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Does the centre of mass of two planets keep following the same course before and after a collision of two planets in a solar system?

Is it possible to use conservation of linear momentum even though there is an external force acting on the planet in some way? If we assume the duration of collision is very short. Another question: Is this also related to the 'reduced mass' approach of planets in a gravity field?

Hopefully someone can clear up my problem.

EDIT 1/2/2022: I wasn't being very clear so my particular question is: Imagine if you have two planets orbiting a sun. So the path of the centre of mass of those two planets is an ellipse.

Then imagine the planets collide (just like billiard balls with no destruction etc). What kind of curve will be the new path of the COM after this collision? Will it be the same ellipse? Does it matter if it is elastic vs inelastic?

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    $\begingroup$ Re. your second question, LM is not conserved when there's an external force in play. $\endgroup$
    – Gert
    Jan 30 at 13:31
  • $\begingroup$ Ambiguity in the question is leading to some confusion in the answers. “Conservation of linear momentum through a collision” is a standard problem in which the collision is instantaneous and all non-collision forces can be neglected; linear momentum of the center of mass is conserved through the collision, such that its velocity is the same immediately before and after the collision. Comments from the OP indicate that the actual question may be “does the path of the COM of two planets experiencing orbital motion match the orbital path of a one planet starting at the initial state of the COM?” $\endgroup$
    – RLH
    Jan 31 at 17:28
  • $\begingroup$ Yes you are right, I have updated the question $\endgroup$ Feb 1 at 22:03

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Disproof by counterexample: imagine two planets of equal mass moving on the same circular orbit around the sun, but in opposite directions (assume they missed each other in the past due to a tiny deviation from congruency). Then their center of mass moves on a line through the sun, back and forth, sinusoidally as a function of time.

Now they happen to collide, so the center of mass at first stops in the inertial system. But instead of continuing to move sinusoidally forever, the center of mass of the new planet falls into the (formal) gravitational singularity of the sun. This is a one-time event, and it is not even remotely being described by a sinusoidal function of time (at first it is approximately free fall at a certain gravitational acceleration, but this g-factor increases as the CM approaches the sun).

Generally, linear momentum conservation only holds for systems that are not exposed to an external force (in this case the gravtiational force of the sun).

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    $\begingroup$ Does your argument depend on the fact that the new "double planet" will crash and burn when it encounters the sun? $\endgroup$
    – garyp
    Jan 30 at 14:45
  • $\begingroup$ No, because the motion is different already far away from the sun. $\endgroup$
    – oliver
    Jan 30 at 17:28
  • $\begingroup$ Instantaneously, however, the sun's gravity has negligible effect. The center of mass will go from "zero velocity wrt the sun because it's at an extremum of an oscillation" to "zero velocity because the two planets are at rest in the frame we're considering". $\endgroup$
    – RLH
    Jan 30 at 22:34
  • $\begingroup$ Ignoring the fact that you're ignoring the fact that the sun would "catch" the newly doubled planet, I believe that this is a special case. As a counterexample to your counterexample, if Earths collided in such a way that a sizable piece of debris passed close to Jupiter, it would get a gravity boost -- then the system of the two planets plus their remnants would have a change in momentum that the two planets, individually, would not have experienced. $\endgroup$
    – TimWescott
    Jan 31 at 0:00
  • $\begingroup$ @RLH: absolutely true. I am not questioning the fact that momentum will be approximately conserved in the vicinity of the impact. I interpreted the original question as to whether the center of mass will follow the same motion over a finite period, depending on whether an impact happened or not. If you are only looking at an infinitesimally small period of time around the impact, you might also argue that the sun is irrelevant even if there is no impact because the planets' circular motion is "tangentially equivalent" to inertial/straight motion. $\endgroup$
    – oliver
    Jan 31 at 9:45
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Does the centre of mass of two planets keep following the same course before and after a collision of two planets in a solar system?

The question that I can answer easily is the one that ends in "two rogue planets in interstellar space", and the answer is "how can it not keep following the same course". There's a collision, and momentum must be conserved.

Is it possible to use conservation of linear momentum even though there is an external force acting on the planet in some way?

I'm pretty sure this is where the "in a solar system" part comes in in your question. If two planets collide in a solar system then in the short term you can ignore the effects of the sun's gravity (for instance, if I'm getting my math right, the Earth's acceleration due to the pull of the sun is under $6\cdot10^{-3}\ \mathrm {m/s^2}$). How short is "short term" depends on what you want to know, but in the case of something in Earth orbit, after a day the difference in velocity will be just $490 \mathrm{m/s}$, which is probably negligible*.

