Central forces are forces between two particles that depend only on the distance between the particles and point from one particle to another. Central forces are conservative (why?). Furthermore, if we look at a central force between two particles we can separate the motion of the center of mass from the relative motion of the particles reducing the complexity of the problem.

## Separation of Motion

We have two particles of masses $m_1$ and $m_2$ located at positions $\vec r_1$ and $\vec r_2$. Furthermore, let's define $\vec r=\vec r_1 - \vec r_2$. Let's assume that the potential depends only on the relative position of the two particles. We can write out the Lagrangian as

$L = T - V = \frac{1}{2} m_1 \dot \vec r_1^2 + \frac{1}{2} m_2 \dot \vec r_2^2 - V \left ( \vec r \right ) + m_1 \vec g \cdot \vec r_1 + m_2 \vec g \cdot \vec r_2$

where $\vec g$ is the direction of the acceleration due to a uniform gravitational field.

Let's define $\vec R$ to be the position of the center of mass

$\vec R=\frac{m_1 \vec r_1 + m_2 \vec r_2}{m_1+m_2}$

so

$\vec r_1 = \vec R + \frac{m_2 \vec r}{m_1 + m_2}$ and $\vec r_2 = \vec R - \frac{m_1 \vec r}{m_1 + m_2}.$

### Lagrangian

Let's write out the Lagrangian using the new coordinates

$L = \frac{1}{2} m_1 \left ( \dot \vec R^2 + 2 \dot \vec R \cdot \dot \vec r \frac{m_2}{m_1+m_2} + \dot \vec r^2 \frac{m_2^2}{\left (m_1+m_2\right )^2} \right ) +$

$\frac{1}{2} m_2 \left ( \dot \vec R^2 - 2 \dot \vec R \cdot \dot \vec r \frac{m_1}{m_1+m_2} + \dot \vec r^2 \frac{m_1^2}{\left (m_1+m_2\right )^2} \right ) -$

$V \left (\vec r \right ) + m_1 \vec g \cdot \left ( \vec R + \frac{m_2 \vec r}{m_1 + m_2} \right ) + m_2 \vec g \cdot \left ( \vec R - \frac{m_1 \vec r}{m_1 + m_2} \right )$

Notice that the middle terms in the first two sets of parentheses and the final terms in the final set cancel each other out and we are left with

$L = \frac{1}{2} M \dot \vec R^2 + \frac{1}{2} \mu \dot \vec r^2 - V \left (\vec r \right ) + M \vec g \cdot \vec R$

where $M=m_1+m_2$ is the total mass and the reduced mass is

$\mu = \frac{m_1 m_2}{m_1 + m_2}$

So we find that for two bodies the dynamics reduces to the motion of the center of mass (possibly in a uniform gravitational field) and the motion of single particle about the center of mass.

### Angular Momentum

Let's calculate the total angular momentum of the system. We know from the results on many-particle systems that the angular momentum of the system will separate into the angular momentum about the center of mass and that of the center of mass. Furthermore, if $V\left ( \vec r \right ) = V \left ( \left | \vec r \right | \right )$, the internal forces do not contribute to the change in the angular momentum. We have

$\vec L = m_1 \vec r_1 \times \vec v_1 + m_2 \vec r_2 \times v_2$

$= m_1 \left ( \vec R + \frac{m_2 \vec r}{m_1 + m_2} \right ) \times \left ( \vec V + \frac{m_2 \vec v}{m_1 + m_2}\right ) + m_2 \left ( \vec R - \frac{m_1 \vec r}{m_1 + m_2} \right ) \times \left ( \vec V - \frac{m_1 \vec v}{m_1 + m_2}\right )$

where $\vec v=\vec v_1 - \vec v_2$ and $\vec V = \left ( m_1 \vec v_1 + m_2 \vec v_2 \right ) / M$.

$\vec L = M \vec R \times \vec V + \left [ m_1 \frac{m_2^2}{\left ( m_1 + m_2 \right )^2} + m_2 \frac{m_1^2}{\left ( m_1 + m_2 \right)^2} \right ] \left ( \vec r \times \vec v \right ) = M \vec R \times \vec V + \mu \vec r \times \vec v.$

We know from the results on many-particle systems in a uniform gravitational field that the internal angular momentum ($\mu \vec r \times \vec v$) of the two-body system is conserved. We could have seen this from separation of the internal and external degrees of freedom in the Lagrangian.

