# How does the orbital motion of reduced mass tell us how the individual planets/stars move?

To give context to my question, here is Kepler's first Law restated in terms of reduced mass. Here, $$\vec{r}$$ is the position vector of the reduced mass whose origin lies at the centre of mass of the binary system. L is the angular momentum of the system, e is eccentricity and $$\mu$$ is reduced mass

In the second sentence that I highlighted, it is mentioned that when equation 29 is translated into a physical standpoint, it implies that "both objects in a binary orbit move about the centre of mass in ellipses, with the centre of mass occupying one focus of each ellipse" however I don't see how the physical implications of equation 29 will translate into this that they claim.

My interpretation of eqn (29) is that the position vector $$\vec{r}$$ of the reduced mass is also the relative distance between the 2 masses, therefore if the reduced mass is undergoing elliptical motion about its centre of mass, it is equivalent to taking the reference frame one of the objects in the binary system and measuring the position vector of the other object since $$\vec{r}$$ is relative distance. Therefore I feel that what this equation is saying, is that in a binary system, the objects will undergo elliptical motion with respect to the other object. I don't see why you can say each object undergoes elliptical motion with respect to the centre of mass, as reduced mass only gives you information about relative motion with respect to one another. Any clarification would help.

## 2 Answers

I think the choice of a specific set of coordinates is somewhat blurring the answer to your question.

Let start from the beginning and let us follow the main steps required to arrive the Kepler's first law in the form of the equation (29) in your figure (the polar representation of a conic).

The usual strategy is to write down the equation of motions for two point-like masses $$m_1$$ and $$m_2$$ in an inertial reference frame:

\begin{align} m_1 {\bf \ddot r}_1 &= G\frac{m_1 m_2}{|{\bf r}_2-{\bf r}_1|^3}({\bf r}_2-{\bf r}_1)\\ m_1 {\bf \ddot r}_2 &= G\frac{m_1 m_2}{|{\bf r}_2-{\bf r}_1|^3}({\bf r}_1-{\bf r}_2) \end{align}

It turns down that in terms of the following new variables the equations are simpler: \begin{align} {\bf r} &= {\bf r}_2-{\bf r}_1\\ {\bf R} &= \frac{m_1 {\bf r}_1 + m_2 {\bf r}_2 }{m_1+m_2} \end{align} Indeed, we get the following equations: \begin{align} {\bf \ddot R} &= 0 \tag{1}\\ \mu {\bf \ddot r} &= -G\frac{m_1 m_2}{|{\bf r}|^3}{\bf r} \tag{2} \end{align} where the reduced mass $$\mu$$ is defined as $$\frac{1}{\mu}=\frac{1}{m_1}+\frac{1}{m_2}$$. The interpretation of equation ($$1$$) is that the center of mass moves with a uniformly rectilinear motion. Equation ($$2$$) can be interpreted as the equation of motion of a body of mass $$\mu$$ subject to the force due to Newton's gravitational law. It is the orbit obtained by solving equation ($$2$$) that is shown in equation ($$29$$) of your text.

The inverse change of coordinates: \begin{align} {\bf r}_1 &= {\bf R}-\frac{m_2}{m_1+m_2}{\bf r} \tag{3}\\ {\bf r}_2 &= {\bf R}+\frac{m_1}{m_1+m_2}{\bf r} \tag{4} \end{align} is the key ingredient to understanding how the trajectory obtained by solving equation ($$2$$) for the relative position vector $${\bf r}$$ can be translated in terms of the two bodies' trajectories in the original inertial frame.

Indeed, equations ($$3$$) and ($$4$$) directly show that at each time $$t$$ the center of mass $$\bf R$$ and the two positions $${\bf r}_1$$ and $${\bf r}_2$$ keep aligned. Moreover, the distance from the center of mass is proportional to the ratio between each body's mass and the total mass.

Therefore, if, by solving the equation ($$2$$), we obtain an ellipse described by the vector $${\bf r}$$, such that the origin of the vector is in one focus, equations ($$3$$) and ($$4$$) show that each body $$i$$ has a trajectory that is an ellipse rescaled by the factor $$\frac{m_i}{m_1+m_2}$$, relatively to that described by $${\bf r}$$, but now the focus is occupied by the center of mass position.

Notice that the introduction of the relative position $${\bf r}$$ is something more than a purely mathematical passage. It is consistent with a description of the problem in a non-inertial reference frame centered on one of the two bodies (body $$1$$, with the definition above). Interestingly, the appearance of the reduced mass can be obtained as the effect of writing down the equation of motion for body $$2$$ in a non-inertial reference frame by taking into account the acceleration of body $$1$$ due to the gravitational force due to body $$2$$. In this way, we can see more clearly the coexistence of two different descriptions of the same phenomenon:

1. in an inertial reference frame, we have two bodies performing each an elliptic motion with the center of mass at the common focus.
2. in the non-inertial reference frame centered on each of the two bodies, the other is describing an elliptic orbit having the one focus at the origin.

A final word of caution is on the geometric description of the orbits. While in the case of the reference frames centered on one of the two bodies, there is a unique description of the orbit as a conic section having the other body at one focus, in the case of the description in an inertial reference frame, it is only in the reference frames where the center of mass is at rest that one can speak about conic sections. In the most general inertial reference frames, the orbit's shape is the composition of a pure translation with the conical trajectory.

• Thank you for that informative reply! I have 1 clarification to make, so the conclusion from equations (3) and (4) if we sub in the equation for an ellipse in $\vec{r}$, would be that the position vector of each mass is an ellipse, scaled down by $-\frac{m_i}{m_1+m_1}$ along with a translation of $\vec{R}$ units along that same axis. Would I be right to say this? I should have awarded the bounty to you, but I forgot to check stackexchange for a while so the bounty was auto awarded, my sincere apologies. Commented Jan 12, 2021 at 4:40
• @LucasTan, No problem. The conclusion is what you have summarized, with the exception that the scaling factor is the absolute value of the ratio between the mass of one body and the total mass, while the sign this scaling factor enters in formulae (3) and (4) depends on the choice of the definition of $\bf r$ between the two possibilities (position 1 - position 2) or (position 2 - position 1). Commented Jan 12, 2021 at 7:04

Here is the equation of an ellipse in polar coordinate WHEN the origin of the polar coordinate is in a FOCAL point.

$$r(\theta) = \frac{a(1-e^2)}{1+e \cos (\theta)} \;\;\;\;\;\;\;\;\;\;\;\;Eq.2$$

This is not the only way to write down the equation of an ellipse. we could have chosen the center of the polar coordinate to be at the center of an ellipse, then

$$r(\theta) = \frac{b}{\sqrt{1-(e \cos (\theta))^2}} \;\;\;\;\;\;\;\;\;\;\;\;Eq.1$$

we see that in our physics equations we ended up with Eq.1 which tell us the origin of the polar coordinate is in a focal point which was our center of mass.

Now if you want to look at the position of each mass all you have to do is use one of the following equations

$$r(\theta) = \frac{m_1 + m_2}{m_2} r_1(\theta) \\ r(\theta) = -\frac{m_1 + m_2}{m_1} r_2(\theta)$$.

You can see that when you only look at the trajectory of one mass, we still have an equation similar to Eq.1. The only difference is that it is multiplied by a factor.

• Thank you, if I'm understanding you correctly, if I wanted to find the position vector of let's say m1, I would need to multiply the formula derived by the textbook with the factor $-\frac{m_2}{m_1+m_2}$, which would result in an elliptical orbit for m1 with respect to the foci. Would I be right to say this? Commented Dec 28, 2020 at 4:34