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EDG
2010-Mar-06, 03:06 AM
I'm trying to figure out how the temperature of a planet orbiting a binary star varies over time, but I'm running into trouble accounting for the eccentricity of the stars' orbits and the planet's orbit.

So far I've figured out how the distance between each star and the planet varies if the stars' orbits around their Centre of Mass (CoM) and the planet's orbit all have zero eccentricity (see attached image). This assumes that the planet is a fixed point (I know it isn't really, but it makes it easier). The stars are orbiting the centre of mass of the system, so theta is going from 0 to 360 degrees. I've shown the equations in the attached image.

Now, if you go to this link: http://csep10.phys.utk.edu/guidry/java/binary/binary.html
(warning: this is a Java app) you'll see a java app that lets you see what the orbits looks like and that I can use to explain it better.

The setup I show in the attached image is a snapshot of what you'd get if you set the eccentricity (the bar at the bottom) in the java app to zero, and then set M2 (bar at lower left) to something like 0.5, and then zoom out (bar on the lower right). So you'll have two nested circular star orbits - treat the the planet as a distant fixed point is directly above the CoM on the screen.

So the problem I have now is this: how do I factor in the stars' orbital eccentricity into the equations that calculate the distance between each star and the planet? (if you adjust the eccentricity on the java app you'll see how the stars' orbits change). And ultimately, once that's figured out, how do I factor the planet's orbital eccentricity as well?

EDG
2010-Mar-06, 03:55 AM
I'm guessing I'm going to have to multiply a and b in those equations by a term that involves the eccentricity e aren't I? And since the length of a and b at any given time will depend on where they are in their orbits around the CoM, I guess that term will involve a sin or cos as well? (and if the planet's orbital eccentricity is in the mix too, then c will have to be multiplied by the planet's e too?)

EDIT: I've attached another diagram I made showing the stars with equal mass and eccentric orbits, when theta=90°. The red dot and ellipse are the planet and its orbit. The point where the green lines cross is the Centre of Mass of the system.

EDG
2010-Mar-06, 04:19 AM
Hm... thinking about it more, do I have to replace all instances of a, b and c in the equations with a(1+sin.e), b(1+sin.e) and c(1+sin.ep) (where e is eccentricity of the stars' orbits, and ep is the eccentricity of the planet's orbit)?

frankuitaalst
2010-Mar-06, 03:48 PM
It is impossible to find a closed relationship for the angle .
One way to solve a GIVEN system is the following :
1. define the binary in terms of Sma and eccentricity and mass . Then you'll get the period of the binary .
From the formula : M = E-e sinE
( see http://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion) you can find iteratively tyhe position of the two bodies .
2. Then define the planet with respect to the center of mass of the binary .
Applying the above formula again you'll find the position of the planet related to the CM of the binary .
What you'll get is a superposition of two sine like functions , one having the period of the binary , the other having the period of the planet around the binary . The period of the planet around the binary will be greater than the period of the binary .

EDG
2010-Mar-06, 04:09 PM
Hrm... is it not even possible to at least just find the closest and furthest distances between each star and the planet? I really just need to figure out how hot and cold it can get at the planet, so I just need to figure this out when the stars and planets are in specific positions relative to eachother.

I've got some specifics here:
--r1 and r2 are the distance of each star from the Centre of Mass, sep is the separation between each star (should be the sum of r1 and r2). peri and aph are Perihelion and Aphelion distances for the planet (measured relative to the Centre of Mass). Planet mass is in Jupiter masses.--

Star orbits:
Ecc Sep(AU) r1(AU) r2(AU) Period(yr)
0.336 0.558 0.273 0.285 0.335

Stars:
Type Size Mass Temp Luminosity Radius Age
F1 V 1.418 6851. 4.8150711 1.561 1276813056.
F2 V 1.398 6810. 4.4820566 1.524 1276813056.

