Orbits for two bodies interacting through a central force
It seems that we are on the verge of being able to solve our 2 differential equations for $r$ and $\phi$, and then apply initial conditions in order to completely solve this problem.
It would be nice to *see* the allowed orbits...
Solve for $r(t)$ and $\phi(t)$?
The E-L equation for $r$ resulted in this "radial" equation of motion:
$$\mu \ddot{r}= -\frac{d}{dr}U_{\text{eff}}=F(r) +\frac{\ell^2}{\mu r^3}\label{ELr}$$
And, the E-L equation for $\phi$ told us that angular momentum is conserved
$$\dot{\phi} = \frac{ \ell}{\mu r^2}.$$
So, maybe we could just solve the radial equation of motion, and then plug it in to the second equation to get $\phi(t)$?
That turns out to be a very hard problem.
Instead, we'll start by finding $r(\phi)$--the trajectory.
Two tricky things to reach the trajectory: An oddball expression for $d/dt$, and the substitution $1/r=u$
Find the trajectory $r(\phi)$ instead
Our starting point is the total angular momentum expression: $$\ell = I\omega = \mu r^2 \dot{\phi} = \mu r^2 \frac{d \phi}{d t}.$$ The chain rule says we can write $\frac{d}{dt}$ as: $$\frac{d}{dt}=\frac{d\phi}{dt}\frac{d}{d\phi}=\frac{\ell}{\mu r^2}\frac{d}{d\phi}.$$
Using this $d/dt$, (and, in the last line, observing that $dr^{-1}/d\phi=-r^{-2}dr/d\phi$), we'll re-write the radial equation of motion (\ref{ELr}): $$\begineq F(r) +\frac{\ell^2}{\mu r^3}=&\mu \ddot{r}=\mu\color{#05b}{\frac{d}{dt}}\color{#bb0}{\frac{dr}{dt}}\\ =&\mu \color{#05b}{\frac{\ell}{\mu r^2} \frac{d}{d\phi} } \left[ \color{#bb0}{\frac{\ell}{\mu} } \color{#880}{\frac {1}{r^2} \frac{dr}{d\phi} }\right]\\ =&-\mu \color{#05b}{\frac{\ell}{\mu r^2} \frac{d}{d\phi} } \left[ \color{#bb0}{\frac{\ell}{\mu} } \color{black}{\frac{d(1/r)}{d\phi}} \right] \endeq$$
It will turn out to be useful to carry out the substitution $u \equiv 1/r$: $$F(1/u) + \frac{\ell^2u^3}{\mu} =-\frac{\ell^2u^2}{\mu} \frac{d^2 u}{d\phi^2}.$$
Dividing by $(l^2u^2)/(\mu)$ and re-arranging a little, we get this very general differential equation for *any* central force:
$$\frac{d^2u}{d \phi ^2}+u = -\frac{\mu}{\ell^2u^2}F(1/u).$$
Kepler orbits
Now we'll specialize to the case of inverse-square forces like gravity and Coulomb forces. To catch both possibilities, let's write our force as $$F(r) = \frac{\gamma}{r^2} = \gamma u^2$$
where $\gamma = -Gm_1m_2$ for gravity, or $\gamma=kq_1q_2$ for the Coulomb force. Our radial equation looks a lot simpler.... $$\frac{d^2 u}{d \phi^2}+u=\gamma \mu/\ell^2$$
There's one pretty simple solution to this diffeq: $$u_p=\gamma\mu/\ell^2\equiv \frac 1c.$$
But recall what we did with the damped, driven harmonic oscillator: The general solution was $x_p+x_h$, the sum of a particular solution, and a solution to the homogeneous equation.
For our differential equation the associated homogeneous equation is: $$\frac{d^2}{d\phi^2}u + u = 0$$
A solution to this homogeneous equation is... $$u_h(\phi) = A \cos(\phi - \phi_0)$$
Let's just pick an orientation of our coordinate system, such that $\phi_0=0$. (It turns out that all the solutions have a symmetry axis, pointing along $\phi_0$. Choosing $\phi_0=0$ just means we have restricted ourselves to solutions which are symmetric about the $x$-axis.)
Our general solution $u = u_h +u_p$ is $$u(\phi) = 1/r(\phi) = A\cos(\phi) + 1/c = \frac{1}{c}(1 + \epsilon \cos (\phi))$$
or...
$$r(\phi) = \frac{c}{1+\epsilon \cos\phi}$$
Graphing exercise
Using Desmos...Make a parametric plot of the trajectories (orbits) $r(\phi)$ from the equation above:
- $\phi$ is the parameter, have it run from $0.2*\pi$ to $1.8*\pi$ (Almost 0 to $2\pi$...).
- The x-coordinate is $r(\phi)\cos\phi$ and $y(\phi)=r(\phi)\sin\phi$,
- Make sliders to let you vary $c$ from 0.1-10.0, and let you vary $\epsilon$ from 0-10.
However, Desmos insists that you may only use "$t$" as your parameter, when making a parametric plot. So in the instructions above, make the substitution $\phi\to t$.
