The Gravitational Two-Body Problem

Section 12.4 The Gravitational Two-Body Problem

We will consider motion of two bodies interacting with gravitational force only. We will show that the motion of the two bodies separates into a motion of a fictitious body of their combined mass at the center of motion of the two bodies and another body of reduced mass at a position corresponding to the relative position of the two bodies.

Let \(m_1\) and \(m_2\) be the masses and \(\vec{r}_1\) and \(\vec{r}_2\) their positions. Their separate equations of motions are

\begin{align} m_1 \vec{a}_1 \amp = G\dfrac{m_1m_2}{|\vec{r}_1 - \vec{r}_2|^2}\, \hat u_{1\rightarrow 2}, \tag{12.15}\\ m_2 \vec{a}_2 \amp = -G\dfrac{m_1m_2}{|\vec{r}_1 - \vec{r}_2|^2}\, \hat u_{1\rightarrow 2}, \tag{12.16} \end{align}

where \(\hat u_{1\rightarrow 2}\) is a unit vector in the direction of \(1\rightarrow 2\text{.}\) The unit vector is needed to express the direction of the vector equation analytically. Here, \(| \vec{r}_1 - \vec{r}_2 |\) is the distance between the two bodies.

Adding the two equations of motion, we get a simple equation.

\begin{equation*} m_1 \vec a_1 + m_2 \vec a_2 = 0. \end{equation*}

Now, when we divide this equation by the total mass \(m_1+m_2\text{,}\) we see that the left side is the acceleration of the center of mass, denoted by \(\vec A_\text{cm}\text{.}\) Thus

\begin{equation} A_\text{cm} = 0.\tag{12.17} \end{equation}

The difference of the two equations of motion, Eqs. (12.15) and (12.16), gives us another useful equationl and equation obeyed by the separation vector between the two masses.

\begin{equation} m_1 \vec{a}_1 - m_2 \vec{a}_2 = 2G\dfrac{m_1m_2}{|\vec{r}_1 - \vec{r}_2|^2}\, \hat u_{1\rightarrow 2}.\tag{12.18} \end{equation}

We will call these equations of motion in the LAB frame. Below, we derive another set of equations of motion in a more convenient frame called the center of mass or CM-frame.

Eq. (12.17) says that the CM has no acceleration. Therefore, if we choose a coordinate system with origin at the CM, it will also be an inertial frame. We say that we are using CM Frame. The coordinates are shown in Figure 12.28. We will find that using the CM-frame also simplifies the treatment of the separation coordinate also. We will denote quantities with respect to the CM coordinate system by placing a prime on the symbols.

Figure 12.28. The motion of two particles \(m_1\) and \(m_2\) interacting with gravitational force only are simpler if described with respect to the center of mass. The position vectors have the following relation: \(\vec r_1 = \vec r^{\,\prime}_1 + \vec R_\text{cm}\) and \(\vec r_2 = \vec r^{\,\prime}_2 + \vec R_\text{cm}\text{.}\) The separation coordinate between the two is \(\vec r = \vec r_1 - \vec r_2 = r^{\,\prime}_1 - r^{\,\prime}_2 \text{.}\) Note the direction of vector \(\vec r\) is from \(2\rightarrow1\text{,}\) the unit vector \(\hat u_{2\rightarrow 1}\text{.}\)

Equations of Motion in the CM Frame:

Let \(\vec r^{\,\prime}_1 \) \(\vec r^{\,\prime}_2 \) be the postions of the two particles with respect to their CM. Then, we have the following relations between the original coordinates (also called LAM frame) and the new ones (the CM frame).

\begin{gather} \vec r^{\,\prime}_1 - \vec r^{\,\prime}_2 = \vec{r}_1 - \vec {r}_2, \tag{12.19}\\ m_1\vec r^{\,\prime}_1 + m_2 \vec r^{\,\prime}_2 = 0. \tag{12.20} \end{gather}

For convenience we give the relative position of particle 1 with respect to particle 2 its own symbol.

\begin{equation} \vec r = \vec r_1 - \vec r_2 = \vec r^{\,\prime}_1 - \vec r^{\,\prime}_2,\tag{12.21} \end{equation}

and acceleration of this coordinate, i.e., the relative acceleration,

\begin{equation} \vec a = \text{ acceleration of relative position } \vec r.\tag{12.22} \end{equation}