Planet:
AU ecc mass density radius peri aph
8.520 0.199 2.427 2587.65 75196. 6.828 10.213

hhEb09'1
2010-Mar-06, 04:39 PM
Hrm... is it not even possible to at least just find the closest and furthest distances between each star and the planet? I really just need to figure out how hot and cold it can get at the planet, so I just need to figure this out when the stars and planets are in specific positions relative to eachother. I was trying to figure out how this was a trigonometry problem! :)

Yes, you can calculate those distances, but I'm not sure if it is that important. By "how hot it can get" I guess you mean surface temperatures of the planets? But that's not just a function of distance from the star(s) is it? Do you have an insolation/temperature rule of thumb that you plan to use?

EDG
2010-Mar-06, 05:08 PM
I was trying to figure out how this was a trigonometry problem! :)

Well, I'm drawing lots of lines and circles, right? ;)

Yes, you can calculate those distances, but I'm not sure if it is that important. By "how hot it can get" I guess you mean surface temperatures of the planets? But that's not just a function of distance from the star(s) is it? Do you have an insolation/temperature rule of thumb that you plan to use?

For my current purposes, it is just a function of distance from the star because I'm just interested in the blackbody temperature. I'll worry about accounting for albedo and greenhouse effects and all that other stuff later.

The way I figure it, if all the orbits of the stars and the planet in my example were circular and in the same plane, then the planet's BB temperature is highest when the brighter star is closest to it (but just before it eclipses the dimmer star behind it). During the stellar eclipses in this system the temperature can drop significantly (the eclipses take about 17 hours from start to finish) as one star is blocked by the other. In this case, the temperature would be about 166K just before or after eclipse when both stars are visible, and 144K or 141K during eclipse (depending on which star is being eclipsed).

So what I'm after here is a way (if possible) to figure out those temperature extremes if the star (and planet) orbits are eccentric.

hhEb09'1
2010-Mar-06, 05:15 PM
So what I'm after here is a way (if possible) to figure out those temperature extremes if the star (and planet) orbits are eccentric.But you know the distances, angles, etc., right? What are you using for the backside temperature? I ask, because at some points the starshine would be offset, right?

EDG
2010-Mar-06, 05:27 PM
But you know the distances, angles, etc., right? What are you using for the backside temperature? I ask, because at some points the starshine would be offset, right?

Usually the stars won't get off-centre (relative to the planet) by far enough to illuminate much of the anti-starward/CoM side of the planet.

hhEb09'1
2010-Mar-06, 05:32 PM
OK! that also means you don't have to be too careful in the other calculations. A rough distance down the middle should do it.

EDG
2010-Mar-06, 05:49 PM
In this specific case it wouldn't make much of a difference, but if the planet's closer (say, 3 AU from the CoM) then the changing distances to the stars can change the planet's BB temperature quite significantly.

For example, if the planet were at 3 AU from the CoM instead of 8.52 (and the orbits were circular and non-inclined), then I figure its temperature would vary from 280K to 283K while both stars are visible, and drop to 249K and 246K while the stars are eclipsed.

I guess that doesn't sound like much in this case, but it seems that average global temperatures rising by half a degree over the past 50 years or so make a big impact on Earth, imagine if the average temperature changes on a habitable world by a few degrees several times over a planetary year!

hhEb09'1
2010-Mar-06, 06:07 PM
I guess that doesn't sound like much in this case, but it seems that average global temperatures rising by half a degree over the past 50 years or so make a big impact on Earth, imagine if the average temperature changes on a habitable world by a few degrees several times over a planetary year!Have you done the same calculations for earth? It varies from 147.5 million kilometers to 152.5 million kilometers over the course of a year.

tony873004
2010-Mar-06, 09:24 PM
You can do this numerically with Gravity Simulator. Set up your system, and have it output the position vectors once per day over the course of an entire orbit to a text file that Excel can read.

