# Central Forces

## Central Forces

Spherically symmetric forces

Gravity is a central force

It acts between 2 objects of masses: $m_1$ and $m_2$ and depends on their distance: $r$.

$$\mathbf{F} = -G\frac{m_1 m_2}{r^2}\mathbf{\hat{r}}$$

The unit vector $\mathbf{\hat{r}}$ points away from the source object.

The gravitational potential is easily found by considering the integral of $F$ over a distance $r$.

$$U(r) = -\int F(r) dr = -G \frac{m_1 m_2}{r}$$

##### Springs (or spring-like)

$$\mathbf{F} = -k \mathbf{r} = -k r \hat{\mathbf{r}}$$

##### Coulomb Force

$$\mathbf{F} = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r^2}\mathbf{\hat{r}}$$

However, we usually assumed the source didn't also move, or if it did, it only moved a little bit and we could ignore it.

Obviously, if we are considering the gravitational interaction between two objects with similar masses, then we can't assume this to be the case.

## The Two-Body Problem

$$\mathbf{R}_{\textrm{cm}} \equiv \frac{m_1 \mathbf{r}_1 + m_2 \mathbf{r_2}}{m_1 + m_2}$$ and $$\mathbf{r} \equiv \mathbf{r_2} - \mathbf{r_1}$$

Solving for $\mathbf{r_1}$ and $\mathbf{r_2}$ in terms of $\mathbf{R}_\textrm{cm}$ and $\mathbf{r}$

\begin{align} \mathbf{r_1} & = \mathbf{R}_\textrm{cm}-\frac{m_2}{M}\mathbf{r}\\ \mathbf{r_2} & = \mathbf{R}_\textrm{cm}+\frac{m_1}{M}\mathbf{r} \end{align}

Using those two coordinates, we can express the kinetic energy of each particle: $$T = \frac{1}{2}m_1 \mathbf{\dot{r}}_1^2 + \frac{1}{2}m_2 \mathbf{\dot{r}}_2^2$$

After some mildly tedious algebra, this becomes: $$T = \frac{1}{2}M \mathbf{\dot{R}}_\textrm{cm}^2 + \frac{1}{2}\mu \mathbf{\dot{r}}^2$$ where $\mu = \frac{m_1 m_2}{m_1 +m_2} = \frac{m_1 m_2}{M}$ is the reduced mass for the two-body system.

This kinetic energy can be conceptualized as just the kinetic energies of two particles. (Neither of which really exist) $$T = \underbrace{\frac{1}{2}M \mathbf{\dot{R}}_\textrm{cm}^2}_{\textrm{Particle 1}} + \underbrace{\frac{1}{2}\mu \mathbf{\dot{r}}^2}_{\textrm{Particle 2}}$$

Let's construct a Lagrangian: $$L = T - U = \frac{1}{2}M \mathbf{\dot{R}}_\textrm{cm}^2 + \frac{1}{2}\mu \mathbf{\dot{r}}^2 - U(r)$$

Looking at this, we can see write away that one of the coordinates is cyclic (or ignorable): $\mathbf{R}_\textrm{cm}$ doesn't appear in the Lagrangian so we know that: $$\mathbf{P} = M \mathbf{\dot{R}}_\textrm{cm}$$ is conserved. (That's the total momentum of the system)

$$L = \underbrace{\frac{1}{2}M \mathbf{\dot{R}}_\textrm{cm}^2}_\textrm{cm} + \underbrace{\frac{1}{2}\mu \mathbf{\dot{r}}^2 - U(r)}_{\textrm{rel. position}}$$

$$L = L_\textrm{cm} + L_\textrm{relative position}$$

Re-writing the second part of the Lagrangian in polar coordinates (and realizing that this is effectivly now a 2d system): $$L = \frac{1}{2}\mu \left(\dot{r}^2 + r^2 \dot{\phi}^2 \right) - U(r)$$

Two things to note:

1. $L$ is not explicitly time dependent, which that the Hamiltonian $H$ is conserved.
2. The generalized coordinate $\phi$ is cyclic, so that means that $$p_\phi \equiv l = \mu r^2 \dot{\phi} = r (\mu r \dot{\phi}) = \textrm{constant}$$

The acheivement?

We've dealt with a non-intertial frame (i.e. the accelerative motion of the source ) by describing the motion of a particle of mass $\mu$ around the source.