## General Results

We can treat the central force between two particles as a force between a fixed center that we take to be the origin and a single particle. We have

$L = \frac{1}{2} \mu \dot \vec r^2 - V \left (\left | \vec r \right |\right ),$

Taking the particular case of the results from the first section where the potential does not depend on direction, we know that the angular momentum of the particle is conserved. This means that the motion will be restricted to a plane that we will take to be the equatorial plane ($\theta=\pi/2$) without a loss of generality. Let's write the Lagrangian in spherical coordinates. First let's write the position of the particle in terms of the spherical coordinates. We have

$\vec r = \left [ \begin{matrix} r \sin \theta \cos \phi \\ r \sin \theta \sin \phi \\ r \cos \theta \end{matrix} \right ]$

and

$\dot \vec r = \left [ \begin{matrix} \dot r \sin \theta \cos \phi + r \dot \theta \cos \theta \cos \phi - r \dot \phi \sin \theta \sin \phi \\ \dot r \sin \theta \sin \phi + r \dot \theta \cos \theta \sin \phi + r \dot \phi \sin \theta \cos \phi \\ \dot r \cos \theta - r \dot \theta \sin \theta \end{matrix} \right ]$

so

$\dot \vec r^2 = \dot r^2 + r^2 \dot \theta^2 + r^2 \dot \phi^2 \sin^2\theta.$

### Lagrangian

Let's write out the Lagrangian in terms of the spherical coordinates

$L = \frac{1}{2} \mu \left ( \dot r^2 + r^2 \dot \theta^2 + r^2 \dot \phi^2 \sin^2\theta\right ) - V(r).$

### Conserved Quantities

First let's notice that $\partial L/\partial \phi = 0$ so $p_\phi$ is conserved. Also $\partial L/\partial t=0$ so $H$ is conserved as well.

Let's calculate the conserved momentum,

$p_\phi = \frac{\partial L}{\partial \dot \phi} = \mu r^2 \dot \phi\sin^2 \theta$

which has units of angular momentum. It is simply the angular momentum about the $z-$axis.

### Equations of Motion

We can also use Lagrange's equation to get the equations of motion

$\frac{d}{dt} \frac{\partial L}{\partial \dot \theta} - \frac{\partial L}{\partial \theta} = 0,$

$\frac{d}{dt} \left ( \mu r^2 \dot \theta \right ) = 2 \mu r \dot r \dot \theta + \mu r^2 \ddot \theta = \mu r^2 \dot \phi^2 \sin \theta \cos \theta.$

If $\dot \theta=0$ and $\theta=0,\pi/2$ or $\pi$ initial, then $\ddot \theta=0$ and $\theta$ stays constant. We take $\theta=\pi/2$ and $\dot \theta=0$ initially, so the particle remains in the equatorial plane as we expected from the conservation of angular momentum.

$\frac{d}{dt} \frac{\partial L}{\partial \dot r} - \frac{\partial L}{\partial r} = 0,$

$\frac{d}{dt} \left ( \mu \dot r \right ) - r \mu \dot \theta^2 - r \mu \dot \phi^2 \sin^2 \theta + \frac{\partial V}{\partial r} = 0 .$

Using the initial conditions we have

$\ddot r = -\frac{1}{\mu} \frac{\partial V}{\partial r} + r \dot \phi^2 = -\frac{1}{\mu} \frac{\partial V}{\partial r} + \frac{p_\phi^2}{\mu^2} \frac{1}{r^3}.$

where the dependence of the acceleration on the angle $\phi$ has been included through the conserved quantity $p_\phi$, separating the radial equation from the angular equation. The first term is the force between the two bodies and the second term is a 'fictitous' force because of the spherical coordinates. We will solve this differential equation for two special cases, but we can get a good idea of the nature of the motion by looking at the conserved quantities.

### Bounded Motion

First, we have

$p_\phi = \mu r^2 \dot \phi = \mu dA/dt$

so the path subtends equal areas in equal times. Given a solution for the radial motion we can determine the angular motion

$\dot \phi = \frac{p_\phi}{\mu} \frac{1}{r^2}.$

Second we have the Hamiltonian,

$H = \dot r \frac{\partial L}{\partial \dot r} + \dot \theta \frac{\partial L}{\partial \dot \theta} + \dot \phi \frac{\partial L}{\partial \dot \phi} - L = \frac{1}{2} \mu \left ( \dot r^2 + r^2 \dot \theta^2 + r^2 \dot \phi^2 \sin^2\theta\right ) + V(r)$

