Hamiltons equations for a satellite

  • Thread starter Thread starter anubis01
  • Start date Start date
  • Tags Tags
    Satellite
Click For Summary
SUMMARY

The discussion focuses on deriving Hamilton's canonical equations for a satellite of mass m orbiting a star of mass M, utilizing both polar and Cartesian coordinates. The kinetic energy in polar coordinates is defined as T=m/2 * (ṙ² + r²θ̇²), while the potential energy is V=-GMm/r. In Cartesian coordinates, the kinetic energy is T=m/2(ẋ² + ẏ²) and the potential energy is V=-GMm/√(x²+y²). The correct Hamiltonian formulation is H=T+V, which leads to four Hamilton's equations.

PREREQUISITES
  • Understanding of Hamiltonian mechanics
  • Familiarity with polar and Cartesian coordinate systems
  • Knowledge of kinetic and potential energy formulations
  • Basic grasp of Lagrangian mechanics
NEXT STEPS
  • Study Hamilton's equations in detail
  • Explore the derivation of the Lagrangian from kinetic and potential energy
  • Learn about the implications of energy conservation in orbital mechanics
  • Investigate the differences between polar and Cartesian coordinate systems in mechanics
USEFUL FOR

Students and professionals in physics, particularly those focusing on classical mechanics, orbital dynamics, and Hamiltonian systems.

anubis01
Messages
149
Reaction score
1

Homework Statement


Find Hamiltons canonical equations for a satellite of mass m moving about a star of mass M at the origin. Consider motion in a plane, using polar coordinates r,\theta.

Repeat the process using cartesian coordinates x,y from the start.


Homework Equations


H=T+V


The Attempt at a Solution


I did both questions but I'm not exactly sure if I defined the kinetic and potential energies correctly for each case. For polar coordinates I found T=m/2 * (r\dot{}2 +r2 \theta\dot{}2) and potential energy V=V(r)

For cartessian coordinates I found the Kinetic T=m/2(x\dot{}2)
and potential energy V=\frac{-GMm}{\sqrt{x^2+y^2}}.

So if anyone can confirm if I set up the kinetic and potential energy equations correctly that would be great.
 
Last edited:
Physics news on Phys.org
That is the correct kinetic energy in the Lagrangian formulation, what did you have for your kinetic energy and how did you form the Hamiltonian?
 
Taking care to rearrange the equation I found H=\frac{1}{2m} [pr2+\frac{p\theta^2}{r^2}]. Using this I got the hamilton equations, which there 4 of in this case. I'm sure i did that part correctly i just wasn't 100% sure about setting up the kinetic and potential energy equations correctly.

Oh and when doing the same problem in cartessian coordinates was I correct in defining the Kinetic energy T=m/2(x\dot{}2)
and potential energy V=\frac{-GMm}{\sqrt{x^2+y^2}}.
 
anubis01 said:
Taking care to rearrange the equation I found H=\frac{1}{2m} [pr2+\frac{p\theta^2}{r^2}]. Using this I got the hamilton equations, which there 4 of in this case.

Not quite. H=T+V, and you only included the kinetic energy. The potential energy, V=-GMm/r, must also be included.

Oh and when doing the same problem in cartessian coordinates was I correct in defining the Kinetic energy T=m/2(x\dot{}2)
and potential energy V=\frac{-GMm}{\sqrt{x^2+y^2}}.

The potential energy is correct, but don't forget the y component of velocity counts towards the kinetic energy!
 
Ah, that makes more sense. Thanks for the help, its much appreciated.
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
Replies
5
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
6
Views
2K
Replies
7
Views
3K
Replies
6
Views
3K