them complete a single orbit. (I was guessing it would be 6 months)

A planet that orbits outside a pair of 1 solar-mass stars will have the same period as if it orbited a single star with 2 solar masses.
This planet, orbiting at about 1.5 billion km from the center of mass of 2 solar-mass stars takes about 25 years to complete an orbit. If the planet orbited any closer than about 3-4 AUs from the pair, its orbit would be unstable.

This planet, orbiting at about 1.5 billion km from a single 2-solar mass star also takes about 25 years to complete an orbit.

Posted: **08.12.2003, 09:39**

Evil Dr Ganymede wrote:Can this orbit be eccentric if both stars have the same mass? If so, does the centre of mass move to one of the foci of the ellipse - so the distance between each star and the centre of mass varies over the orbit, even though the masses of the stars are the same?

Each star has its own elliptical orbit with the centre of gravity of the system at one focus. The eccentricity of both orbits is the same, and they're orientated in the same plane (same ascending node and inclination). The periods are the same, but the arguments of the pericentres are opposite each other, and the mean anomalies are the same, so that each object passes through pericentre at the same moment, on opposite sides of the centre of gravity of the system, and then they stay opposite each other throughout the entire period. I made a little model of mutually orbiting asteroids which you can find here, which might make the situation clear - effectively, you'll see an animation of the elliptical case that Tony illustrates above, except with unequal masses. Try tracking the barycentre, and watching the two bodies move around it, and then try following one body while watching the other. (I use mean longitude in that model, though: the mean longitudes of the two orbits are opposed, since they're measured from the node, not the pericentre.)
You then need to work out the semimajor axes - the mean distance between the two objects is divided between them in inverse proportion to their mass. So if one object is twice as massive as the other, it will be twice as close to the centre of gravity of the system. If we use M for the large mass and m for the smaller mass, then M's distance from the centre of mass will be:

a(M) = m/(M+m) * d

and m's distance will be:

a(m) = M/(M+m) * d

where d is the mean separation of m and M.

Evil Dr Ganymede wrote:Also, how long would it take each star to complete an orbit?

You use the standard Kepler equation:

P^2 = 4*pi^2*a^3/[G[M+m])

using what I've called d, the mean separation of M and m, for the value of a. In your example, since a is 1AU, but M+m is two solar masses, you'd expect the period to be 1/sqrt[2] = 0.707 years, or ~8.5 months as Tony found above.

Evil Dr Ganymede wrote: And how do you figure out the orbital period of a planet that orbits the centre of mass outside the pair - do you assume the central mass is the same as the combined mass of the stars?

That's right.

Grant

Posted: **08.12.2003, 14:55**

Tony,

What orbit simulation program are you using?

Posted: **08.12.2003, 17:37**

Thanks guys

. Tony, are those images from the gravitysimulator on your webpage? That looks rather handy...

I must confess I did find a webpage (you need java) that showed the orbital motions of the binary stars at:
http://instruct1.cit.cornell.edu/course ... binary.htm

But that gravity simulator looks rather fine

.

I got confused about the orbital periods because as far as I could tell, the stars weren't orbiting anything other than the barycentre of the orbit, so I couldn't decide what the central mass was. But the equation Grant showed is just the normal one for finding orbital periods isn't it - so you really just assume that one star is the central masss and the other star is orbiting around it for those purposes?

Still, thanks for the assist, that's made me less confused

.

Next question is how does one figure out where the stable orbits are around the stars? I've always assumed that planets can orbit each star within a distance of 1/3rd of the separation (i.e. within 0.33 AU of each star), or beyond a distance three times the separation from the centre of mass (i.e. beyond 3 AU from the centre of mass). I've seen this sort of thing quoted for the main Alpha Centauri binary. Is there a way to actually calculate this more accurately though? How would eccentricity affect that? And although orbits might be stable in the longterm in those regions, would planets actually be able to form there?