I tried it with the system you describe in post #5. Here's how I did it:

File > New
Objects > Edit Objects
Change name to Star 1, set mass to 1.418 solar masses, set color, click OK
Objects > Create Objects, set mass to 1.398 solar masses, set SMA to 0.558, set eccentricity to 0.336, set name to Star 2, set color, click Create
Objects > Create Objects, verify that the reference object is set to Star 1 and click the 'barycenter' option, set mass to 2.427 Jupiter masses, sma to 8.52 AU, ecc to 0.199, name to Planet, set color, press Create
File > Save As, give the simulation a name

Here's the simulation file and a screen shot
http://orbitsimulator.com//gravity/simulations/edgBinary.gsim
http://orbitsimulator.com//gravity/images/EDGbinary.GIF

To create a data file:
File > Output File
click "Create Data File"
sample every 1 days
maximum data samples 5411 (the planet's period is 5411 days)
Select Star 1, Star 2, Planet
Select Rx, Ry, Rz
Press OK
Let the planet complete a full orbit and your file will be complete

Modify your Excel program to compute distances and flux
In Excel (I'm using Excel 2003)
File > Open, browse to the same folder where you saved your simulation. Under types of files, choose Text Files. Select the text file with the same name as your simulation file.
The 'Text Import Wizard' opens. Choose "Delimited" and press "Next", then choose "Comma" and press "Finish"
In cell Q2, type =SQRT((C2-M2)^2+(D2-N2)^2+(E2-O2)^2)
This gives you the distance between the planet and star 1

In cell R2, type =SQRT((H2-M2)^2+(I2-N2)^2+(J2-O2)^2)
This gives you the distance between the planet and star 2

In cell S2 type =4.8150711*3.846E+26/(4*PI()*Q2^2)
This gives you the flux arriving at your planet in Watts / square meter from star 1

In cell T2 type =4.4820566*3.846E+26/(4*PI()*R2^2)
This gives you the flux arriving at your planet in Watts / square meter from star 1

In cell U2 type =S2+T2
This gives you the total flux from the pair of stars.

Extend all 5 of these new cells down to the last row.
Now you can graph your flux over the course of one orbit.
You'll have to relate the received flux to the temperature of your planet based on albedo, greenhouse effect, etc. As Astromark points out in his split thread, the flux received is not very high out in the region far enough for stable orbits. We're getting values around 200 W/m^2 . Earth by comparison receives about 1400 W/m^2 from the Sun.

Here's my output file and a screen shot of the graph.
You can see the small cycles caused by the orbits of the binary stars orbiting eachother, superimposed on top of the much larger cycle caused by the planet's changing distance due to its eccentricity.
http://orbitsimulator.com//gravity/images/edgExcel.txt
http://orbitsimulator.com//gravity/images/EDGexcel.GIF

EDG
2010-Mar-06, 09:33 PM
Oh cool. I thought gravsim didn't do barycentres, for some reason. I'll give that a go later on - thanks!

and hheb - Yeah, for Earth the variation is about 5K due to orbital eccentricity. But I guess the thing about global warming is that it's extra warming on top of all the other effects (eg axial tilt variation, orbital eccentricity etc). But maybe I'm thinking of this from too earth-centric a position - after all, the climate of a world orbiting a binary pair would probably have reached some kind of stable equilibrium that accommodates all the variability there.

As it is, in this case the orbital eccentricity effects are much bigger than the effect of the variation in distance to the binary (as tony pointed out).

EDG
2010-Mar-07, 01:25 AM
Tony, I don't get the image that you get when I tried to create the system myself - I get the attached image instead. There doesn't seem to be an option to just view it with the barycentre as the focus object (which is what your graphic seems to be showing).

The only object that I set with the reference object as 'barycentre' instead of 'object' was the planet. Star 2 just had Star1 as the reference object. Is that right?

(I posted this on the gravsim forum too)

EDIT: Though if I download the gsim you posted, I get the picture you showed. I must have typed something wrong I guess? I'll try it again.

tony873004
2010-Mar-07, 02:02 AM
...The only object that I set with the reference object as 'barycentre' instead of 'object' was the planet. Star 2 just had Star1 as the reference object. Is that right?...
Yes, that is correct.

Open the Graphics Options interface by pressing F8 or F9. Then press the button that is labeled "F". That puts it in floating mode centered on the barycenter. Or you can simply toggle the floating mode on and off with the F button on your keyboard.

