# Tag Info

## New answers tagged celestial-mechanics

1

There are surely solutions for any $Z$ for which the electrons will remain bound forever. For example, the electrons may orbit the nucleus exactly on the opposite sides and the same distance. On the other hand, for any $Z$, almost all initial states will lead to the escape of one of the electrons because the motion is chaotic and will have a nonzero ...

0

In the Earth-Moon system, the mass of the Moon is sufficiently large and it is sufficiently close to raise tides in the matter of Earth. Because of this tidal friction can occur which dissipates energy. Enormous amount of power are involved in this process (of $\approx$ TW) and part of this causes the Moon to constantly being promoted to a slightly higher ...

2

According to Newton's theory of gravity, the orbit would be an ellipse, which could take the form of a perfect circle. However, Einstein's theory of General Relativity tells us that this elliptical orbit would very gradually decay due to the emission of gravitational waves, and perhaps also precess. The Wikipedia page on the two-body problem in general ...

-1

It will depend on the initial conditions and I personally think that all you will need to answer your question can be found in any book on classical mechanics ("Kepler problem") or on Wikipedia

2

The stable orbits around a star are given by the Kepler's laws oft planetary motion. In general these are ellipses with the center star in one of the two foci. Circular orbits are the special case when there is only one focus. For a orbit with a given radius, there is only one speed which allows a circular orbit. If you have a starting condition where the ...

0

Geometry says no. Since the orbital (DISREGARD - direction reverses) (INSERT - period changes) in a horseshoe orbit, the rotation of the body around its axis would have to change as well, and this is not going to happen over the course of a single orbit. Tidal locking takes a long time.

1

Direct integration schemes give bad results. You can do much better by using the exact solution in the absence of gravitational interactions between the planets. You can then set up a variation of constants approach where you take the integration constants (which are the orbital parameters) as dynamical variables and write the differential equations (where ...

3

The standard way to choose a time-step is to run a test simulation with multiple bodies and plot the total energy of the system versus time. The total energy should remain (roughly) constant. If your step size is too large then you will get energy drift. So simply find the largest time-step that does not produce energy drift. In the case of modelling the ...

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