If you wanted to know what the post-collision effect would be for some significant fraction of the orbital period after two planets collide, then yes, gravity would make a big difference, and given that the differential equations involved are nonlinear and multi-bodied, I would guess that you'd need to simulate a bazillion collisions and see what happened, rather than just being able to figure it out with pencil and paper.

* Even though this value is about 1000 times bigger than my pre-edit value -- it's still negligible given the sizes and speeds involved. Or so I claim.

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  • $\begingroup$ Oops -- divided by $2 \pi$ when I should have multiplied -- and that error got raised to the power of 4, and voila! Thanks. $\endgroup$
    – TimWescott
    Jan 31 at 5:50
  • $\begingroup$ Thanks for your response, I think I worded my question not particularly right. I meant to say: Imagine if you have two planets orbiting a sun. Then the path of the centre of mass of those two planets is an ellipse. Then the planets collide (just like billiard balls with no destruction etc). What kind of curve will be the new path of the COM after this collision? $\endgroup$ Jan 31 at 12:50
  • $\begingroup$ This being stackexchange, why don't you edit your question to reflect that detail? Short answer -- if all the pieces stay pretty close to each other, momentum will be preserved -- but if bits end up slingshotting around other planets or the central star(s), then no. $\endgroup$
    – TimWescott
    Jan 31 at 21:45
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The short answers to the original question is no (title question), yes, and yes: basically, the CM follows the original trajectory and the conservation of momentum applies. But the question and the other answers are not directly addressing the most important element of this problem: in orbital mechanics the gravitational field is not uniform and therefore the assumptions that make the CM a useful concept no longer generally apply, although they will apply in the large vicinity of the collision.

The CM is not a useful concept when the field is not uniform, and therefore the CM does not apply for most orbital mechanics. For the collision then, the CM applies at the time of the collision, but long before and long after when the objects are in different fields, the concept of the CM is not useful. Therefore, in a collision between two planets in space, the two planets enter their pre-collision-trajectories using trajectories that have no relation to a useful CM concept. As they approach, they start to enter a more uniform field where the CM now becomes relevant, and if they stay stuck together after the collision, this immediate-pre-collision CM trajectory will be the trajectory after the collision and establish long-term trajectory of the new planet.

As for conservation of momentum, it is a useful concept here (and much more so than one commenter implied stating that it's not useful because there's an applied force). To start, think of "conservation of momentum" as the result of $F = 0$ in the equation $F = dp/dt$, so when there's no applied force, momentum is conserved. But one idea of the CM is that when there's a uniform field, the CM follows a trajectory as though it were a single mass, independent of the internal forces and dynamics, and what often happens is that the internal forces lead to dynamics that are separable from the CM trajectory. This essentially leaves two separate problems: the CM trajectory and the internal trajectories relative the CM. This is the case, for example, in collisions and explosions, where something will explode in mid-air where the motion of the parts can be calculated using conservation of momentum relative to the CM. That is, if there are no internal forces, then momentum is conserved relative to the motion of the CM. And it's also useful, for example, with masses and springs, or maybe with planets and the internal gravitational forces of the planets. This is the beauty of the CM. And this is relevant in space too, so long as the CM is relevant, which will be for a long time since it will take the particles a long time to drift out of (or have arrived from) a part of space with sufficiently different gravitational field to make the CM not relevant. When the fields are different, the CM can still be calculated, but it no longer provides anything useful.

(As is probably obvious from reading this answer, the assumption in the edit to the question, which states: "So the path of the centre of mass of the two planets is an ellipse." This is not generally the case because the gravitational fields at objects in orbit are different. Consider, for example, Mercury and Neptune, where Neptune is about $300\times$ the mass of Mercury. The CM of these will basically be Neptune's orbit with a small and fast wobble created by Mercury -- so not an ellipse. That is, for most collisions of planets, the CM is very useful in the vicinity of the collision, but not useful long before or long after the collision because at these times the planets are generally in different gravitational fields. Whereas this is the case for stones in the gravitational field of the earth: the CM will always be applicable and CM of two objects following parabolas will also be a parabola.)

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In the radial field from the star, the center of gravity of the two planet system is not in the same location as the center of mass. A rearrangement of the mass distribution will probably cause a change in the orbit being followed by the center of mass (while conserving angular momentum).

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Lots and none at all and what did your own calcs show?

Until they collide, no two bodies share either a centre of mass or a path, except by irrelevant co-incidence. To suggest they did would bring us perilously close to the three-body problem…

Consider two billiard balls and first, explain how it matters whether both, either or neither is in motion, relative to each other or to any external point of reference…

If the balls collide, does it matter whether either of them bursts apart?

I suggest that bursting or remaining intact, the impact will not change the centre of mass or gravity of either ball, but it should change the path of both.

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