$\epsilon < 1$ - bounded orbits
The denominator $1+\epsilon \cos\phi$ is never zero. There are extreme values at $$r_{max} = \frac{c}{1-\epsilon};\ \ r_{min} = \frac{c}{1+\epsilon}.$$
Also, it is easy to see that $r(\phi=0) = r(\phi=2\pi)$, so apparently the orbits are closed.[graph]
We can graph $x(\phi)$ and $y(\phi)$ as a function of the parameter $\phi$ by using: $$x = r\cos\phi$$
and $$y = \pm\sqrt{r^2-x^2} = \pm r\sqrt{1-\cos^2\phi}$$
This graph was made with $c=1$ and $\epsilon=0.5$, which leads to $ r_{min}=2/3$ and $r_{max}=2$.
Problem 8.16
Let's show that this form of the equation really is an ellipse by re-writing it in Cartesian coordinates. $$r=\sqrt{x^2+y^2};\ \ \cos \phi = x/r$$
Putting only the second of these coordinate transforms into (15):
$$r = \frac{c}{1+\epsilon x/r}$$
$\Rightarrow r+\epsilon x=c$
$\Rightarrow r^2 = (c-\epsilon x)^2$
Now, substituting in the Cartesian expression for $r$:
$x^2+y^2 = c^2 -2c\epsilon x + \epsilon^2 x^2$
$\Rightarrow (1-\epsilon^2)x^2 +2c\epsilon x +y^2 = c^2$
$\Rightarrow x^2 + \frac{2c\epsilon x}{1-\epsilon^2} + \frac{y^2}{1-\epsilon^2} = \frac{c^2}{1-\epsilon^2}$
Completing the square of the first two terms...
$\left(x^2 + \frac{2c\epsilon x}{1-\epsilon^2} + \frac{c^2\epsilon^2}{(1-\epsilon^2}^2\right)+
\frac{y^2}{1-\epsilon^2} = \frac{c^2}{1-\epsilon^2} +\frac{c^2\epsilon^2}{(1-\epsilon^2)^2}$
$\left(x+\frac{c\epsilon}{1-\epsilon^2}\right)^2 +\frac{y^2}{1-\epsilon^2} = \frac{c^2}{(1-\epsilon^2)^2}$
Now using the constants...
$a=c/(1-\epsilon^2)$, $b=c/\sqrt{1-\epsilon^2}$, and $d=a\epsilon$
The equation becomes...
$(x+d)^2 +y^2(a^2/b^2)=a^2$
Which finally ends up as $$\frac{(x+d)^2}{a^2} +\frac{y^2}{b^2}=1$$
This is the equation for an ellipse centered at $(x,y)=(-d,0)$ with major axis $a$ and minor axis $b$.
This parameter, $\epsilon$, is known as the eccentricity. See the How Earth's eccentricity compares to other familiar planetary bodies.
There is a related parameter called the ellipticity which is 1 for a circle,
Our model only takes into account 2 bodies. But of course in our solar system there are more than 2. This has produced changes in ellipticity / eccentricity in the planetary orbits, and other planetary parameters. Some of these are approximately periodic. See the Milankovitch cycles and compared with paleo temperatures.
Precession of the perihelion
Our theory predicts that the position of the perihelion is constant.
But a variety of effects can conspire to make the perihelion move.
- gravity of other planets
- the sun's less-than-perfect spherical symmetry
Urbain Le Verrier predicted the existence of Neptune based on deviations of Uranus' orbit from pure ellipticity.
After taking into account the effects of other planets, and the asphericity of the sun, he still found (1859) an unexplained 43 arcsecond per century precession of Mercury's perihelion.
Unsuccessful attempts to explain this included...
- an unknown planet, called Vulcan!
- taking into account the finite speed with which gravity propagates,
- deviations of the exponent in the gravitational force law away from an exact value of exactly 2
Einstein was able to show that general relativity (GR) predicted a shift of this magnitude, and this was a key factor (along with predictions of the bending of light by gravity) in GR gaining credibility.
See the two-body problem in general relativity.
Unbounded orbits: $\epsilon \geq 1$
The relation between $r$ and $\phi$ in (15) is $$r(\phi) = \frac{c}{1+\epsilon \cos \phi}.$$
The denominator vanishes when $$\epsilon \cos \phi_{max} = -1$$
That is, the orbits are confined to $-\phi_{max} \lt \phi \lt \phi_{max}$
Actually, if $\epsilon \gt 1$, this means $\cos\phi_{max}\lt 0$. So $\phi_{max} = \pi -\eta$ where $\eta$ is an angle in the first quadrant, such that $\epsilon = 1/\cos\eta$
Here's a graph for $c=1$ and $\epsilon = 1/(\cos(\pi/2))$, that is $\phi_{max}=(3 \pi)/2$ - drawn as the gray line.
In Cartesian coordinates, this works out to... $$\frac{(x-\delta)^2}{\alpha^2} - \frac{y^2}{\beta^2} = 1$$
which is the equation of a hyperbola.
For $\epsilon$ = 1, $\phi_{max}=\pi$, and the corresponding figure is a parabola with the equation: $$y^2 = c^2-2cx$$
Homework
Chapter 8, problems: 19, 28, 29