Because the Hamiltonian is conserved, let's take its initial value to be

$E = \frac{1}{2} \mu \dot r^2 + \frac{1}{2} \frac{p_\phi^2}{\mu} \frac{1}{r^2} + V(r).$

We can solve this for $\dot r^2$ to get

$\dot r^2 = \frac{2 E}{\mu} - \frac{p_\phi^2}{\mu^2} \frac{1}{r^2} - \frac{2 V(r)}{\mu}$

which is very similar to the one-dimensional case but with an added term for the centrifugal force. Unless $V \sim r^n$ as $r\rightarrow 0$ with $n\leq -2$ and $V$ is negative, motion of the the particle is bounded if $p_\phi \neq 0$. This point of closest approach is called the pericenter. Depending on the potential and the initial conditions, the motion may be bounded at large $r$ as well. In particular if the potential is attractive and positive there is always a point of maximum distance. The point of maximum radius is called the apocenter.

If the motion is bounded at both large and small radius, we can calculate the time for the particle to go from the minimum to maximum radius -- this is one half of the period of the complete radial motion as we did for the one-dimensional pendulum. The angular motion as we shall find can have a different (and not usually comeasureable) period. We have

$\frac{d r}{d t} = \sqrt { \frac{2 E}{\mu} - \frac{p_\phi^2}{\mu^2} \frac{1}{r^2} - \frac{2V(r)}{\mu} }$

and rearranging we have

$P = 2 \int_{r_{\rm min}}^{r_{\rm max}} \frac{dr}{ \sqrt { \frac{2 E}{\mu} - \frac{p_\phi^2}{\mu^2} \frac{1}{r^2} - \frac{2 V(r)}{\mu} } }$

## Specific Results

### Spherical Harmonic Oscillator

#### Central Force Solution

For a spherical harmonic oscillator, $V = k r^2/2 .$

##### Bounds of Motion

Let's first calculate the bounds of the motion we have

$0 = \frac{2 E}{\mu} - \frac{p_\phi^2}{\mu^2} \frac{1}{r^2} - \frac{k r^2}{\mu}$

which yields a quadratic equation for $r^2$,

$r^4 - \frac{2 E}{k} r^2 + \frac{p_\phi^2}{\mu k} = 0$

which yields the solutions

$r_{\pm}^2 = \frac{E}{k} \pm \sqrt{\frac{E^2}{k^2} - \frac{p_\phi^2}{\mu k}}$

We will first use the conservation of energy to calculate the radial motion, we have

$t_1 - t_0 = \int_{r_0}^{r_1} \frac{dr}{\sqrt{\frac{2E}{\mu}-\frac{k}{\mu} r^2 - \frac{p_\phi^2}{r^2 \mu^2} }} = \sqrt{\frac{\mu}{k}} \int_{r_0}^{r_1} \frac{r dr}{\sqrt{\left (r_+^2-r^2\right )\left (r^2-r_-^2\right ) }}$

$= \frac{1}{2} \sqrt{\frac{\mu}{k}} \int_{r_0^2}^{r_1^2} \frac{d \left ( r^2 \right )}{\sqrt{\left (r_+^2-r^2\right )\left (r^2-r_-^2\right ) }}.$

To perform this integral let's make the substitution

$s=\frac{2 r^2 - \left (r_+^2 + r_-^2 \right )}{r_+^2 - r_-^2}$

to give

$t_1 - t_0 = \frac{1}{2} \sqrt{\frac{\mu}{k}} \int_{r=r_0}^{r=r_1} \frac{d s}{\sqrt{1-s^2}} = \left . \frac{1}{2} \sqrt{\frac{\mu}{k}} \arcsin s \right |_{r=r_0}^{r=r_1}$

so

$r^2 = \frac{r_+^2 + r_-^2}{2} + \frac{r_+^2-r_-^2}{2} \sin \left [ 2 \sqrt{\frac{k}{\mu}} \left ( t - t_0 \right ) + \psi_0 \right ]$

$= \frac{E}{k} + \sqrt{\frac{E^2}{k^2} - \frac{p_\phi^2}{\mu k}} \sin \left [ 2 \sqrt{\frac{k}{\mu}} \left ( t - t_0 \right ) + \psi_0 \right ]$

with

$\psi_0 = \arcsin \left [ \frac{2 r_0^2 - \left (r_+^2 + r_-^2 \right )}{r_+^2 - r_-^2} \right ] = \arcsin \left [ \frac{r_0^2 - \frac{E}{k}}{\sqrt{\frac{E^2}{k^2} - \frac{p_\phi^2}{\mu k}}} \right ]$

