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I have two stars, with known masses and known orbital radius. How do I calculate the orbital periods of both stars?

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Have you searched the web? I think you have not investigated enough. There are plenty of answers out there, like answers.yahoo.com/question/index?qid=20071225133716AA5gj9p and voyager.egglescliffe.org.uk/physics/gravitation/binary/… –  Envite Feb 21 at 14:19
    
@Envite I wouldn't blame the OP for not trusting yahoo answers, but the Egglescliffe School resource is a good one, even if they have poor site design. –  called2voyage Feb 21 at 14:25
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Just two results from the first page of Google for "Binary Stars Orbital Period" –  Envite Feb 21 at 14:28
    
@Envite I did try searching the web several times; perhaps I used the wrong queries, and I habitually don't click yahoo answers links. –  Jakob Weisblat Feb 21 at 14:56
    
Ok, but now that you have Egglescliffe document, please read it and answer your own question. Or if there is something specifically you do not understand of it, ask for that specific issue. –  Envite Feb 21 at 16:28

1 Answer 1

If you need a rough estimation, you can sample the positions of the stellar objects from their acceleration, using Newton's law. A full picture is drawn on this Wikipedia page, but basically, given N stellar objects at position P(i), with their respective mass M(i):

$$ M_i.\vec{acc_i} = -G\sum_{n \neq i}^N \frac {M_i.M_n.(\vec{pos_i}-\vec{pos_n})}{\begin{vmatrix} \vec{pos_i}-\vec{pos_n}\end{vmatrix}^3 } $$

You can then derivate acceleration into speed, then into position using a small enough time delta. With initial positions and speed (which are the trickiest and also funniest to choose), you can simulate a rough N-body system.

On the fun and simulation side, this is what Universe Sandbox aims to present (in advance, sorry for linking to a commercial store, I'm not related to this studio).

Lagrangian points are also interesting to look at when simulating.

However, to simplify things, you may like to think nearly all stellar objects "close" to the binary system would have been eaten by the massive couple over time, and then consider only the barycenter of the pair of stars.

EDIT: although OP was clear, my answer is for N objects. The simplification to N=2, with A and B being the positions of the two objects, results in: $$ \vec{acc_A} = \frac{G.M_B}{AB^3}\vec{AB} $$ $$ \vec{acc_B} = \frac{G.M_A}{BA^3}\vec{BA} $$ And with an iterative process, once acceleration is computed for step k, using known initial conditions for speed and position: $$ \vec{spd_{k+1}} = \vec{acc_k}.\Delta{t} + \vec{spd_k} $$ $$ \vec{pos_{k+1}} = \vec{spd_k}.\Delta{t} + \vec{pos_k} $$

EDIT2: well, another rough estimation, for the period duration (which I misunderstood as "trajectory" from the question). I'd use cylindrical coordinates as reference, for their clean spatial representation, and for the single-angle conversion:

$$ \left( \begin{array}{c} R_k \\ \theta_k \\ h_k \end{array} \right) = \left( \begin{array}{c} \sqrt{pos_{x,k}^2+pos_{y,k}^2} \\ \arctan \left( \frac{pos_{y,k}}{pos_{x,k}} \right) \\ pos_{z,k} \end{array} \right) $$

...with adequate precaution. Introduce part of that conversion at each iteration, and wait for theta to complete a rotation.

(in advance, sorry for the weird notation, which I just tried to make consistent from my previous posts)

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That's a general answer for an n-body system. For the special case of a binary, the sum/integral can be solved explicitly, leading to orbits of the same period around the common barycenter. May be you could add a solution for this special case. –  Gerald Feb 21 at 18:31
    
You're right, OP clearly stated the requirement but I provided a broader answer nonetheless. I'll edit the answer to match N=2. –  TallFurryMan Feb 24 at 8:37
    
Ok, this isn't quite easy now: Can you turn the $\Delta$ to infinitisimals to get a differential equation, and find a connection to the 3rd Kepler law, like here: en.wikipedia.org/wiki/Gravitational_two-body_problem ? –  Gerald Feb 24 at 21:27
    
@Gerald I'm afraid I won't :) my proposal is nothing but a simple iterative process, where additional perturbations may be introduced at each step. I avoided differentials because the problem doesn't require an exact solution (and partly because that would require me to reopen some books...). But based on your comment, it appears I misread the OP: I thought the question was about trajectories, but might be about the "period", as duration? –  TallFurryMan Feb 25 at 13:12
    
It was worth a try :) . I've been understanding the question in the sense of duration. Nevertheless, your iterative solution is more robust regarding perturbations. By comparing the actual positions with the starting positions, it's easy to find out the period (duration) during a simulation run, if needed. –  Gerald Feb 25 at 17:57

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