# Lagrange and Hamiltonian question

I cant understand what the question is asking~ hope somebody can help me~
A particle of mass m moves in a plane under the influence of Newtonian gravitation force, described by the potential V(r) = - GmM/r (symbol in conventional meaning)

Now introduce a new variable u(theta) = 1/r(theta) and obtain an equation satisfied by u. Find the equation of the orbit in polar coordinates. What are the allowed trajectories of the particle?

yukyuk

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Well, I guess the easiest way to go about figuring it out would be to start out by writing the lagrangian, but in polar cooridinates (kinetic energy, potential energy parameterized by radius, theta). Once you have done this, you should be able to see where it goes. Of course you could also write down the Hamiltonian as well and go from there. Pay attention to what happens to the kinetic energy term when you change momentum in cartesian coordinates to polar coordinates.

A particle in an inverse-square force field~ what will be the potential then?
yukyuk

HallsofIvy
Homework Helper
Since force is the (negative) gradient of the potential energy, what is the
integral of 1/r2?

reilly
See Goldstein's Classical mechanics, or any text that deals with orbits from a central force. Going to 1/r is a trick that has been used for many years. With 1/r as the variable, Newton's 2nd Law gives a straightforward differential equation. Google porbably will lead you to what you want.

Regards,
Reilly Atkinson

O~~ please help me!!! I don't know how to deal with the new variable!! Does the question want me to replace all the r in the original Lagrange equation by u or use u to write out another Lagrange equation??

also in the polar coordinate given in the case, r and theta are functions of t, then does it mean that
dr/dt now become -1/u^-2 (theta). d(theta)/dt,as r=1/u~~~

so confusing yukyuk

Additionally, you may want to write the derivative with respect to time in the final differential equation (containing r's and theta's) as:

$$\frac{d}{dt} = \frac{d\theta}{dt} \cdot \frac{d}{d\theta}$$

Nothing more than a chain rule here. However, you're probably going to glean a lot of information from looking at a mechanics text book.