##### Angular Motion

What remains is to solve for the motion in the $\phi-$direction. Using the conservation of angular momentum we have

$\frac{\partial \phi}{\partial t} = \frac{p_\phi}{\mu} \frac{1}{r^2} = \frac{p_\phi}{\mu} \frac{1}{\frac{r_+^2 + r_-^2}{2} + \frac{r_+^2-r_-^2}{2} \sin \left [ 2 \sqrt{\frac{k}{\mu}} \left ( t - t_0 \right ) + \psi_0 \right ]}.$

We can integrate this directly to yield

$\phi = \frac{p_\phi}{\sqrt{k\mu} r_+ r_-} \arctan \left \{ \frac{1}{2 r_+ r_-} \left [ \left (r_+^2 + r_-^2 \right ) \tan \left ( \sqrt{\frac{k}{\mu}} \left ( t-t_0 \right ) + \psi_0 \right ) + r_+^2 -r_-^2 \right ] \right \}.$

Notice that the angular frequency is $\sqrt{k/m}$, one-half of the radial frequency.

If we substitute the values of $r_+$ and $r_-$ we get

$\phi = \arctan \left \{ \sqrt{\frac{\mu E^2}{p_\phi^2 k}} \left [ \tan \left ( \sqrt{\frac{k}{\mu}} \left ( t-t_0 \right ) + \psi_0 \right ) + \sqrt{ 1 - \frac{p_\phi^2 k}{\mu E^2}} \right ] \right \}.$

It is possible to write $r$ as a function of $\phi$ and find that it is the equation for an ellipse centered at the origin. This is left as an exercise for the reader.

##### The Shape of the Orbit

Let's step back and try to determine $r(\phi)$ instead of $r(t)$. First we define $u=1/r^2$ so we have

$\frac{du}{d\phi} = -\frac{2}{r^3} \frac{dr}{d\phi}$

and using the chain rule we get

$\dot r = \frac{dr}{d\phi} \dot \phi = -\frac{r^3}{2} \dot \phi \frac{du}{d\phi} = - \frac{r}{2} \frac{p_\phi}{\mu} \frac{du}{d\phi}.$

We can substitute this into the energy equation to get

$\dot r^2 = \frac{r^2}{4} \frac{p_\phi^2}{\mu^2} \left ( \frac{du}{d\phi} \right )^2 = \frac{1}{4 u} \frac{p_\phi^2}{\mu^2} \left ( \frac{du}{d\phi} \right )^2 =\frac{2 E}{\mu} - \frac{p_\phi^2}{\mu^2} u - \frac{k}{\mu u}$

and rearranging

$\left ( \frac{du}{d\phi} \right )^2 + 4 u^2 - \frac{8 E \mu}{p_\phi^2} u = -\frac{4 k \mu}{p_\phi^2}.$

We add a constant to both sides of the equation to complete the square

$\left [ \frac{d}{d\phi} \left ( u - \frac{E \mu}{p_\phi^2} \right ) \right ]^2 + 4 \left ( u - \frac{ E \mu}{p_\phi^2} \right )^2 = -\frac{4 k \mu}{p_\phi^2} + 4 \left ( \frac{E \mu}{p_\phi^2} \right )^2$

We find that the derivative of a function squared plus four times the function itself yields a constant, so the function must be a sine or cosine. We have $u - \frac{E \mu}{p_\phi^2} = \left [ \left ( \frac{E \mu}{p_\phi^2} \right )^2 - \frac{ k \mu}{p_\phi^2} \right ]^{1/2} \cos 2 \left ( \phi - \phi_0 \right ).$

We can write this in terms of the radius

$r^2 =\frac{p_\phi^2}{E\mu} \frac{1}{ \left ( 1 - \frac{k p_\phi^2}{E^2 \mu} \right )^{1/2} \cos 2 \left ( \phi - \phi_0 \right ) + 1}.$

This is the equation for an ellipse centered on the origin. The constant in front of the cosine is related to the shape of the ellipse

$\left ( 1 - \frac{k p_\phi^2}{E^2 \mu} \right )^{1/2} = \frac{e^2}{2-e^2}$

where $e^2=1-b^2/a^2$ is the square of the eccentricity of the ellipse ($a$ and $b$ are the semimajor and semiminor axes). If and only if the spring constant is negative (repulsive spring), then the eccentricity can be greater than one (up to $\sqrt{2}$ and the curve is a hyperbola centered on the origin. An interesting case is $k=0$ so $e=1$. This gives a straight line passing a distance $p_\phi/\sqrt(2E\mu)$ from the origin.