Now, notice the following for \(\vec a\text{:}\)

\begin{align*} \vec a \amp = \frac{d^2 \vec a}{dt^2} = \frac{d^2 (\vec r_1 - \vec r_2) }{dt^2}\\ \amp = \frac{d^2}{dt^2} \left(\vec r_1 - \frac{M}{m_2}\vec R + \frac{m_1}{m_2} \vec r_1 \right), \ \ \ \text{using } m_1\vec r_1 + m_2 \vec r_2 = M \vec R, \\ \amp = \frac{d^2}{dt^2}\left( \frac{m_1 + m_2}{m_2} \vec r_1 \right), \ \ \ \text{using } \frac{d^2\vec R}{dt^2} = \vec A_\text{cm} = 0, \\ \amp = \frac{m}{\mu} \vec a_1, \end{align*}

where I inroduced a combination of masses, called the reduced mass, denoted by \(\mu\text{.}\)

\begin{equation} \mu = \dfrac{m_1 m_2 }{m_1 + m_2},\tag{12.23} \end{equation}

and total mass by \(M\)

\begin{equation} M = m_1 + m_2.\tag{12.24} \end{equation}

From this relation, we get

\begin{equation*} \mu\,\vec a = m_1 \vec a_1. \end{equation*}

The right side can be replaced by the force on 1 my 2,

\begin{equation*} \mu\,\vec a = G\dfrac{m_1m_2}{|\vec{r}_1 - \vec{r}_2|^2}\, \hat u_{1\rightarrow 2}, \end{equation*}

where we can replace the separation vector by \(\vec r\text{,}\) and \(m_1m_2\) by \((m_1 + m_2)\,\mu\text{.}\) After some algebra, we get the following for the magnitude of the relative acceleration,

\begin{equation} \mu\,\vec a = -G_N \dfrac{ M\,\mu}{r^2}\,\hat u,\tag{12.25} \end{equation}

where \(\hat u\) is a unit vector pointed in direction 2 to 1, which is the direction of the origin of the CM system, i.e., the CM. Or, as differential equation for the relative coordinate alone,

\begin{equation} \mu\,\frac{d^2\vec r}{dt^2} = -G_N \dfrac{ M\,\mu }{|\vec r|^2}\,\hat u,\tag{12.26} \end{equation}

That is, the original coupled systems of equation of the two-body system in (12.15) and (12.16), has been decoupled into two separate one-body problems: (1) Eq. (12.17) of the CM and (2) Eq. (12.25) for the relative coordinate in the center of frame.

If you were to look at the original two particles from CM, you would find their relative motion to be described by a single particle of mass \(\mu\) moving in the field of a particle of mass \((m_1 + m_2)\) placed at the origin as illustrated in Figure 12.29. If \(m_1\lt\lt m_2\text{,}\) then, the CM will be very near \(m_2\) and \(\mu\approx m_1\text{,}\) i.e., in this case, it would appear that \(m_1\) is moving in the field of \(m_2\) while \(m_2\) is fixed, as is the case for Earth-Sun system.

Figure 12.29. The problem of two particles \(m_1\) and \(m_2\) interacting with gravitational force only in CM frame becomes study of motion of one particle of mass \(\mu\) moving in the gravitational field by an object of mass \(M\) at the origin, which is at the CM.

Going Between CM and LAB Frames:

Using the definition of CM in terms of the coordinates of the individual objects in the CM frame, we find the following relation between the coordinates of the two bodies in the CM frame. We have the following starting equation.

\begin{align*} \amp m_1\vec r_1 + m_2\vec r_2=M\vec R\\ \amp m_1\vec r^{\,\prime}_1 + m_2\vec r^{\,\prime}_2=0. \\ \amp \vec r = \vec r_1 - \vec r_2 = \vec r^{\,\prime}_1 - \vec r^{\,\prime}_2 \end{align*}

Once we solve Eq. (12.26), we will get \(\vec r\text{.}\) From this we can get the individual coordinates using these relations.

\begin{align*} \amp \vec r^{\,\prime}_1 = \frac{\mu}{m_1}\,\vec r\\ \amp \vec r^{\,\prime}_2 = -\frac{\mu}{m_2}\,\vec r \end{align*}

Now, if we also solve

\begin{equation*} \frac{d^2 R}{dt^2} = 0, \end{equation*}

we will also have \(\vec R\) and then, we get the LAB coordinates to locate the particles in Lab frame.

\begin{align*} \amp \vec r_1 = \vec R + \vec r^{\,\prime}_1. \\ \amp \vec r_2 = \vec R + \vec r^{\,\prime}_2. \end{align*}

A Preferred Analytic Approach:

Although, Eq. (12.26) simplifies the original LAB frame equations quite a bit, it is not the simplest way to study a two-body system. We will present a better approach based on energy and angular momentum conservation. While Eq. (12.26) is a three-dimensional vector equation, spherical symmetry of the force actualy helps them simplify to a one-dimensional problem when we look at conservation of energy in these problems. That would be preferred method!

Exercises Exercises

1. Radius of Star’s Orbit from Motion of a Campnion Planet.

A planet of mass \(m\) moves around a star of mass \(M\text{.}\) For an inertial observer the planet and the star appear to move in circles of radii \(r\) and \(R\) respectively. For \(m = M/4\) and \(r = 1.0\times10^{10}\ \textrm{m}\text{,}\) find radius \(R\) of the orbit of the star.

Answer.

\(2\times 10^9\ \textrm{m}\text{.}\)

Solution.

The radius of the orbit of the star will be the distance from the center of the star to the CM of the planet/star system. Let \(d\) be the distance from the star to the CM.

\begin{equation*} d = \frac{m r}{M+ m} = \frac{1}{5} r = 2\times 10^9\ \textrm{m}. \end{equation*}

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