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I am modeling a two binary star system, and I am wondering if this is the case.

The way I have it right now is that I first figure out the mass center, and then the radius from each of the planet to the mass center.

I then figure out the acceleration for the first planet with:

$acceleration_1=\frac{\frac{G*m_1*m_2}{(r_1+r_2)^2}}{m_1}$ where $r_1$ and $r_2$, together make the full distance between the planets.

I then figure out the speed by $\frac{v^2}{r_1}=a_1$

The same could be said for the other planet:

$acceleration_2=\frac{\frac{G*m_1*m_2}{(r_1+r_2)^2}}{m_2}$ where $r_1$ and $r_2$, together make the full distance between the planets.

I then figure out the speed by $\frac{v^2_2}{r_2}=a_2$

The distance around the whole circle (The path they will they travel if they travel in a circle) is $r_1π2$ and $r_2π2$

I then figure out the time it takes to rotate one period by dividing distance by speed and then rotate the planets round the radius(from the planet to the mass center) in that time manner.

I now read that because their masses are unequal they are not supposed to rotate in a circle. Have I done this incorrectly?

This is how it looks at the moment:

https://gyazo.com/5c08ed6eec4f6af5758c83baaf7572be

David Lund
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2 Answers2

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Revised Answer in response to your comments :

Your calculations are correct.

Generally the orbits will be ellipses, but the 2 planets can each orbit the CM in circles with constant speed around the circle. (The acceleration is a change in direction rather than a change in speed.) If the masses are equal they will orbit on the same circle; if they are different the circles will be concentric. Either way, they always keep on opposite sides of the CM and therefore have the same period.

Your animation gives the impression that the planets are the same size (and therefore mass). If so, their orbits are incorrect, because the CM rotates.

sammy gerbil
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Force, in your case you have written down the equation for the force for Newtonian gravity, is not the same as acceleration. So either equate forces, in which case you would have to use the centripetal force and not acceleration, or equate accelerations, in which case you can use Newton's second law to find the acceleration from the force obtained using Newtonian gravity. Both methods will result in the same answer.

And in general binary stars will orbit each other in elliptical orbits, but a circular orbit is a special case of an elliptical orbit, namely with an eccentricity of zero. Since for elliptical orbits in general there is not closed form solution for the position as a function of time (it would require you to iteratively solve Kepler's equation), therefore in your case it might be easier to just stick to the circular orbits.

fibonatic
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