# Homework Help: Hamiltonian formulation and the Kepler problem

1. Oct 19, 2016

### gammon54

This isn't exactly homework, but something which you'd get as an assignment, so I'll still post it here in order to reach the right people..

I'm attempting to freshen up my knowledge on Hamiltonian systems, so I've tried to formulate the Kepler problem in Hamiltonian dynamics. I[/PLAIN] [Broken] got results similar to the one from this thread:

In polar coordinates ($\varphi$, r), the Lagrangian is $L = \frac12 m(r^2\dot \varphi^2 + \dot r^2) + \frac{\mu}r$, where μ is the standard gravitational parameter, and m the mass of the object in orbit. The generalized momenta are $p_\phi = \frac{\partial L}{\partial \dot \varphi} = mr^2 \dot \varphi$ and $p_r = m\dot r$. The associated Hamiltonian reads

$H = \frac12 \frac{p_\phi^2}{m r^2} + \frac12 \frac{p_r^2}{m} - \frac{\mu}r$

The equations of motion should hence be

$\frac{dr}{dt} = \frac{\partial H}{\partial p_r} = \frac{p_r}m$
$\frac{d\varphi}{dt} = \frac{p_\phi}{m r^2}$
$\frac{dp_r}{dt} = -\frac{\partial H}{\partial r} = \frac{p_\phi^2}{m r^3} - \frac{\mu}{r^2}$
$p_\phi = const$

At this point I thought I'd validate this by calculating the orbit of the ISS around earth. It's nearly circular, so I should be able to set $\dot p_r = 0$ and then check that the resulting equation

$\frac{p_\phi^2}{m r^3} - \frac{\mu}{r^2}$

holds for the orbit parameters (which are available e.g. from Wolfram Alpha). Problem is: The result is off by a factor of the mass $m$. This can be seen from the formula as well: I arrive at the expression

$m \dot \varphi^2 r^3 = \mu$

for the orbit parameters, whereas according to Wikipedia on "Circular orbit", $\dot \varphi^2 r^3 = \mu$ would be correct.

I am hence assuming that I made a mistake with the mass somewhere in the derivation, but I can't find it. I'd appreciate a hint! :-)

Last edited by a moderator: May 8, 2017
2. Oct 19, 2016

### phyzguy

Shouldn't your Lagrangian be $L = \frac12 m(r^2\dot \varphi^2 + \dot r^2) + \frac{\mu m}r$ ?

3. Oct 20, 2016

### gammon54

Right, thanks.. I was too worried about the details and forgot to double check the potential I guess.

$\int \frac{GMm}{r^2} dr = -\frac{GMm}{r} = -\frac{\mu m}{r}$

of course. Thanks a lot, now everything works out :-)

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