EDIT: Oh yeah... if a planet orbits the pair at 4 AU from the centre of mass and in the same plane as the stars, then there'll be a time when one of the stars is eclipsed behind the other? If so, for that period the planet's temperature should plummet because while the eclipse is in progress it's effectively the same as orbiting a single star 3.5 AU away, right? So how does one figure out how long these stellar eclipses last? And is there an easy way to figure out what orbital inclination the planet needs to have to be able to always see both stars no matter where they are in their orbits?

Posted: **08.12.2003, 18:52**

Evil Dr Ganymede wrote:Next question is how does one figure out where the stable orbits are around the stars? I've always assumed that planets can orbit each star within a distance of 1/3rd of the separation (i.e. within 0.33 AU of each star), or beyond a distance three times the separation from the centre of mass (i.e. beyond 3 AU from the centre of mass). I've seen this sort of thing quoted for the main Alpha Centauri binary. Is there a way to actually calculate this more accurately though? How would eccentricity affect that? And although orbits might be stable in the longterm in those regions, would planets actually be able to form there?

Szebehely provides a graph and some datapoints, based on numerical simulations - he also gives references for analytical solutions that you might care to chase if you're really into this. His results are for circular orbits, but I've read elsewhere that what matters is the closest approach of the perturbing body, so you could perhaps plug in the value of its pericentre instead of its semimajor axis, if the orbit is eccentric. Again I have a jotting that planet formation is perturbed at greater distances than stable orbital conditions - a factor of 10 rather than the factor of 3 you mention, but unfortunately I didn't record a source for that bit of informatic flotsam, sorry.

Grant

I'll get back to you on the eclipse thing ... seems interesting. But planetary temperature wouldn't necessarily plummet - there's a fair old thermal inertia to something the size of a planet, and we don't all freeze to death during the night. :wink:

Posted: **09.12.2003, 02:12**

selden wrote:Tony,

What orbit simulation program are you using?

Gravity Simulator
http://www.gravitysimulator.com

granthutchison wrote:where d is the mean separation of m and M.

How do you compute the mean separation of an elliptical orbit? is it (Semi-major axis + Semi-major axis) / 2, or is it a more complicated formula that takes into account that the object spends more time near its aposis when they are orbiting slower?

Posted: **09.12.2003, 03:52**

granthutchison wrote:Szebehely provides a graph and some datapoints, based on numerical simulations - he also gives references for analytical solutions that you might care to chase if you're really into this. His results are for circular orbits, but I've read elsewhere that what matters is the closest approach of the perturbing body, so you could perhaps plug in the value of its pericentre instead of its semimajor axis, if the orbit is eccentric.

Aha. That's a handy paper (I didn't know you could get full papers from the ADS!)... thanks!

EDIT: I found a better one though - a 1999 paper by Holman and Weigert that gives an empirical formula for figuring out the limits for a much wider range of mass ratios and eccentricity.

Again I have a jotting that planet formation is perturbed at greater distances than stable orbital conditions - a factor of 10 rather than the factor of 3 you mention, but unfortunately I didn't record a source for that bit of informatic flotsam, sorry.

I was wondering about that... I'd assumed that you'd end up with asteroids belts around the orbit beyond which a planet can theoretically orbit both stars. Though I'd imagine that factor you mention probably varies depending on the mass of the stars, since the other limits vary like that too.

I'll get back to you on the eclipse thing ... seems interesting. But planetary temperature wouldn't necessarily plummet - there's a fair old thermal inertia to something the size of a planet, and we don't all freeze to death during the night.

True. But then I have no idea how long these eclipses last.

(though I think I goofed up the temperature calculation, so it looked like the temperature dropped by about 100 degrees when it actually drops only by a few degrees. So that might explain it!).

Posted: **09.12.2003, 09:30**

Evil Dr Ganymede wrote:EDIT: I found a better one though - a 1999 paper by Holman and Weigert that gives an empirical formula for figuring out the limits for a much wider range of mass ratios and eccentricity.

