Questions about Lagrange points in the Pluto / Charon system

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SUMMARY

The discussion centers on the dynamics of the Pluto-Charon system, specifically regarding its barycentre and the stability of Lagrange points. It is established that the barycentre does not represent a stable equilibrium point, as any slight disturbance will cause a point mass to drift towards the more massive body. The relationship between the barycentre and the classical Lagrange points (L1 to L5) is clarified, noting that while the distances to these points depend on the masses of the bodies, their relative positions remain consistent regardless of mass variations.

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Cerenkov
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Hello.

With the recent interest in the JWST orbiting at the L2 Lagrange point of the Earth - Moon system, I was wondering about the dynamics of the Pluto - Charon system. Specifically, the barycentre of that system.

This barycentre lies at a point in space between these two bodies. Does this mean that it can be considered as a stable or semi-stable region, similar to the classical Lagrange points, L1 to L5? Effectively an L6?

Also, since the system's barycentre is displaced from the interior of Pluto, does this mean that points L1 to L5 are proportionally displaced too? Or is there some else going on that I'm unaware of?

Thank you for any help given.

Cerenkov.
 
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Cerenkov said:
This barycentre lies at a point in space between these two bodies. Does this mean that it can be considered as a stable or semi-stable region, similar to the classical Lagrange points, L1 to L5? Effectively an L6?
No, the barycentre of two bodies is, in general, not a point where nett forces are zero.
It's easy to see why, if we first imagine the one case where the forces are indeed zero at that point - the case of two bodies of equal mass. At barycentre we have no centrifugal forces, just gravity. From Newton's law of gravity: ##F_g=GMm/r^2##, the forces acting on a point mass ##m## placed at the barycentre of two massive bodies ##M1=M2## are ##F_{nett}=F_{g_{M_1}}-F_{g_{M_2}}=GM_1m/r_1^2-GM_2m/r_2^2##. And since both masses ##M_1## and ##M_2## are the same, and the barycentre is equally distant from each (##r_1=r_2##), these add up to zero.
But it is an unstable equilibrium. If you imagine the point mass getting a tiniest push in the direction of either body, the distance r to that body will reduce, which will increase the force towards it while reducing the force towards the other body. The point mass then ends up dragged away from the barycentre.
Now, if you do not have two bodies of equal mass, the barycentre will be such that ##r_1<r_2## if ##M_1>M_2##. If you place a point mass there, it will be attracted more towards the more massive body not only due to the smaller distance, but also due to the larger mass.
Intuitively, if you imagine an object at e.g. the barycentre of the solar system, which is sometimes just above the surface of the Sun, you would not expect it to be able to hover in place. Would you? The entire mass of the Sun just next to it, and the tiny planets AUs away.

Cerenkov said:
Also, since the system's barycentre is displaced from the interior of Pluto, does this mean that points L1 to L5 are proportionally displaced too? Or is there some else going on that I'm unaware of?
The distances to the Lagrange points depend on the masses of the bodies, yes. But not in the same way as the position of the barycentre. Similar to the barycentre and point of zero nett force, described above - there is no simple proportionality.
However, no matter how you change the masses of the two main bodies, it will always be true that L1, L2, and L3 are collinear with the line connecting the two bodies. While L4 and L5 are always at 60 degrees away from either massive body. These don't change, even if you vary the masses.

edit: maths error corrected
 
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Thank you Bandersnatch. :smile:

Intuitively, if you imagine an object at e.g. the barycentre of the solar system, which is sometimes just above the surface of the Sun, you would not expect it to be able to hover in place. Would you? The entire mass of the Sun just next to it, and the tiny planets AUs away.

This was helpful. I can visualise what you mean here.

However, no matter how you change the masses of the two main bodies, it will always be true that L1, L2, and L3 are collinear with the line connecting the two bodies. While L4 and L5 are always at 60 degrees away from either massive body. These don't change, even if you vary the masses.

As was this.

When I was writing my question I'd forgotten that the barycentre of the solar system sometimes lies outside of the sun's surface - so thanks for reminding me of that.

Cheers.

Cerenkov.
 

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