# Find semi-minor axis of elliptical binary orbit given semi-major axis and mass ratio

I'm working on a simulation of binary stars. I'm using Kepler's Laws to solve for the angular velocity given the semi-major axis $$a$$ and the masses $$m_1$$ and $$m_2$$ of the two objects in an elliptical orbit. First, I'm using Kepler's 3rd Law to solve for the period: $$T^2 = \frac{a^3}{m_1 + m_2}$$ where $$T$$ is the orbital period in years, $$a$$ is the semi-major axis of the orbit in AU, and $$m_1$$ and $$m_2$$ are the masses of the two objects in orbit in units of solar masses. Next, I'm using Kepler's 2nd Law to solve for the angular velocity: $$\frac{dA}{dt} = \frac{r^2}{2}\frac{d\theta}{dt}$$ where $$r$$ is the instantaneous magnitude of the position vector, and $$\frac{d\theta}{dt}$$ is the angular velocity of the object in orbit. Since the areal velocity is constant, we can replace it with the area swept out in one period divided by the length of the period: $$\frac{A}{T} = \frac{r^2}{2}\frac{d\theta}{dt}$$ and rearrange to solve for the angular velocity: $$\frac{d\theta}{dt} = \frac{2A}{Tr^2}$$ Now I need to know the area $$A$$ of the elliptical orbit in order to solve for the angular velocity. In this case $$A = \pi a b$$, so I am trying to solve for $$b$$ with the parameters I'm given: $$a$$, $$m_1$$, and $$m_2$$. I went back to the original definitions for $$a$$ and $$b$$, $$a = \frac{r_{max} + r_{min}}{2}$$ and $$b = \sqrt{r_{min} {r_{max}}}$$, where $$r_{max}$$ is the maximum distance from a focus of the ellipse to the endpoint of its major axis and $$r_{min}$$ is the minimum distance. I think I can use these relationships to solve for $$b$$ using the mass ratio of the objects, because I think $$r_{max}$$ and $$r_{min}$$ are related to this ratio.

However, I'm having trouble with deriving this because I don't understand the relationship between the parameters of each individual elliptical orbit (absolute orbit-- see the attached image) and the orbit where the larger mass $$(m_1)$$ is the reference point (relative orbit) which is the orbit for which the semi-major axis $$a$$ is given. As I understand it, the center of mass is one of the foci of the relative orbit (and both absolute orbits) and the distances of each mass from the CM are $$r_1$$ and $$r_2$$. These relative distances are equal to the mass ratio by the definition of the center of mass. I also have the relationship $$m_1 a_1 = m_2 a_2$$ and $$a_1 + a_2 = a$$ where $$a_1, a_2$$ are the individual semi-major axes of the absolute orbits, but I wasn't able to use these equations to solve for $$b$$. I tried the relationship: $$m_1 r_{min} = m_2 r_{max}$$ Using that and the definitions for $$a$$ and $$b$$ came up with the result: $$b = \frac{2a^2}{1 + \frac{m_2}{m_1}} \sqrt{\frac{m_2}{m_1}}$$ which I checked against data of planets' orbits in the solar system and found that it is wrong. It would be greatly appreciated if anyone could help me solve this problem or point out to me which incorrect assumption I made in my attempt. Thank you!

• this is two body problem with $r={\frac {a \left( 1-{e}^{2} \right) }{1+e\cos \left( \varphi \right) }}$ you obtain $r_1=\frac{m_2}{M}\,r$ and $r_2=-\frac{m_1}{M}\,r$ the center of mass condition $m_1\,r_1+m_2\,r_2$ must be equal to zero. with this equations you can't solve your problem – Eli May 28 at 15:01
• Agreed. You don't have enough information. Given just elliptical semimajor axis $a$ and mass ratio, all semi-minor axes $b$ such that $0<=b<=a$ are valid solutions. (previous version of my comment called $b$ a semimajor axis) – notovny May 28 at 15:41