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If two bodies are close, both will get attracted to each other and collide. Under what conditions will the two bodies start revolving around their common center of mass? I understand that such bodies represent the gravitional two-body problem, but I want to know what are the initial conditions that allow it so settle into such a configuration (without going into the maths)

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    $\begingroup$ I've removed a number of comments that were attempting to answer the question and/or responses to them. Please keep in mind that comments should be used for suggesting improvements and requesting clarification on the question, not for answering. $\endgroup$ – David Z Apr 22 at 9:01
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Unless other objects are "near enough" to complicate the motion, the bodies can always be described by a two-body solution whenever both are in freefall (nothing is pushing one of them). The only difference that being close makes is that forces from other objects become less significant.

You could then describe their motion as one of two groups: a hyperbolic pass if speed is great enough ($\mathit{KE} + \mathit{GPE} > 0$), or an elliptical orbit ($\mathit{KE} + \mathit{GPE} < 0$). A collision can happen in either case, it just means the orbital paths intersect the surface of the objects.

Falling straight toward each other is just a degenerate case, where the ellipse minor axis (if a bound state) or the hyperbola transverse axis (if an unbound state) is zero length.

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  • $\begingroup$ So, two objects will go into an elliptic orbit around their barycentre if they were initially moving with respect to each other (and KE+GPE < ), and this is always guaranteed to happen (even though the path might make the two objects collide)? Under which condition will they never collide and keep moving in the same elliptical orbit? And if they were not moving with respect to one another, they'll just plain collide, right? $\endgroup$ – Daud Apr 22 at 9:31
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    $\begingroup$ @Daud The condition is simply that the sum of the bodies' radii must be less than their distance at periapsis – which is almost a tautology. They don't collide if, when they're the closest to each other in their orbit, they don't touch. And that condition is fulfilled if they have enough relative sideways velocity. But see Roche limit for the case that one or both of the bodies don't have enough internal strength to withstand the tidal forces caused by passing close to each other. $\endgroup$ – JohannesD Apr 22 at 15:14
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    $\begingroup$ @Daud Simply: Any two objects in space will always co-orbit around their shared center of mass, unless (A) the orbits are small enough that they actually slam into each other, or (B) some other object's gravity interacts with them. $\endgroup$ – Mooing Duck Apr 23 at 0:01
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It depends on their initial velocity. If they start off not moving relative to each other, they will fall together and collide. But if they have a (sufficient, depending on their size) component of velocity perpendicular to the line between them they will orbit, or for a large enough component, fly apart.

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    $\begingroup$ Wether they will fly apart or orbit depends on the total velocity, not the component that is non radial. $\endgroup$ – Taemyr Apr 22 at 7:42
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    $\begingroup$ @Taemyr That's what this answer says. But before any of that is relevant the two bodies must be not on a collision course, otherwise you're reading their obit, not their orbit. $\endgroup$ – corsiKa Apr 22 at 14:56
  • $\begingroup$ @corsiKa The answer says "or for a large enough component" and not "or for a large enough velocity" $\endgroup$ – Taemyr Apr 22 at 17:43
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    $\begingroup$ @Taemyr, the answer is true as written, but would not be true if you replace "component" with "velocity", since that would also allow collision. The questioner did not want a full mathematical analysis. If you want to do one, write your own answer. Please do not use comments for stating the obvious. $\endgroup$ – Charles Francis Apr 22 at 18:39
  • $\begingroup$ @CharlesFrancis " but would not be true if you replace "component" with "velocity", since that would also allow collision." - That is incorrect, because you have already accounted for the possibility of a collision in the first part of that sentence. $\endgroup$ – Taemyr Apr 23 at 8:03
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Orbits are like a game of chicken.

Say you're walking down the street and there's someone straight towards you. If you both keep walking straight or don't make enough effort to side step, you'll end up colliding.

You don't want to collide, so you move to the side as you approach... if you're moving too fast you might side step too much, or walk too far at an angle, and then you end up flying off into the busy street or the field on the other side. That's probably going to hurt too.

So, if you manage to side step enough but not too much, you can miss colliding and walk around them. As you do, they get your attention and you slow down to talk... and end up stepping in behind them to not block people.

Orbits in classical mechanics are pretty much like that. For a two body situation:

  • Two bodies lacking sufficient side ways velocity will collide.
  • Two bodies with far too much side ways velocity will miss and only be deflected.
  • Two bodies with sufficient but not excessive side ways velocity will continue to "miss" each other as gravity pulls their vector around the common center of gravity.
  • So long as the two bodies don't attain escape velocity, the shape and size of the orbit of two given objects could be considered as derived from the variables of velocity decomposed to a scalar variable that in a radial direction and a vector variable in a tangential direction.

Hopefully I've got all that right, since I'm writing off the top of my head.

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  • $\begingroup$ It's not completely wrong, but I'd say the emphasis on sideways velocity is a bit misleading. While it's true that two macroscopic bodies approaching each other need to have some sideways relative velocity to avoid colliding, otherwise whether or not they will orbit each other or just fly past each other on hyperbolic trajectories depends on the sum of their potential and kinetic energy, of which the former does not depend on velocity and the latter only depends on the magnitude of their relative velocity, not on its direction. $\endgroup$ – Ilmari Karonen Apr 22 at 18:03
  • $\begingroup$ (Specifically, this holds for a system of two pointlike or spherically symmetric rigid bodies under Newtonian gravity. With more than two bodies, of course, all bets are off, since it's possible for gravitational interactions to transfer energy between the bodies. There are even pathological solutions to Newton's equations with pointlike bodies where the bodies, initially all with negative total energy and thus in bound orbits, end up shooting off to infinity in finite time.) $\endgroup$ – Ilmari Karonen Apr 22 at 18:13
  • $\begingroup$ @IlmariKaronen I would consider the case of having enough velocity to just fly past to be a case of having escape velocity, hence the qualifier on the fourth bullet point. Guess that could have been more clear though. As for n-body orbits ... yeah no, "there be edge case dragons" lol $\endgroup$ – Kaithar Apr 23 at 0:20

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