#### Cartesian Solution

We can solve the spherical harmonic oscillator in Cartesian coordinates as well. We have for the Lagrangian

$L = \frac{1}{2} \mu \left ( \dot x^2 + \dot y^2 + \dot z^2 \right ) - \frac{1}{2} k \left ( x^2 + y^2 + z^2 \right )$

which yields the following equation of motion for the $x-$coordinate

$\mu \ddot x + k x = 0$

and similarly for the other directions. We find the following solution for the motion in the $x-$direction

$x(t) = A \cos \left ( \omega t + a \right )$

with $\omega^2=k/\mu$ and similarly for the other coordinates. Since the value of $\omega$ is the same for all three directions we find that the figure is an ellipse. We can also calculate the value of $r^2$

$r^2 = A^2 \cos^2 \left ( \omega t + a \right ) + B^2 \cos^2 \left ( \omega t + b \right )$

where for simplicity and without loss of generality we have restricted the motion to the $x-y-$plane. We have

$r^2 = \frac{A^2}{2} \left [ 1 + \cos \left (2 \omega t + a \right ) \right ] + \frac{B^2}{2} \left [ 1 + \cos \left (2 \omega t + b \right ) \right ]$

$= \frac{A^2+B^2}{2} + \frac{A^2}{2} \cos \left (2 \omega t + a \right ) + \frac{B^2}{2} \cos \left (2 \omega t + b \right )$

We can rewrite these two cosines as a single cosine,

$r^2 = \frac{A^2+B^2}{2} + \frac{1}{2} \sqrt{ A^4 + B^2 + 2 A^2 B^2 \cos \left (a - b\right ) } \cos \left [ 2\omega t + \psi_0 \right ]$

where

$\psi_0 = \arccos \frac{A^2 \cos a + B^2 \cos b}{\sqrt{ A^4 + B^2 + 2 A^2 B^2 \cos \left (a - b\right ) }}$

which has the same form as the result for the spherical analysis.

### Gravity

For gravity we have

$V = - \frac{G m_1 m_2}{r} = - \frac{G \mu M}{r} =$

yielding the following Lagrangian,

$L = \frac{1}{2} \mu \left ( \dot r^2 + r^2 \dot \theta^2 + r^2 \dot \phi^2 \sin^2\theta\right ) + \frac{G \mu M}{r}.$

#### Bounds of the Motion

Let's first calculate the bounds of the motion we have

$0 = \frac{2 E}{\mu} - \frac{p_\phi^2}{\mu^2} \frac{1}{r^2} + \frac{2 G M}{r}$

which yields a quadratic equation for $r$

$r^2 + \frac{G M \mu}{E} r - \frac{p_\phi^2}{E \mu} = 0$

which yields the solutions

$r_{\pm} = -\frac{G M\mu}{2 E} \pm \sqrt{ \left (\frac{G M \mu}{2 E}\right)^2 + \frac{p_\phi^2}{E \mu}}.$

For the case of an attractive force $G>0$ we find that if $E>0$, only $r_+$ is greater than zero and this represents the point of closest approach. On the other hand if $E<0$ we have $r_-< r < r_+$. For a repulsive force $G<0$ and again only $r_+$ is greater than zero if $E>0$. If both $G$ and $E$ are negative, we cannot find a positive value of $r$ with $\dot r^2>0$.

We can also solve for the radial velocity

$\dot r^2 =\frac{2 E}{\mu} - \frac{p_\phi^2}{\mu^2} \frac{1}{r^2} + \frac{2 G M}{r} = \frac{2 E}{\mu r^2} \left ( r^2 + \frac{G M \mu}{E} r - \frac{p_\phi^2}{E \mu} \right ) = \frac{2 E}{\mu r^2} \left ( r - r_+ \right ) \left ( r - r_- \right ).$

