Kepler Problem

1 Central Force and Central Potential

The central force on an object is a force directed along the line joining the object and the origin. We often describe such forces as

$\mathbf{F}(\mathbf{r})=F(\mathbf{r})\hat{\mathbf{r}},$ …(1)

where $\mathbf{r}$ is the position, or called radius vector, with $\hat{\mathbf{r}}$ being its unit vector (or called unit radial vector) and $r$ being its magnitude. $\mathbf{F}$ is the force as a vector-valued function of position. The (vector) field derived from a central force is called a central field.

Here we restrict our attention to conservative systems (i.e. without energy dissipation due to non-conservative forces, such as friction), then the magnitude $F(r)$ of a central force can always be expressed as the derivative of a time-independent potential energy function $U(r)$ :

$F(r)=-\dfrac{\text{d}U(r)}{\text{d}r}.$ …(2)

2 Two-Body Problem

In classical mechanics, the two-body problem deal with system containing two objects, viewed as point particles.

3 Two-Body Central Force Problem

For a two-body central force problem, we have $T=T_1+T_2$ and $U=U(r)$ , so the Lagrangian may be written as

$L=T-U=\dfrac{1}{2}m_1\left|\dot{\mathbf{r}}_1\right|^2+\dfrac{1}{2}m_2\left|\dot{\mathbf{r}}_2\right|^2-U(r).$ …(3)

3.1 Reduced Mass

In three-dimensional case, describing a two-body system requires the specification of six quantities. Consider two particles, with masses $m_1$ and $m_2$ , respectively. Let $\mathbf{r}_1$ amd $\mathbf{r}_2$ be their position vectors. We see that then there are six components here.

Yet, we have an alternative way to treat these six quantities. First, we want to re-formulate this two-body problem into two independent one-body problems. Second, we switch ourselves to another frame of reference, leaving out one one-body problem that is not of our interest.

In the original frame of reference, let $\mathbf{R}$ be the center-of-mass position vector, and $\mathbf{r}\equiv\mathbf{r}_1-\mathbf{r}_2$ be the displacement or separation vector (from Particle 2 to Particle 1). Then

$m_1\mathbf{r}_1+m_2\mathbf{r}_2=\mathbf{R}.$ …(4)

This equation, combined with $\mathbf{r}\equiv\mathbf{r}_1-\mathbf{r}_2$ , gives

$\left.\begin{array}{l}\mathbf{r}_1=\mathbf{R}+\frac{m_2}{m_1+m_2}\mathbf{r},\\ \mathbf{r}_2=\mathbf{R}-\frac{m_1}{m_1+m_2}\mathbf{r}.\end{array}\right\}$ …(5)

We have $\mathbf{R}\equiv\mathbf{0}$ in center-of-mass frame (COM frame, i.e. the one in which the center of mass remains at the origin)

$\left.\begin{array}{l}\mathbf{r}_1=\frac{m_2}{m_1+m_2}\mathbf{r},\\ \mathbf{r}_2=-\frac{m_1}{m_1+m_2}\mathbf{r}.\end{array}\right\}$ …(5′)

Put Equation (5′) into Equation (3) yields

$\begin{aligned} L&=\frac{1}{2}m_1\left|\frac{m_2\dot{\mathbf{r}}}{m_1+m_2}\right|^2+\frac{1}{2}m_2\left|-\frac{m_1\dot{\mathbf{r}}}{m_1+m_2}\right|^2-U(r)\\&=\frac{1}{2}\frac{m_1m_2(m_1+m_2)}{\left(m_1+m_2\right)^2} \left|\dot{\mathbf{r}}\right|^2-U(r)\\&=\frac{1}{2}\frac{m_1m_2}{m_1+m_2} \left|\dot{\mathbf{r}}\right|^2-U(r)\\&=\boxed{\frac{1}{2}\mu\left|\dot{\mathbf{r}}\right|^2-U(r)},\end{aligned}$ …(6)

where $\mu$ is the reduced mass,

$\mu\equiv\dfrac{1}{\frac{1}{m_1}+\frac{1}{m_2}}=\dfrac{m_1m_2}{m+1+m_2}.$ …(7)

We have therefore reduced the two-body central force problem to an equivalent one-body central force problem, in which we must determine only the motion of a “particle" of mass $\mu$ in the central field described by the potential function $U(r)$ .

3.2 Conservation Theorems

In this subsection, we shall use one of the Euler-Lagrange equations to evaluate the first integrals of the motion and prove that a specific quantity $\ell$ is constant throughout the motion.

First, we argue that the one-body central field system is two-dimensional.

The potential energy depends only on the separation of the particle from the force center $r$. (We can call such system spherical symmetrical.)
The system’s rotation about any axis throught the force center do not affect the motion.
Under the above conditions, the angular momentum vector of the system is conserved: $\mathbf{L}=\mathbf{r}\boldsymbol\times\mathbf{p}=\text{constant}$ .
Since $\mathbf{L}$ is fixed in space, $\mathbf{r}$ and $\mathbf{p}$ always lie in a fixed plane normal to $\mathbf{L}$ , that is to say, the problem is of two dimension.

Second, we express the Lagrangian in plane polar coordinates:

$\boxed{L=\frac{1}{2}\mu\left(\dot{r}^2+r^2\dot{\theta}^2\right)-U(r)}.$ …(8)

Applying the Euler-Lagrangian equations with respect to $\theta$ to Equation (8), we have

$\begin{aligned}&\frac{\partial L}{\partial \theta}-\frac{\text{d}}{\text{d}t}\frac{\partial L}{\partial \dot{\theta}}=0\\\implies&\dot{p}_\theta\equiv\frac{\partial L}{\partial \theta}=\frac{\text{d}}{\text{d}t}\frac{\partial L}{\partial \dot{\theta}}=0.\end{aligned}$ … (9)

(Note that $\dot{p}_\theta$ is the polar component of generalized force.)

