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3

According to my understanding, for an orbital resonance to be stable, there must be at least three circumstances present: The forces on the bodies in resonance have to cancel out over time, The momentary forces shouldn't be that high that the orbits are changed significantly, after a perturbation, the bodies should pull each other back into resonance. For ...


1

I see one possible way for this to sort-of happen. Anders Sandberg did a good job of showing why a simple capture is impossible, but there's an extremely unlikely but not impossible scenario he missed. Consider Jupiter and Saturn. They're on the same side of the sun, Saturn is ahead of Jupiter. The black hats kidnap the sun and they're free flying. ...


2

I love Anders's answer — well done. But you could have got the idea already by looking at orbit perturbations the planets inflict on each other: They are mostly negligible. Come to think of it — it cannot be otherwise in a stable planetary system: During the evolution of planetary systems the orbits of planets which are too close to each other will be ...


-2

Without the sun... at current orbital velocities... The planets would fly apart... But exactly how far would they fly? Would they continue indefinitely into interstellar space? Or would they fly out into the Kuiper belt or Ooort cloud and subsequently fall back toward some barycenter. Removing the sun removes 99% of the mass of the solar system... so this ...


6

The stability of this system depends the ratio of masses of the two stars. If the larger star is more than 25 times more massive than the smaller star, then L5 is potentially stable, and this remains the case even if the planet does not have negligible mass The calculation of the value is done in detail on physics.stackexchange and there you can establish ...


6

I gave a try at refining Anders Sandberg's answer with the equations for a two body system. Basically, a two body system is bound if the sum of the total kinetic energy and the (negative) gravitational potential energy is still negative. (To escape, kinetic energy needs to cancel out the potential energy.) Gravitational potential energy: $$U = -{GMm\over{R}}$...


7

No. The 8 planets would go into 8 different directions. It is because their relative velocity to each other is much higher than the escape velocity, even from their smallest distance. If it would not be so, their orbits could disturb each other significantly, even today with the Sun. Thus, the Solar System would become chaotic. However, the moon systems of ...


31

The issue here is whether pairs of planets can become gravitationally bound to each other. In the two-body problem the trajectories or orbits are ellipses (bound orbits), parabolas and hyperbolas (unbound). For all practical purposes, an encounter looks like they start out at infinity with some finite speed, approach each other, and then maybe fly away or ...


1

Scenario: This looks specifically at a near-miss scenario with an Earth-like intruder passing by at a distance of 1,000,000 km at a speed of 100 km/s. This is about three times the distance to the Moon and at about three times Earth’s orbital velocity. The 100 km/s seems a reasonable number given the relative velocities of our stellar neighbors, the solar ...


3

Sometime before now and "tens of millions of years". Phobos currently orbits at a radius of about 9400km The theoretical Roche limit is different for rigid and fluid bodies. If Phobos were a fluid body, it would already have passed the Roche limit at 10500km. This is because fluid bodies will deform into an ellipse pointing towards the planet, ...


3

Partial answer about the retrograde motion only: Yes, the Sun has yearly retrograde motion as seen from Mercury I've written a simple script to generate the orbital motion of Mercury based on its eccentricity. I've used reduced units with $GM = a = 1$ and period $T = 2 \pi$. Mercury's rotational period is 2/3 of its orbital period. I've plotted the ...


2

Interesting question. The terms day and year become funny when you move to Mercury: Mercury is tidally locked to the Sun and has a 2:3 spin-orbit-resonance. That means that a complete day-night cycle takes two Mercury year, thus one year / orbital period around the Sun you have completely day, the other completely night). Du to this spin-orbit resonance, you ...


2

It depends how you define 'Keplerian orbit'. A precessing orbit can obviously not be represented by a stationary ellipse anymore, but it is essentially still the same ellipse, merely uniformly rotating about its focus. The main features of the orbit, like semi-major axis and period of revolution (as defined by the time interval between two perihelion ...


0

It depends! If we plotted the orbits of the innermost planets in our solar system, they would be closer to ellipses with the Sun at one focus. If we plotted the orbit of Jupiter, the biggest "gravitational bully" in the solar system (it messes with everything!) it would be closer to an ellipse with the Sun-Jupiter barycenter at one focus. It gets a ...


3

A point not mentioned above is that the primordial swarm of material that would eventually form the Moon would, on average, have the same direction of orbit around the Earth as the Moon does today. In that case, we shouldn't ignore collisions between that material where the radial momenta are cancelled and leave only, or mostly, the tangential momenta (...


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