## General Keplerian Motion

Previous section: General Kinematic Theorem.

Next sectionPlanar Rigid Body Motion and Reference Frames.

#### Johannes Kepler

(1571 - 1630)

General Keplerian motion under an inverse square force.

Reference:

NFCM {= New Foundations in Classical Mechanics, 2nd Ed., 1999}, Chapter 4.

(1) Dynamical model:

$\Large{m\,\,\frac{{d{\bf{v}}}}{{dt}}\,\,\, = \,\,\, - \,\,\frac{k}{{{r^2}}}\,\,{\bf{\hat r}}}$

(2) Problem: Reduce to algebraic model by finding constants of motion.

(a) Angular momentum:

$\Large{{\bf{L}}\,\,\, = \,\,\,m\,{\bf{r}} \wedge {\bf{v}}\,\,\, = \,\,\,m{r^2}\,{\bf{\hat r}}\,\,\frac{{d{\bf{\hat r}}}}{{dt}}}$

(b) Eccentricity vector
ε in:

$\Large{{\bf{Lv}}\,\,\, = \,\,\,k\left( {{\bf{\hat r}}\, + \,{\boldsymbol{\varepsilon}} } \right)}$

Hints [See link under part (3), below, for more detail.]:

$\Large{(i)\,\,\,\,m\,{\bf{r}} \wedge \frac{{d{\bf{v}}}}{{dt}}\,\,\, = \,\,\,\frac{d}{{dt}}\left( {m\,{\bf{r}} \wedge {\bf{v}}} \right)\,\,\, = \,\,\,{\bf{r}} \wedge {\bf{f}}\left( {\bf{r}} \right)\,\,\, = \,\,\,0\,\,\,\,\,\,\,\,({\rm{Central}}\,\,\,\,{\rm{force}})}$
$\Large{(ii)\,\,\,\,{\bf{L}}\,\,\frac{{d{\bf{v}}}}{{dt}}\,\,\, = \,\,\, - \,\,\frac{{k\,{\bf{L}}}}{{m{r^2}}}\,\,{\bf{\hat r}}\,\,\, = \,\,\,k\,\,\frac{{d{\bf{\hat r}}}}{{dt}}}$

(3) Model analysis. Derive the following algebraic features of the model.
[Details are fully worked out in
NFCM and see link below for more detail and an example.]

(a) Energy

$\Large{E\,\,\, = \,\,\,\frac{1}{2}\,\,m{v^2}\,\, - \,\,\frac{k}{r}}$

is constant and related to eccentricity and angular momentum by

$\Large{{\varepsilon ^2}\,\, - \,\,1\,\,\, = \,\,\,\frac{{2{L^2}E}}{{m{k^2}}}}$

(b) Orbit:

$\Large{\begin{array}{*{20}{l}}{r\,\,\, = \,\,\,\left| {\bf{r}} \right|\,\,\, = \,\,\,r\left( {{\bf{\hat r}}} \right)\,\,\,\,\,~~\,\,\,{\rm{(for}}\,\,\,\,{\bf{L}} \ne 0)}\\\\{r\,\,\, = \,\,\,\frac{{ \pm \,\,\ell }}{{1 + {\rm{ }}{\boldsymbol{\varepsilon}} \cdot {\bf{\hat r}}}}\,\,\,\,\,\,\,\,where\,\,\,\,\,\,\,\,\ell \,\,\, = \,\,\,\frac{{{L^2}}}{{m\,\,\left| k \right|}}\,\,\,\,\,\,\,\,( \pm \,\,\,\,{\rm{for}}\,\,\,\,{\rm{attractive/repulsive}}\,\,\,\,{\rm{force}})}\end{array}}$

[
Note: orbit is not a trajectory, which gives position as a function of time.]

(c) Hodograph:

$\Large{{\bf{v}}\,\,\, = \,\,\,{\bf{v}}\left( {{\bf{\hat r}}} \right)\,\,\, = \,\,\,\frac{k}{{\bf{L}}}\,\,\left( {{\bf{\hat r}}\, + \,{\boldsymbol{\varepsilon}} } \right)\,\,\,\,\,\,\,~~\,({\rm{orbit}}\,\,\,\,{\rm{in}}\,\,\,\,{\rm{velocity}}\,\,\,\,{\rm{space}})}$

Show this is a circle by deriving the non-parametric equation (
u is the hodograph center, see below)

$\Large{{\left( {{\bf{v}} - {\bf{u}}} \right)^2}\,\,\, = \,\,\,\frac{{{k^2}}}{{{L^2}}}}$

(d) Classification of orbits (by eccentricity, energy & hodograph center u):

Circle:          ε = 0
E = - mk2/(2L2)          u = 0

Ellipse:       0 < ε < 1          E < 0                      u < |k|/L

Parabola:          ε = 1             E = 0                      u = |k|/L

Hyperbola:          ε > 1             E > 0                      u > |k|/L

(e) Initial value problem: Determine L and ε given m, k, and one concurrent value of v and r.

