# Mathematical model for this graph of a simplified binary star system?

Unnecessary background for question: I had a school assignment asking us to relate a quadratic equation to a real life example relating to our future dream career, making sure to express the accuracy of the equation (knowing it would be practically impossible to find an exact match considering parabolas go on infinitely both ways on a graph). I want to be a theoretical physicist and so the best I could come up with was modeling the velocity of a star of a simplified binary star system.

Here is an animation of the simplified model in question. It assumes inertia is never lost. Here is a rough graph modeling a star from the system. So, is there a good mathematical equation for this graph? The equation should model the spikes out infinitely.

This problem can be treated as an elliptical Kepler orbit. But for a Kepler orbit it is assumed that one mass is much more massive than the other: $m_1>>m_2$, which means that $m_1$ basically sits still at their center of mass.

But both masses will move symmetrically around their center of mass (if one moves closer the other will move closer as well, inversely proportional to their masses), which allows you to write their attractive force as a function of their distance towards their center of mass ($r_{COM}$). $$F_1=\frac{Gm_1m_2}{\left(\frac{m_1+m_2}{m_2}r_{COM}\right)^2}$$ This only scales the force and is equal to an object orbiting a much more massive object (located at the previous center of mass), like a Kepler orbit, but with different masses.

However Kepler orbits do not have an explicit solution for the position as a function of time, these are often calculated numerically. But you can calculate it explicitly the other way around, time as a function of position and the exact trajectory an object will take can also be found explicitly.

Edit:
You can also approximate an orbit with a Fourier series which has the advantage that it goes on infinitely, but will contain small errors. I did some testing with this and got the following results:
$$\theta(t)=\bar{\omega}t+\sum_{n=1}^\infty{A(n)\sin\left(n\bar{\omega}t\right)}$$ for $e=0.5$ (orbital eccentricity) $A(n)\approx 0.9757633n^{-1.944954}$. $$v(t)=\sqrt{\frac{\mu}{a(1-e^2)}(1+e(2\cos{\theta(t)}+e))}$$ When choosing $\frac{\mu}{a}=1$, using the approximation and a limited sum of $n=50$, the graph looks like this: I replicated your sketch using the equation $$f(x) = 10e^{-|5-x|}+10e^{-|-5-x|}+2$$

The $\pm5$ in the exponential changes the position of the spike, $10$ controls the upper limit while the $2$ controls its lower limit. ## But...

This isn't really right for a highly eccentric orbit. Your animation clearly shows the stars traveling in opposite directions in a sinusoidal motion and your sketch does not replicate this. I suggest taking a look at something like Nightfall, which models binary eclipses.

• The equation should model the spikes out infinitely. – David Ball Oct 10 '13 at 21:27
• @Mr.Fate: That is completely unphysical then. – Kyle Kanos Oct 10 '13 at 21:57
• That's why I said simplified model assuming no inertia is lost. – David Ball Oct 10 '13 at 22:31
• The unphysical behaviour is the velocity spiking to infinity: they wouldn't be bound to each other & would fly away after the first orbit. – Kyle Kanos Oct 10 '13 at 23:34
• If you want to model this for an infinite periods you can only use a summation of (co)sines (or a sum of $\left(e^{i\omega t}+e^{-i\omega t}\right)$, but that is just equal to $2\cos{\omega t}$). – fibonatic Oct 11 '13 at 8:16