Yielding the following equation for the time evolution

$t_1 - t_0 = \sqrt {\frac{\mu}{2E} } \int_{r_0}^{r_1} \frac{r}{\sqrt{\left ( r - r_+ \right ) \left ( r - r_- \right )} }d r$

and let's make the substitution

$s=\frac{2 r - \left (r_+ + r_- \right )}{r_+ - r_-}$

to give the following integral

$t_1 - t_0 = \sqrt {-\frac{\mu}{2E} } \frac{r_+-r_-}{2} \int_{r=r_0}^{r=r_1} \frac{s}{\sqrt{1-s^2}} + \frac{r_+ + r_-}{\sqrt{1-s^2}} d s$

where we have assumed that $E<0$. This gives

$t_1 - t_0 = \left . \sqrt {-\frac{\mu}{2E} } \frac{r_+-r_-}{2} \left [ \left ( r_+ + r_- \right ) \arcsin s - \sqrt{1-s^2} \right ] \right |_{r=r_0}^{r=r_1}$

and for $E>0$ we have

$t_1 - t_0 = \left . \sqrt {\frac{\mu}{2E} } \frac{r_+-r_-}{2} \left [ \left ( r_+ + r_- \right ) {\rm arcsinh} s - \sqrt{s^2-1} \right ] \right |_{r=r_0}^{r=r_1}$

Unfortunately we cannot invert this to find the radius as a function of time, but we can determine, the time to go from the minimum to the maximum radius in the $E<0$ case

$P = \pi \sqrt{-\frac{\mu}{2E}} \left ( r_+^2 - r_-^2 \right ) = \pi \sqrt{-\frac{\mu}{2E}} \frac{G M \mu}{-E} \sqrt{\left(\frac{G M \mu}{E}\right)^2 + \frac{4 p_\phi^2}{E \mu}}$

#### The Shape of the Orbit

Let's step back and try to determine $r(\phi)$ instead of $r(t)$. First we define $u=1/r$ so we have

$\frac{du}{d\phi} = -\frac{1}{r^2} \frac{dr}{d\phi}$

and using the chain rule we get

$\dot r = \frac{dr}{d\phi} \dot \phi = -r^2 \dot \theta \frac{du}{d\theta} = - \frac{p_\phi}{\mu} \frac{du}{d\phi}.$

We can substitute this into the energy equation to get

$\dot r^2 = \frac{p_\phi^2}{\mu^2} \left ( \frac{du}{d\phi} \right )^2 = \frac{2 E}{\mu} - \frac{p_\phi^2}{\mu^2} u^2 + 2 G M u$

and rearranging

$\left ( \frac{du}{d\phi} \right )^2 + u^2 - \frac{2 G M \mu^2 u}{p_\phi^2} = \frac{2E\mu}{p_\phi^2}$

Let's define the length $l=p_\phi^2/(G M \mu^2)$ and simplify the equation

$\left ( \frac{du}{d\phi} \right )^2 + u^2 - \frac{2}{l} u = \frac{2E}{G M \mu l}.$

Let's add $1/l^2$ to both sides of the equation and get

$\left [ \frac{d}{d\phi} \left( u - \frac{1}{l} \right ) \right ]^2 + \left ( u - \frac{1}{l} \right )^2 = \frac{2E}{G M \mu l} + \frac{1}{l^2}.$

We have a function squared plus its first derivative squared being a constant. The function is clearly a trigonometric function we have

$u = \frac{1}{r} = \frac{e}{l} \cos \left ( \phi- \phi_0 \right ) + \frac{1}{l}$

where

$e^2 = \frac{2E l}{G M \mu} + 1.$

Rearranging the equation gives

$r \left [ e \cos \left ( \phi - \phi_0 \right ) + 1 \right ] = l$

which is the polar equation for an ellipse with its focus at the origin for $0. For $e>1$ the curve is a hyperbola. $e=1$ is a parabola and $e=0$ is a circle. (Contrast this with the spherical harmonic oscillator where the center of the ellipse was at the origin).

If we have a repulsive force (e.g. the force between two positive charges) $e \geq 1$ because $E \geq 0$. The resulting curve is a hyperbola.

The figure shows the various conic sections as a function of the value $e$, the eccentricity. The blue parabola has $e=1$, the aqua ellipse has $e=0.5$ and the green circle has $e=0$. The red hyperbola has $e=2$. All of the curves have a focus at the origin and $l=1$. The semi-latus rectum is the official name for the quantity $l$ which sets the size of the curve. In all cases the curve intersects the $y-$axis at $y=\pm 1$. In a coordinate free way, the semi-latus rectum is the distance between the curve and the focus in the direction perpendicular to the point of closest approach.

What is the meaning of the two branches of the hyperbola? The left branch is for an attactrive force and the right branch is for a repulsive force.