(Click here for more derivation details and an example of elliptical orbit with associated hodograph.)

(4) Scattering problem: Given asymptotic initial velocity and angular momentum, find vf = vfinal.

(a) Asymptotic conditions:

$\Large{\begin{array}{l}E\,\,\, = \,\,\,\frac{1}{2}\,\,m{v^2}\,\, - \,\,\frac{k}{r}\,\,\,\, \xrightarrow[{r \to \infty}]{}\,\,\,\,\frac{1}{2}\,\,m{v_0}^2\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{v_0}\,\,\, = \,\,\,\,\left| {{{\bf{v}}_0}} \right|\,\,\, = \,\,\,\,\left| {{{\bf{v}}_f}} \right|\,\,\, = \,\,\,\sqrt {\frac{{2E}}{m}} \\\\{\bf{\hat r}} \wedge {\bf{v}}\,\,\, = \,\,\,\frac{{\bf{L}}}{{mr}}\,\,\,\, \xrightarrow[{r \to \infty}]{}\,\,\,\,0\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{{{\bf{\hat r}}}_0}\,\,\, = \,\,\, - \,\,{{{\bf{\hat v}}}_0}\,\,,\,\,\,\,\,\,{{{\bf{\hat r}}}_f}\,\,\, = \,\,\,{{{\bf{\hat v}}}_f}\end{array}}$

(b) Eccentricity conservation:

$\Large{\begin{array}{l}{\bf{L}}{{\bf{v}}_2}\,\, - \,\,k\,{{{\bf{\hat r}}}_2}\,\,\, = \,\,\,{\bf{L}}{{\bf{v}}_1}\,\, - \,\,k\,{{{\bf{\hat r}}}_1}\\\\\\\,\,\,\,~~~~~\,\,\,\, \Rightarrow \,\,\,\,\,\,\left( {{\bf{L}}{v_0}\,\, - \,\,k} \right){{{\bf{\hat v}}}_f}\,\,\, = \,\,\left( {{\bf{L}}{v_0}\,\, + \,\,k} \right){{{\bf{\hat v}}}_0}\\\\\,\,\,\,~~~~~\,\,\,\, \Rightarrow \,\,\,\,\,\,{{\bf{v}}_f}\,\,\, = \,\,\,\left( {\frac{{{\bf{L}}{v_0}\,\, + \,\,k}}{{{\bf{L}}{v_0}\,\, - \,\,k}}} \right)\,\,{{\bf{v}}_0}\end{array}}$

This solves the scattering problem completely!!

The remaining problem is to reformulate it in terms of observed quantities:

Scattering angle
Θ and impact parameter b are defined and related as follows:

$\Large{\begin{array}{l}{{\bf{v}}_f}\,\,\, = \,\,\,{{\bf{v}}_0}\,\,{e^{{\bf{i}}\Theta }}\,\,\, = \,\,\,{e^{ - {\bf{i}}\Theta }}\,\,{{\bf{v}}_0}\\\\\\L\,\,\, = \,\,\,bm{v_0}\,\,\, = \,\,\,\frac{{2bE}}{{{v_0}}}\,\,\,\,\,\,\,\,and\,\,\,\,\,\,\,\,{\bf{\hat L}}\,\,\, = \,\,\,\frac{k}{{\left| k \right|}}\,\,{\bf{i}}\\\\\\{e^{{\bf{i}}\Theta }}\,\,\, = \,\,\,\frac{{{\bf{L}}{v_0}\,\, - \,\,k}}{{{\bf{L}}{v_0}\,\, + \,\,k}}\,\,\, = \,\,\,\frac{{2{\bf{i}}\,b\,E\,k/\,\left| k \right|\, - \,\,k}}{{2{\bf{i}}\,b\,E\,k/\,\left| k \right|\,\, + \,\,k}}\,\,\, = \,\,\,\frac{{2{\bf{i}}\,b\,E\,\, - \,\,\left| k \right|}}{{2{\bf{i}}\,b\,E\,\, + \,\,\left| k \right|}}\\\\\,\,\,\,\,~~~\,\,\, \Rightarrow \,\,\,\,\,\,b\,\,\, = \,\,\,\left| {\frac{k}{{2E}}} \right|\,\,\cot \left( {\frac{1}{2}\Theta } \right)\end{array}}$

which is the relationship known from Rutherford scattering (Coulomb scattering)!

Previous sectionGeneral Kinematic Theorem.

Next section: Planar Rigid Body Motion and Reference Frames.