||Message]
I Lesson Overview
Learning Goals
In this lesson you will learn the following.
- Conservation of angular momentum (direction) implies orbits lie in a plane So the problem gets reduced to solving for two coordinates in a plane.
- The conservation of angular momentum (absolute value) implies Kepler's second law holds for any spherically symmetric potential.
- Conservation of energy and angular momentum (magnitude) helps in reducing the problem to quadratures.
Prerequisites
- Basic knowledge of differential and integral calculus and vectors.
- Solving the equations of motion of a planet around the sun can be is equivalent to the solution of a single body of reduced mass \(\mu\) where \[\mu = \frac{mM}{M+m}.\] and \(M, m\) are masses of the sun and the planet respectively.
- For a particle in a spherically symmetric potential, the angular momentum is conserved.
II Main Lessons
\(\S\) 1 Kepler Orbits
\(\S\) 2 Question for You to try it now, in two different ways
Show that the latus rectum of the elliptic Kepler orbit is just the radius of circular orbit for energy \(E\) and angular momentum \(L\).
\(\S\) 3 Equation of Planetary Orbits
\(\S\) 4 Orbit Parameters
\(\S\) 5 Hyperbolic Orbits
\(\S\) 6 Question for You to try it now
Are there orbits for zero angular momentum? Is the orbit, if it exists, is bounded or unbounded? Answer different ranges of energies.
III EndNotes
Key concepts used
Spherically symmetric potentials; Effective potential, Bounded motion; Areal velocity; Conservation of energy and angular momentum;Polar form equations of ellipse, parabola and hyperbola.
Summary of properties of Kepler orbits
Notation Mass = \(\mu\), \hspace Energy = \(E\) \hspace , Angular Momentum \(=\ell\)
| {ll} | |
| Potential | \( V(r)=-\dfrac{k}{r}\) |
| Effective Potential | \( V_\text{eff}(r)=-\dfrac{k}{r} + \dfrac{\ell^2}{2\mu r^2}\) |
| Radius of circular orbit with same \(\ell,\mu,k\) | \(r_0=\dfrac{\ell^2}{\mu k}\) |
| Equation of orbit | \(\dfrac{r_0}{r}=1-\epsilon\cos\theta\) |
| Eccentricity | \( \epsilon= \sqrt{1+\dfrac{2E \ell^2}{\mu k^2}}\) |
| Minimum distance | \(r_\text{min}=\dfrac{r_0}{1+\epsilon}\) |
| Maximum distance | \(r_\text{max}=\dfrac{r_0}{1-\epsilon}\) |
| Semi major axis | \(a= \dfrac{(r_\text{min} + r_\text{max})}{2}=\dfrac{r_0}{1-\epsilon^2}=\dfrac{k}{2|E|}\) |
| Semi minor axis | \(b=a\sqrt{(1-\epsilon^2)}= \dfrac{r_0}{\sqrt{(1-\epsilon^2)}} =\dfrac{\ell}{2\mu |E|} \) |
| Time period | \(T^2=\dfrac{\pi^2\mu k^2}{(-2E^2)}= \dfrac{4\mu\pi^2a^3}{k}\) |
Runge Lenz Vector
For the Kepler problem there is an additional constant of motion, given by \begin{equation} \vec{N} = \vec{v} \times \vec{L} \frac{k\vec{r}}{r} \end{equation} Also the orbit equation can be derived using conservation of Runge-Lenz vector.\\ \paragraph*{Learn More} For a range of topics related to Runge Lenz vector,including its role in quantum mechanics, see \\{\scriptsize \url{http://www.cds.caltech.edu/~marsden/wiki/uploads/projects/geomech/Alemicds205final.pdf
Bohlin Transformation
Bohlin transformation connects the Keplar problem to harmonic oscillator problem. Learn more from\\ Maria Luisa Saggio, Eur. J. of Physics {\bf 34} (2013) 129-137 }
Exclude node summary :
Exclude node links:
4727:Diamond Point