Ha. We must have followed the same chain of references - I was just going to post a link to that one myself.

Evil Dr Ganymede wrote:But then I have no idea how long these eclipses last.

But in a hand-waving sort of way, it's going to be on the order of a day, yes?
The width of each of your sunlike stars must occupy about a degree of its circular orbit (Sun subtends half a degree at 1AU, stars are 0.5AU from the centre of gravity).
With an orbital period of 0.7 years, they're going to be shifting through rather more than a degree per day. So after a day, they'll have shifted relative to each other by more than their own widths.
The movement of your planet in orbit is going to prolong the alignment a bit, of course, but we're talking soming like a single day with a bit less than half the usual amount of sunshine - hey, we get that in Scotland all the time :wink:

I'll get back to you with a proper formula for eclipse duration, though.
BTW: there's no inclination from which the planet can see both stars all the time, unless you deliberately create some sort of resonance between the period of the stars and the planets. The planet's orbit will always have to pass through the plane of the stars' mutual orbit at two nodes, and that's where eclipses can happen.

Grant

Posted: **09.12.2003, 10:13**

granthutchison wrote:Ha. We must have followed the same chain of references - I was just going to post a link to that one myself.

Well, I went the roundabout route via a load of papers by D. Benest.

Evil Dr Ganymede wrote:With an orbital period of 0.7 years, they're going to be shifting through rather more than a degree per day. So after a day, they'll have shifted relative to each other by more than their own widths.

That's a degree in their orbit per day, yes? Or do you mean a degree in the sky as seen from the planet?

The movement of your planet in orbit is going to prolong the alignment a bit, of course, but we're talking soming like a single day with a bit less than half the usual amount of sunshine - hey, we get that in Scotland all the time

Well, if we have two stars at 1 solar mass each, separated by 1 AU, with circular orbits around the centre of mass (orbital radius = 0.5 AU), then according to the H&W paper the absolute closest stable orbit beyond the pair is at about 2.4 AU from the CoM of the system.

If we have a planet at 2.5 AU - let's assume it move there from further out if it couldn't form there - what would the planet see? When the two stars are lined up one behind the other, it'll just see one star (the nearest one, which is 2 AU away from the planet at that point). Every other time, it'll see the stars getting further apart, reach a maximum separation from the CoM, and then get closer together. So... it's coldest when it has only one star in the sky during an eclipse, and it's hottest when the stars are oriented at 90 degrees to the line connecting the planet to the Centre of Mass of the system (each star is 2.55 AU from the planet), and at all other points during the stars' orbit the temperature is between, yes?

If that's true, I figure that the blackbody temperature of the planet during an eclipse is about 197K, and when the stars are at widest separation it goes up to 207K. Hmm, it's a bit too cold .

If I upped the luminosities to 1.7 sols each (which means the stars are probably 1.1 solar masses, and they're both either G0 V or F9 V stars that are a few billion years old) then the blackbody temperatures go to 257K during eclipse and 270K at widest separation, which means that if we have earthlike albedo and greenhouse effect then our final temperatures are somewhat nippy during an eclipse but generally earthlike around widest separation.

Does that sound about right? Of course, the problem is that the planet probably wouldn't have formed in that orbit. I found a paper on terrestrial planet formation in the Alpha Centauri system that might have something useful in it on the planet formation front though...

I'll get back to you with a proper formula for eclipse duration, though.
BTW: there's no inclination from which the planet can see both stars all the time, unless you deliberately create some sort of resonance between the period of the stars and the planets. The planet's orbit will always have to pass through the plane of the stars' mutual orbit at two nodes, and that's where eclipses can happen.

D'oh, you're right of course. I told you I had trouble visualising this

.
This is where I wish Celestia could handle gravity...

