Astrophysics Series: Derivation of the Total Energy of a Binary Orbit

SOURCE FOR CONTENT: An Introduction to Modern Astrophysics, Carroll & Ostlie, Cambridge University Press. Ch.2 Celestial Mechanics

Here is my solution to one of the problems in the aforementioned text. I derive the total energy of a binary system making use of center-of-mass coordinates. In order to conceptualize it I have used the binary Alpha Centauri A and Alpha Centauri B. While writing this I stumbled upon the Kepler problem, the two-body problem, and the N-body problem. Leave a comment if you would like me to consider that in another post.

Clear Skies!

Derivation of the Total Energy of a Binary Orbit:

Setup: Consider the nearest binary star system to our solar system: Alpha Centauri A and Alpha Centauri B. These two stars orbit each other about a common center of mass; a point called a barycenter. The orbital radius vector of Alpha Centauri A is $\textbf{r}_{1}$ and the orbital radius vector of Alpha Centauri B is $\textbf{r}_{2}$ . The masses of Alpha Centauri A and B are $m_{1}$ , and $m_{2}$ , respectively. The total mass of the binary orbit $M$ is the sum of the individual masses of each component. In the context of this system, we encounter what is called the two-body problem of which there exists a special case known as the Kepler Problem (by the way let me know if that would be something that you guys would want to see…). We can simplify this two-body problem by making use of center-of-mass coordinates wherein we define the reduced mass $\mu$ . Therefore, the derivation of the total energy of the binary system of Alpha Centauri A and B will be carried out in such a coordinate system.

To derive this energy equation, one would typically make use of center-of-mass coordinates in which

$\displaystyle \textbf{r}_{1}=-\frac{\mu}{m_{1}}r, (0.1)$

and

$\displaystyle \textbf{r}_{2}=\frac{\mu}{m_{2}}r, (0.2)$

where $\mu$ represents the reduced mass given by

$\displaystyle \mu\equiv \frac{m_{1}m_{2}}{m_{1}+m_{2}}=\frac{m_{1}m_{2}}{M}. (0.3)$

Recall from conservation of energy that

$\displaystyle E = \frac{1}{2}m_{1}\dot{r}_{1}^{2}+\frac{1}{2}m_{2}\dot{r}_{2}^{2}-G\frac{m_{1}m_{2}}{|\mathcal{R}|}, (1)$

where $|\mathcal{R}|$ represents the separation distance between the two components. Let us take the derivative of Eqs.(0.1) and (0.2) to get

$\displaystyle \dot{r}_{1}=-\frac{\mu}{m_{1}}v, (2.1)$

and

$\displaystyle \dot{r}_{2}= \frac{\mu}{m_{2}}v. (2.2)$

Substitution yields

$\displaystyle E = \frac{1}{2}\frac{\mu^{2}}{m_{1}}v^{2}+\frac{1}{2}\frac{\mu^{2}}{m_{2}}v^{2}-G\frac{m_{1}m_{2}}{|\mathcal{R}|}. (3)$

Upon making use of the definition of the reduced mass (Eq. (0.3)) we arrive at

$\displaystyle E = \frac{1}{2}\mu v^{2}-G\frac{M \mu}{|\mathcal{R}|}. (4)$

If we solve for $m_{1}m_{2}$ in Eq.(0.3) we get the total energy of the binary Alpha Centauri A and B. This is true for any binary system assuming center-of-mass coordinates.

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