## Effective Potential

Two conservation equations: $$E = \frac{1}{2}\mu \left(\dot{r}^2 + r^2 \dot{\phi}^2 \right) + U(r)$$ and

$$l = \mu r^2 \dot{\phi}$$

To solve these analytically, we have to eliminate either $t$ or $\phi$

We can try to obtain two different descriptions: $$r(t) \;\; \textrm{or} \;\; r(\phi)$$

#### Start with getting rid of $\phi$

Angular momentum conservation let's us rewrite $\dot{\phi}$ in terms of $r$: $$\dot{\phi} = \frac{l}{mr^2}$$

Which we can then use to recast the Energy as: \begin{aligned} E & = \frac{1}{2}\mu \left(\dot{r}^2 + r^2 \dot{\phi}^2 \right) + U(r) \\ & = \frac{1}{2}\mu \left( \dot{r}^2 + r^2 \left( \frac{l}{\mu r^2} \right) ^2 \right) + U(r) \end{aligned}

$$E = \frac{1}{2}\mu \dot{r}^2 + U_\textrm{eff}$$ with $$U_\textrm{eff} = \frac{l^2}{2 \mu r^2} + U(r)$$

Now we consider the first derivative of $U_\textrm{eff}$ w.r.t $r$. $$\left. U'_\textrm{eff} \right|_{r = R} = -\frac{l^2}{\mu r^3} + U'(r) = 0$$

For the case of gravity: $$U_G(r) = - \frac{G m_1 m_2}{r}$$ and so our effective potential is: $$U_\textrm{eff} = \frac{l^2}{2 \mu r^2} -\frac{G m_1 m_2}{r}$$

Different $l$ values

Different $E$ for an orbiter

Find $r(t)$ or $t(r)$

$$\frac{1}{2}\mu \dot{r}^2 + U_\textrm{eff}(r) = E$$ with $$U_\textrm{eff} = \frac{l^2}{2 \mu r^2} - \frac{G m_1 m_2}{r}$$

Solve for $\dot{r}$...

$$\frac{dr}{dt} = \pm \sqrt{\frac{2}{\mu}\left(E + \frac{G M \mu}{r} - \frac{l^2}{2 \mu r^2} \right)}$$

$$t(r) = \pm \sqrt{\frac{\mu}{2}}\int_{r_0}^r \frac{r \; dr}{\sqrt{Er^2 + G M \mu r - \frac{l^2}{2 \mu}}}$$

#### Solve this →

Time from periapse to apoapse (closest to farthest) is: $$t(p\rightarrow a) = G M \left(\frac{\mu}{2(-E)} \right)^{3/2} \pi$$

#### Now eliminate $t$ to get $r(\phi)$

$$E = \frac{1}{2}m \dot{r}^2 + \frac{l^2}{2 m r^2} + U(r)$$ and $$l = m r^2 \dot{\phi}$$
$$\frac{dr}{d\phi} = \frac{dr /dt}{d\phi/dt}$$
$$\frac{dr}{d\phi} = \pm \sqrt{\frac{2m}{l^2}}r^2\sqrt{E - \frac{l^2}{2mr^2} - U(r)}$$
$$\phi = \int d\phi = \pm \frac{l}{\sqrt{2m}}\int^r \frac{r^{-2}}{\sqrt{E- \frac{l^2}{2mr^2} -U(r)}}dr$$

#### For Gravitational Forces:

$$U(r) = -\frac{G M m}{r}$$

This leads to a similar integral: $$\phi = \int d\phi = \pm \frac{l}{\sqrt{2m}}\int \frac{dr}{r \sqrt{Er^2 + GM m r - \frac{l^2}{2 m}}}$$

after some algebra: $$r = \frac{l^2 / G M m^2}{1+ \epsilon \cos(\phi)}$$ with $$\epsilon \equiv \sqrt{1 + \frac{2 E l^2}{G^2 M^2 m^3}}$$

## Conic Sections

### Ellipses

An ellipse satisfies this equation: $$$$r + r' = 2a \label{eq:basic-ellipse}$$$$

• $a$ is the semi-major axis
• $r$ and $r'$ are the distances to the ellipse from the two focal points, $F$ and $F'$.
• $e$ represents the eccentricity of the ellipse ($0 \leq e \leq 1$)

#### Polar Coordinates

$$$$r = \frac{a\left( 1-e^2\right)}{1+e \cos \theta} \label{eq:polar-equation-ellipse}$$$$

Now we have: $$r = \frac{a \left(1-\epsilon^2 \right)}{1+ \epsilon \cos \phi}$$ and $$r = \frac{l^2 / G M m^2}{1+ \epsilon \cos(\phi)}$$

What is $a$ in terms of $E$?

Recall the time from $r_p$ to $r_a$: $$t(p\rightarrow a) = G M \left(\frac{\mu}{2(-E)} \right)^{3/2} \pi$$

Now, with $$-E = \frac{G M \mu}{2 a }$$

$$t = \frac{2 \pi a^{3/2}}{\sqrt{GM}}$$