Posted: **09.12.2003, 11:34**

Evil Dr Ganymede wrote:That's a degree in their orbit per day, yes?

Yes.

OK, here's how to get the eclipse duration for a prograde planet in the same orbital plane as the stars. First you need to work out the view angle along the planet's orbit from which an eclipse is visible: from first contact to last contact. Imagine scooting around the orbit until the stars line up, then shifting back and forth to delimit the range over which their discs overlap to some extent. It's a slightly messy equation, but I've battered it into the simplest formulation I can:

theta = 2*atan{[r1*(a-a2)+r2*(a+a1)]/[a*sqrt((a1+a2)^2-(r1+r2)^2)]}

where r1 is the radius of star 1, r2 is the radius of star 2, a1 is the orbital radius of star 1, a2 is the orbital radius of star 2 and a is the orbital radius of your planet (all around the system CofG). All these distances have to be in the same units. Star 1 is the star being eclipsed by star 2.
Now you need to work out the angular rate at which your planet is going to traverse that eclipse zone:

omega = 360deg/Ps - 360deg/Pp

where Ps is the orbital period of the stars, and Pp is the orbital period of the planet. Omega is just the planet's angular velocity in the rotating reference frame of the stars.

So the planet has to traverse theta degrees at omega degrees per day, and the eclipse therefore lasts theta/omega days.

For your example above, using masses and radii of 1.1 solar, I make the eclipse duration 1.1 days (no, that's just a coincidence! :wink:

)

Grant

Posted: **09.12.2003, 21:00**

So the planet has to traverse theta degrees at omega degrees per day, and the eclipse therefore lasts theta/omega days.

But that sounds like your equation is solely based on the angular velocity at which the planet is travelling around its orbit, not the angular velocity at which the stars are travelling around their orbits. The stars should be moving faster than the planet... or do you not need to worry about this because you're looking at it in the frame of reference rotating with the stars?

Posted: **09.12.2003, 21:24**

Evil Dr Ganymede wrote:But that sounds like your equation is solely based on the angular velocity at which the planet is travelling around its orbit, not the angular velocity at which the stars are travelling around their orbits.

No, if you check the formula you'll see that the calculation of omega involves the angular velocity of the stars, minus the angular velocity of the planet.
If you were stationary and observing the stars, they would sweep from first contact to last contact at a rate determined purely by their orbital angular velocity. But the planet is orbiting in the same direction as the stars are rotating, so it prolongs the eclipse by reducing the apparent angular velocity of the stars by an amount equal to its own orbital angular velocity - hence the subtraction. Although I wrote that "the planet traverses the eclipse zone" in the reference frame of the stars, it does so backwards relative to its orbital direction - you can also think of a distant stationary rest frame in which the eclipse zone sweeping up from behind the planet and whooshes past it, since the stars are revolving faster than the planet.
Yes?

Grant

Posted: **09.12.2003, 21:30**

OK. I think

.

BTW I tried making a rendition of this in Celestia, and came up with this:

The stc file:

#######################
#CoM.stc
#######################

# CoM (HIP 400000)
400000 {
RA 180
Dec 0
Distance 100
SpectralType "M0V"
AppMag 50.00 # Radii: 1.47270096432767
}
#######################

The ssc file:

# Data for our binary system.

"Star1" "HIP 400000"
{
Texture "gstar.jpg"
Radius 696265

EllipticalOrbit {
Period 0.707
SemiMajorAxis 0.5
Eccentricity 0.0
Inclination 0.0
AscendingNode 0.0
LongOfPericenter 0.0
MeanLongitude 0.0
}
}

"Star2" "HIP 400000"
{
Texture "kstar.jpg"
Radius 696265

EllipticalOrbit {
Period 0.707
SemiMajorAxis 0.5
Eccentricity 0.0
Inclination 0.0
AscendingNode 0.0
LongOfPericenter 0.0
MeanLongitude 180.0
}
}

"Planet" "HIP 400000"
{
Texture "earth.png"

EllipticalOrbit {
Period 2.79508
SemiMajorAxis 2.5
Eccentricity 0.00
Inclination 0.00
AscendingNode 0.0
LongOfPericenter 0.0
MeanLongitude 100.464
}

Albedo 0.30
}

I forgot how to make the stars look like stars though... but it's kinda handy.