We may integrate Equation (9) to obtain

$p_\theta\equiv\dfrac{\partial L}{\partial \dot{\theta}}=\mu r^2\dot{\theta}=\text{constant}.$ … (10)

It is the symmetry of the system that permits us to integrate directly one of the Euler-Lagrangian equations. The quantity $p_\theta$ is thus a first integral of the motion and we denote its constant value by the symbol $\ell$ :

$\ell\equiv \mu r^2\dot{\theta}=\text{constant}.$ … (11)

Third, we interpret this new quantity $\ell$ as follows: Referred the diagram below, we see that in describing the path $\mathbf{r}(t)$ , the radius vector sweeps out an area $\frac{1}{2}r^2\text{d}\theta$ in a time interval $\text{d}t$:

$\text{d}A=\dfrac{1}{2}r^2\text{d}\theta.$ … (12)

On dividing by the time interval, the areal velocity is shown to be

$\begin{aligned}\dfrac{\text{d}A}{\text{d}t}&=\dfrac{1}{2}r^2\dfrac{\text{d}\theta}{\text{d}t}=\dfrac{1}{2}r^2\dot{\theta}\\&=\dfrac{\ell}{2\mu}=\text{constant}.\end{aligned}$ … (13)

The result that the areal velocity remains constant in time is known as Kepler’s Second Law, which was inferred empirically by Johannes Kepler (1571-1630) after an exhaustive study of compilations of observational data.

Note that the conservation of areal velocity and of $\ell$ is a necessary condition for central-force motion.

How about other conservation theorems? The conservation of linear momentum is a natural result that add nothing new to the description of motion as long as we discuss the whole problem in COM frame. In addition, the conservation of energy is automatically ensured, for we have limited ourselves to nondissipative systems. Thus we have

$\begin{aligned}E&=T+U\\&=\dfrac{1}{2}\mu\left(\dot{r}^2+r^2\dot{\theta}^2\right)+U(r)\\&=\boxed{\dfrac{1}{2}\mu\dot{r}^2+\dfrac{1}{2}\dfrac{\ell^2}{\mu r^2}+U(r)}=\text{constant}\end{aligned}$ …(14)

3.3 Trajectory

Once the form of potential energy $U(r)$ is given, one can further determine the trajectory of motion in terms of $E$ and $\ell$ . We will present two methods to do so.

Method 1

Solving the third line Equation (14) for $\latex \dot{r}$, we have

$\dot{r}=\dfrac{\text{d} r}{\text{d} t}=\pm\sqrt{\dfrac{2}{\mu}(E-U(r))-\dfrac{\ell^2}{\mu^2r^2}}$ …(15)

Equation (15) is a (separable) differential equation of $r$ whose solution $r=r(t)$ is the radial equation motion.

But currently we are more interested in the equation of trajectory $r=r(\theta)$ or its inverse function $\theta=\theta(r)$ . To do so, we have to figure out the “relation between differentials of $\theta$ and $r$ “. With some inspection, the relation is shown to be

$\text{d}\theta=\dfrac{\text{d}\theta}{\text{d}t}\dfrac{\text{d}t}{\text{d}r}\text{d}r=\dfrac{\dot{\theta}}{\dot{r}}\text{d}r$ …(16)

Then we substitute $\latex \dot{\theta}=\ell/\mu r^2$ from Equation (11) and the expression of $\dot{r}$ form Equation (15) in Equation (16) and later integrate it to obtain

$\displaystyle\begin{aligned}\theta(r)&=\int \frac{(\ell/\mu r^2)\text{d}r}{\pm\sqrt{\frac{2}{\mu}(E-U(r))-\frac{\ell^2}{\mu^2r^2}}}\\&=\int \frac{\pm(\ell/r^2)\text{d}r}{\sqrt{2\mu(E-U(r))-\frac{\ell^2}{r^2}}}\\&=\int \frac{\pm(\ell/r^2)\text{d}r}{\sqrt{2\mu\left(E-U(r)-\frac{\ell^2}{2\mu r^2}\right)}}\end{aligned}$ …(17)

From Equation (17), we see that $\dot{\theta}(t)=\frac{\pm(\ell/r^2)}{\sqrt{2\mu\left(E-U(r)-\ell^2/2\mu r^2\right)}}$ must be monotonical increasing or decreasing becauce $\ell$ is constant in time.

Equation (17) is the explicit formula for trajectory. The very remaining problem is to discuss the certain forms of the force law: what specifically is $F(r)$ like?

For instance, we may assume the force is proportional to some power of the radius, $F(r)\propto r^n$ . Actually, the most important physical cases are ascribed to such force law. The solutions of Equation (17) can be expressed in terms of elliptical integrals for certain integer and fractional values of $n$ . And only for $n=1,-2,-3$ are the solutions expressible in terms of circular functions (sines and cosines).

Reference

Thorton, S, T., & Marion, J. B. (2004). Classical Dynamics of Particles and Systems (fifth edition). Belmont, CA: Brooks/Cole—Thomson Learning.
Wikipedia: Central force, Two-body Problem, Classical central-force problem, Kepler problem…

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Kepler Problem

1 Central Force and Central Potential

2 Two-Body Problem