The orbiting planet produces a potential V(r,ϑ,φ−Ωt). It has explicit time dependence and exerts an azimuthal torque, so generally E˙=Vt and L˙z=−Vφ are nonzero. Nevertheless a simultaneous time-translation symmetry δt=λ and rotation δφ=Ωλ leaves the Lagrangian and dt unchanged. Noether's theorem gives Q=Ωpφ−E and henceEquivalently Vt=−ΩVφ implies d(E−ΩLz)/dt=0. This Jacobi integral assumes a rigidly rotating potential with constant Ω and negligible back-reaction from the test particle.
EJ=E−ΩLz=constant.
Ancestors (11)
- 1
- S7
- 2019
- Past exam of the short options paper shared by Parts A and B of the physics course of the University of Oxford
- Short options paper shared by Parts A and B of the physics course of the University of Oxford
- Physics course of the University of Oxford
- Undergraduate course of the University of Oxford
- Course of the University of Oxford
- University of Oxford
- List of universities
- Home
Discussion (0)
New discussionThere are no discussions about this article yet.
Articles by others on the same topic (0)
There are currently no matching articles.