EDG
2010-Mar-07, 09:05 PM
Interesting... I tried using gravsim to generate graphs for a planet orbiting at 3 AU from the barycentre (roughly in the habitable zone of the binary). I plotted it for 5000 simulated days, which is several orbits of the planet at 3 AU.

The first graph is for a planet with eccentricity 0.199:
http://members.shaw.ca/evildrganymede/temp/central1.gif

The second graph is for a planet with 0 eccentricity (circular orbit):
http://members.shaw.ca/evildrganymede/temp/central2.gif

Note that the vertical scale on the second graph is much smaller than for the first graph - for the eccentric orbit the flux varies between about 1000 and 2250 W/m², whereas for the circular orbit the flux varies between about 1395 and 1520 W/m².

I guess the asymmetries in the graph are due to the planet's orbit around the barycentre - the stars are orbiting the barycentre too, so they're moving relative to the planet.

EDG
2010-Mar-07, 09:18 PM
Just to be sure here...

The blackbody temperature of a planet is:

Blackbody Temperature = [(Stellar flux)*(1/(4*sigma))]^(0.25)

Where Stellar flux is W/m², and sigma is the Stefan-Boltzmann constant (5.67e-8 W/m²/K).

And Stellar flux = Luminosity/(4*pi*distance²)

where Luminosity is in Watts, and distance is distance of the planet from the star in metres.

That's correct, right?

If so, then the BB temp of the eccentric planet varies between about 259K and 318K over a year, while the circular planet varies between 280K and 286K. So the planet on the circular orbit around the binary gets about as much variation in flux (and BB temperature) over a year just from the changing distances to the two stars as the Earth does on its own eccentric orbit around a single star.

Tobin Dax
2010-Mar-07, 10:09 PM
Just to be sure here...

The blackbody temperature of a planet is:

Blackbody Temperature = [(Stellar flux)*(1/(4*sigma))]^(0.25)

Where Stellar flux is W/m², and sigma is the Stefan-Boltzmann constant (5.67e-8 W/m²/K).

And Stellar flux = Luminosity/(4*pi*distance)

where Luminosity is in Watts, and distance is distance of the planet from the star in metres.

That's correct, right?

Stellar flux = Luminosity/(4pi*r2)
Your formula doesn't show distance squared.

For the temperature, the planet's luminosity is sigma*4pi*R^2*T^4 and the incoming flux*area is (stellar flux)*pi*R^2. Set these equal to each other:

(stellar flux)*pi*R^2 = sigma*4pi*R^2*T^4

(stellar flux) = (sigma*4)*T^4 [pi*R^2 cancels out]

T^4 = (stellar flux)/(4*sigma)

Looks good to me. (It's been way too long since I've had to derive that formula, so I started from scratch.)

If so, then the BB temp of the eccentric planet varies between about 259K and 318K over a year, while the circular planet varies between 280K and 286K. So the planet on the circular orbit around the binary gets about as much variation in flux (and BB temperature) over a year just from the changing distances to the two stars as the Earth does on its own eccentric orbit around a single star.Just for kicks, consider the temperature change on Earth if it had an orbital eccentricity of 0.20:

T^4 = L/(16*sigma*r^2) => T~r^0.5

R_max/R_min = (1.20 AU)/(0.80 AU) = 1.5 so T_max/T_min = 1.22

(318 K)/(259 K) = 1.23, so the annual temperature variation seems to come mainly from the eccentricity of the orbit. I don't find this surprising since this "annual" flux variation is the dominating wave in your first graph. The smaller variation is due to the motion of the stars. Are those fluctuations about the same size in both graphs? It looks like they could be, and it wouldn't surprise me if they were.

Edit: Also remember that the eccentricity of Earth's orbit is very small. The orbit is practically circular.

EDG
2010-Mar-07, 11:11 PM
Stellar flux = Luminosity/(4pi*r2)
Your formula doesn't show distance squared.

Ooops, forgot to show that. It is in the formula I use, I just didn't write the squared in here - I'll add it in now. So good, I'm all correct :).