The General two Body Problem I -- Elementary Conservation Laws

In this section we will investigate the two body problem in more general terms. To do this consider the situation in the figure below. There two bodies of masses $m_1$ and $m_2$ are at locations ${\bf r}_1$ and ${\bf r}_2$. We also assume that the only acting force is the gravitational attraction between the two bodies:

$${\bf F}_{21}=G\frac{m_1m_2}{\vert{\bf r}_1-{\bf r}_2\vert^3... ... r}_1-{\bf r}_2)=G\frac{m_1m_2}{r^3}({\bf r}_1-{\bf r}_2),$$

is the force acting on the body $m_2$, and ${\bf F}_{12}=-{\bf F}_{21}$ is the force acting on $m_1$.

Figure 1: The geometry of the general two body problem

h

Newton's law implies

$$m_1\frac{d^2{\bf r}_1}{dt^2} = {\bf F}_{12}$$ (9)

$$m_2\frac{d^2{\bf r}_2}{dt^2} = {\bf F}_{21}$$ (10)

Adding these two equations we get:

$$0=m_1\frac{d^2{\bf r}_1}{dt^2}+m_2\frac{d^2{\bf r}_2}{dt^... ...t(m_1\frac{d{\bf r}_1}{dt}+m_2\frac{d{\bf r}_2}{dt}\right),$$

or

$${\bf P}=m_1\frac{d{\bf r}_1}{dt}+m_2\frac{d{\bf r}_2}{dt}$$

is constant. This is known as conservation of momentum. It can be further interpreted as $\frac{d{\bf\rho}}{dt}=0$, where

$${\bf\rho}=\frac{m_1}{m_1+m_2}{\bf r}_1+\frac{m_2}{m_1+m_2}{\bf r}_2,$$

is the location of the center of mass of the two bodies. We have thus established

Conservation of Linear Momentum:

The center of mass of a two body system with no exterior forces moves at a uniform speed.

Since the motion of the center of mass is trivial we will ignore it for the remainder of this derivation. We will put the the origin of our reference frame at the center of mass.

To continue we multiply the equations motion by $m_2$ and $m_1$ respectively and subtract the from each other to get

$$\frac{d^2}{dt^2}\left({\bf r}_1-{\bf r}_2\right)= -\frac{G... ...\bf r}_1-{\bf r_2}\vert^3}\left({\bf r}_1-{\bf r}_2\right).$$

We can further simplify this by introducing $r_1$ and $r_2$, where $r_j$ is the distance between $m_j$ and the center of mass and set

$${\bf r}_1-{\bf r_2}=\left(r_1+r_2\right){\bf\hat{r}}$$

Furthermore, we introduce the reduced masses:

$$M_R(m_1) = \frac{m_2^2}{m_1+m_2}$$ (11)

$$M_R(m_2) = \frac{m_1^2}{m_1+m_2}$$ (12)

If we only consider the motion of the body $m_1$, we can write:

$$m_1\frac{d^2{\bf r}_1}{dt^2} =-\frac{G m_1 M_R(m_1)}{r_1^2}{\bf\hat{r}}$$

and if we consider the motion of $m_2$:

$$m_2\frac{d^2{\bf r}_2}{dt^2} =-\frac{G m_2 M_R(m_2)}{r_2^2}{\bf\hat{r}}$$

If one considers the motion of a small satellite of mass $m$ around earth,. The reduced mass $M_R\approx M_E$ is nearly equal to the mass of the earth and the problem reduces to a one body problem:

$$m\frac{d^2{\bf r}}{dt^2} =-\frac{G m M_E}{r^2}{\bf\hat{r}},$$ (13)

where $M_E$ is conveniently chosen to sit at the origin. A similar simplification can be done when one considers the motion of a small planet around a large sun.

To continue we consider the motion of the mass $m_1$:

$$m_1\frac{d^2{\bf r}_1}{dt^2} =-\frac{G m_1 M_R}{r_1^3}{\bf r}_1$$

This equation can be skalar multiplied by $\frac{d{\bf r}_1}{dt}$ to get

$$\begin{eqnarray*} \frac{m_1}2\frac{d}{dt}\left(\frac{d{\bf r}_1}{dt}\cdot\frac{... ...{\bf r}_1\\ &=& Gm_1 M_R\frac{d}{dt}\left(\frac1{r_1}\right). \end{eqnarray*}$$

or

$$\frac12 m_1 v_1^2-\frac{Gm_1M_R}{r_1} = K.$$ (14)

The quantity on the left of this equation is called the total energy, and we have established:

Conservation of Total Energy:

The total energy of the two body problem is constant.

For this last result, we computed the skalar product of the equation of motion and the velocity. To continue we compute the vector product of the equation of motion and ${\bf r}_1$. We will have ${\bf r}_1\times {\bf r}_1=0$ on the right and get:

$${\bf r}_1\times m_1\frac{d^2{\bf r}_1}{dt^2} =0$$

Observe that

$$\frac{d}{dt}\left({\bf r}_1\times\frac{d{\bf r}_1}{dt}\righ... ...d{\bf r}_1}{dt} + {\bf r}_1\times \frac{d^2{\bf r}_1}{dt^2},$$

hence we have

$${\bf h}=m_1 {\bf r}_1 \times \frac{d{\bf r}_1}{dt},$$ (15)

is constant.

Conservation of Angular Momentum:

The Angular Momentum ${\bf h}$ as defined in (15) is constant.

This is the third so-called conservation law. These laws are also called first integrals of the equation of motion.

The conservation of angular momentum has an important consequence. Observe that the vector ${\bf h}$ is perpendicular to the pair of vectors ${\bf r}_1$ and ${\bf v}_1=\frac{d{\bf r}_1}{dt}$. Since ${\bf h}$ is a constant vector, it is normal to a constant plane containing ${\bf r}_1$ and ${\bf v}_1$, i.e. the two body problem is a planar (two-dimensional) problem.