# Tag Info

3

I actually answered a related question a couple of days ago on Astronomy. Small world! One of the important properties of the Oort Cloud is that objects in it are not strongly influenced by the Sun. After all, its inner edge is roughly 2,000 AU away - 300 billion kilometers from the Sun. The Sun's gravitational influence in that region is rather weak, so ...

0

I see no reason why it should be impossible. A binary planet system should exist in a reasonably stable orbital configuration. As Javier pointed out, Pluto and Charon are similar in size and mass, like a small-scale version of what you describe. The important thing to note here is that the barycenter of the Pluto-Charon system is outside either planet. In ...

2

On the off chance there was any doubt about the numerics, I too wrote a code to follow these orbits. I use RK4 to take the first half timestep (a full timestep here is always $0.001$), and then perform the remaining integration via leapfrog. Below is the evolution of an asymmetric orbit. The potential is given by $v_0 = 0.9$, $q = 0.7$, $L = 0.05$. I set ...

6

A simpler example is Kepler's laws of planetary motion. In a spherically-symmetric gravitational field, the planets follow elliptical orbits. The orbits certainly do not have the full spherical symmetry of the potential. They may be extremely lopsided. :-D

1

Answer to your old question: Right now, the perihelion currently occurs around the same time as (2 weeks after) the northern winter solstice, but 10000 years from now it will occur around the northern summer solstice. So, like other people have said, it's just coincidence that right now they both happen at about the same time. Answer to your new question: ...

5

How far ahead can we predict solar and lunar eclipses? NASA has uncertainty calculations that show how certain we are about when eclipses happen. From a back of the envelope, the eclipses will likely vary by a full day 35 thousand years from now. That said, we have eclipse seasons, so we know eclipses will continue to happen, and at roughly which time of ...

49

On predicting planetary orbits A number of studies have shown that the inner solar system is chaotic, with a Lyapunov time scale of about 5 million years. This 5 million year time scale means that while one can somewhat reasonably create a planetary ephemeris (a time-based catalog of where the planets were / will be) that spans from 10 million years into ...

1

Look I know link only answers are terrible, but I don't want to ruin the surprise. Check out this link. That's on Wikipedia! It's safe to say that if wikipedia knows something, the experts know quite a bit more. We know the dynamics of the Sun-Earth-Moon system really well including the perturbations and the most likely sources of future upsets, so barring ...

-2

Given that we exist at all, we can confidently back-track the positions of the planets (and our moon) for a couple billion years. I see no reason, barring rather massive exo-system-sourced objects showing up unexpectedly, that the positions will go chaotic enough to be unpredictable any time in the next couple billion years. I suppose it might depend a bit ...

0

I would agree that a circular orbit with full tidal lock is more "in equilibrium" (everything is time independent in the rotating frame) than one in which the orbiting object had zero rotation in an absolute Cartesian reference. There is however an important notion for determining whether something is rotating, and that is its angular momentum. An object ...

0

Even without setting $\sin \theta=1$ you have a cubic equation for $m_2$. Such things can be solved, like the quadratic formula but messier. Various computer algebra packages incorporate the solution. For graphing, you can just pick a bunch of $m_2$ points, solve for $m_1$, and plot the points.

3

The first definition of $\mu=GM$ is the standard definition of the SGP. The second one comes from the velocity of a circular orbit. If you have an object in a circular orbit of radius $r$ and velocity $v$ around a body of mass $M$, then the velocity is given by $$v=\sqrt{\frac{GM}{r}}$$ From this you can see that $rv^2=GM$ for circularly orbiting objects. ...

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