Posted: **09.12.2003, 21:50**

Evil Dr Ganymede wrote:I forgot how to make the stars look like stars though...

"Emissive true" is the best you can do at present, I think.
I did something similar with Celestia to make sure my equation wasn't completely crazy, and got a decent fit with a couple of different mass ratios for the stars. How about your system?

Grant

Posted: **09.12.2003, 23:58**

granthutchison wrote:How about your system?

OK, I installed it and tried for myself (couldn't resist ... :wink:

) My calcs above give an eclipse duration of 1.0241 days, and my best eyeball at extracting first and last contact from Celestia comes out at 1.0246 days. The error falls below the limits of my screen resolution, so I doubt if I'm going to get a better match!

Grant

Posted: **10.12.2003, 00:55**

Evil Dr Ganymede wrote:If I upped the luminosities to 1.7 sols each (which means the stars are probably 1.1 solar masses, and they're both either G0 V or F9 V stars that are a few billion years old) then the blackbody temperatures go to 257K during eclipse and 270K at widest separation, which means that if we have earthlike albedo and greenhouse effect then our final temperatures are somewhat nippy during an eclipse but generally earthlike around widest separation.

Does that sound about right?

I'm not seeing this. Total luminosity 3.4 at ~2.55AU, right? Using an equation I know you've used previously, that's a blackbody temperature of:

T = 278.3 * 3.4^0.25/2.55^0.5 = 237K

During eclipse you've got 1.7 suns at 2AU, which is

T = 278.3 * 1.7^0.25/2^0.5 = 225K

This makes sense, because multiplying the distance by 2.55 (in the first calc) reduces the flux by a factor of 6.5, which is nowhere nearly compensated by the increase in luminosity of only 3.4. Incoming radiation falls to 0.523 Earth-normal. Temperature scales with the fourth root of insolation, though, so is 0.85 of Earth's.

You'll also find that insolation is maximal either side of eclipse, and falls to a minimum when the two stars are at their most widely separated, in the 90-degree position.
Imagine the two stars at maximum separation, both ~2.55AU from the planet. Let's call the overall insolation 1+1 = 2 at this point. One star now moves towards the planet, to 2AU, one moves away to 3AU. Insolation from the approaching star rises by a factor of (2.55/2)^2 = 1.63. Insolation from the retreating star is reduced to (2.55/3)^2 = 0.72. Overall insolation is now 1.63+0.72 = 2.35, representing a 17.5% increase. (But of course it falls to 1.56 during maximal eclipse.)

Hmmm. Might be worth coming up with an equation for the duration of maximal eclipse - from second contact to third contact, which is the only time the insolation is really this low.

Grant

Edited from my original quick estimate to match your more precise reckoning of the distance during the 90-degree phase.

Posted: **10.12.2003, 07:22**

granthutchison wrote:OK, I installed it and tried for myself (couldn't resist ... ) My calcs above give an eclipse duration of 1.0241 days, and my best eyeball at extracting first and last contact from Celestia comes out at 1.0246 days. The error falls below the limits of my screen resolution, so I doubt if I'm going to get a better match!

Yeah, I checked and got 1.0228 days, which is in the right ballpark - I didn't bother zooming right in, but that's close enough I think

.

granthutchison wrote:I'm not seeing this. Total luminosity 3.4 at ~2.55AU, right?

Gah! I multiplied by the luminosity twice, that's why my answers were wrong! Well spotted :oops:

.

