Chapter 7, Problem 7.1P

To determine

The proof that the semi major axis of the orbit of the reduced mass is given by $a=a_1+a_2$.

Answer to Problem 7.1P

The semi major axis of the orbit of the reduced mass is $a_1+a_2$.

Explanation of Solution

Write the expression for general relation between the distance and semi major axis of the star orbit.

    $r=\frac{a\left(1-e^2\right)}{1+e\cos \theta }$        (1)

Here, $r$ is the distance, $e$ is the eccentricity of its orbit and $a$ is the semi major axis.

Write the expression for distance when two stars are in different direction.

    $\begin{array}{c}r=r_1-\left(-r_2\right)\\ =r_1+r_2\end{array}$        (2)

Here, $r_1$ is the position vector of first star and $r_2$ is the position vector of second star.

Write the expression for relation between distance of the stars and their semi major axis.

    $\begin{array}{c}\frac{r_2}{r_1}=\frac{a_2}{a_1}\\ r_2=\frac{a_2}{a_1}{r}_1\end{array}$        (3)

Here, $a_1$ is the semi major axis of first star and $a_2$ is semi major axis of second star.

Write the expression for semi major axis of reduced mass.

    $a=\frac{r_p+r_a}{2}$        (4)

Here, $r_p$ is the perihelion distance and $r_a$ is the aphelion distance.

Conclusion:

Substitute $\frac{a_2}{a_1}{r}_1$ for $r_2$ equation (2).

    $\begin{array}{c}r=r_1+\frac{a_2}{a_1}{r}_1\\ =r_1\left(\frac{a_1+a_2}{a_1}\right)\\ =\left(a_1+a_2\right)\left(\frac{r_1}{a_1}\right)\end{array}$        (5)

Substitute $\frac{\left(1-e^2\right)}{1+e\cos \theta }$ for $\frac{r_1}{a_1}$ in equation (4).

    $r=\left(a_1+a_2\right)\frac{\left(1-e^2\right)}{1+e\cos \theta }$        (6)

Substitute $0°$ for $\theta$ in equation (6).

    $\begin{array}{c}{r}_p=\left(a_1+a_2\right)\frac{\left(1-e^2\right)}{1+e\cos \left(0°\right)}\\ =\left(a_1+a_2\right)\frac{\left(1-e^2\right)}{1+e}\\ =\left(a_1+a_2\right)\left(1-e\right)\end{array}$

Substitute $180°$ for $\theta$ in Equation (6).

    $\begin{array}{c}{r}_a=\left(a_1+a_2\right)\frac{\left(1-e^2\right)}{1+e\cos \left(180°\right)}\\ =\left(a_1+a_2\right)\frac{\left(1-e^2\right)}{1-e}\\ =\left(a_1+a_2\right)\left(1+e\right)\end{array}$

Substitute $\left(a_1+a_2\right)\left(1-e\right)$ for $r_p$ and $\left(a_1+a_2\right)\left(1+e\right)$ for $r_a$ in Equation (4).

    $\begin{array}{c}a=\frac{\left(a_1+a_2\right)\left(1-e\right)+\left(a_1+a_2\right)\left(1+e\right)}{2}\\ =2\frac{\left(a_1+a_2\right)}{2}\\ =a_1+a_2\end{array}$

Thus, the semi major axis of the orbit of the reduced mass is $a_1+a_2$.