During eclipse you've got 1.7 suns at 2AU, which is

T = 278.3 * 1.7^0.25/2^0.5 = 225K

This I agree with, now I've corrected my spreadsheet!

Using an equation I know you've used previously, that's a blackbody temperature of:

T = 278.3 * 3.4^0.25/2.55^0.5 = 237K

And I also get this now, though I did it by adding the 'flux' from both stars and then taking the fourth root of the total, like you said on this thread here. But the answer's the same.

It was a silly mistake in my spreadsheet, my excuse is that it was late last night when I wrote it

.

You'll also find that insolation is maximal either side of eclipse, and falls to a minimum when the two stars are at their most widely separated, in the 90-degree position.

I told you I was a dunce when it came to binaries

. That's odd, but I see how it works like that now you've explained it... I'd just assumed that at the 90 degree position you'd get two stars equally apart, whereas near the eclipse you'd get one star a lot further away than the other, which would be less insolation than at the 90 degree position. Evidently I was mistaken there!

Though this wouldn't necessarily be true if the luminosities of the stars were different, would it?

Hmmm. Might be worth coming up with an equation for the duration of maximal eclipse - from second contact to third contact, which is the only time the insolation is really this low.

The further star is completely occulted by the nearer star for very close to 5 hours, according to Celestia. Quite a long time, compared to our own solar eclipses!

Oh yeah, turn the orbits on, move the viewpoint above the system so you can see a plan view of all the orbits (including the planet's), Track and Lock onto Star1, and speed up the simulation to see how the planet moves around the binary in the reference frame of Star1

. I wish there was some way to view the orbital tracks around the point you've locked onto, that'd make a nice simulation of the planet's motion in a rotating reference frame.

Posted: **10.12.2003, 08:52**

Hmm. I just figured out the maximum angular separation each star in the ssc/stc file I posted earlier should have relative to the centre of mass, as seen from the planet - I reckon it should be 11.536 degrees, when the angle between line connecting the star to the CoM and the line connecting the CoM to the planet is 78.407 degrees.

I looked in Celestia (by doing this: go to HIP 400000, go to planet, follow planet, look backwards (* on numpad), centre on HIP 400000 [the centre of mass], lock on HIP 400000, speed up time), and zoomed in the field of view til I could just see both stars at their widest separations at the edge of the screen, and the field of view was 14 degrees and 32 minutes - so the maximum separation was 7 degrees and 16 minutes, which is somewhat smaller than by 11.536 value.

I suspect I've done something wrong there again... the 11.536 degrees is what I think is the value of maximum eastern/western elongation of the star as seen from the planet, and I gather that occurs when the angle (through the star) between the planet and centre of mass is 90 degrees. (see here for details, replace the sun with 'centre of mass', Venus with 'Star1', and Earth with 'Planet').

Or is Celestia doing something odd? I notice that the stars don't reach their maximum elongation at the same time, is that because of the motion of the planet?

Posted: **10.12.2003, 09:37**

Evil Dr Ganymede wrote:And I also get this now, though I did it by adding the 'flux' from both stars and then taking the fourth root of the total, like you said on this thread here. But the answer's the same.

Yes. The shorter method works in this instance because both stars are at the same distance, so you can lump their luminosities together without pre-processing.

Evil Dr Ganymede wrote:I notice that the stars don't reach their maximum elongation at the same time, is that because of the motion of the planet?

They can't reach maximum elongation at the same time, because they're constrained to be on opposite sides of the centre of mass - when one reaches max elongation, the other is either well past it or still approaching it. You've worked out sin(0.5/2.5), which is correct for the max elongation of one star, but their farthest separation is going to be closer to 2*tan(0.5/2.5) - ie, when they're in the 90-degree position.

Still doesn't explain what you're seeing in Celestia, though - I'm away from my Celestia machine at present, so can't help right now.

Grant

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Orbital periods in binary systems?

Orbital periods in binary systems?

Re: Orbital periods in binary systems?