9
$\begingroup$

Every explanation of Lagrange points I've seen refers to it as a sort of three-body solution, where one of the body's masses is taken to be comparatively negligible to make the solution work.

However, I haven't been able to find a description of at what mass a body is considered negligible, or even a ratio.

For example, if I wanted to add a second moon to the Earth-Moon system by placing it at an L4 or L5 point, how big could that secondary moon be before it no longer fits the 'comparatively negligible mass' category?

Thanks!

$\endgroup$
0

2 Answers 2

9
$\begingroup$

The following is quoted from Reference [1, p. 65] by Kemp (note: the original work goes back to 1843, but I am not able to trace this reference because [1] does not appear to include a Reference list):

$\dfrac{(m_1+m_2+m_3)^2}{m_1m_2+m_1m_3+m_2m_3}>27$

This may be rearranged to give

$\dfrac{m_1^2+m_2^2+m_3^2}{m_1m_2+m_1m_3+m_2m_3}>25$

Suppose, for instance, that object 1 is Earth and object 2 is the Moon, thus in rounded numbers $m_1=81$ and $m_2=1$ taking the Moon's mass as a unit. Therefore

$\dfrac{6562+m_3^2}{81+82m_3}>25$

$m_3^2-2050m_3+4537>0$

$m_3<2.215...$ (for masses smaller than the Earth)

This says that to make a stable system with the Earth and Moon, the dust at one Lagrange point (not both) must measure no more than about twice the Moon's mass.

This result is roughly in line with the total mass of the secondary and Lagrange-point objects being limited to about 4% of the primary.

Naturally occurring Lagrange-point systems, of course, come nowhere near this limit, for two reasons. First, external disturbances such as other planets in a planet-based system tend to reduce the stability range. Second, and more important, the mass ratios of naturally occurring objects do not allow the limit to be approached. For instance, the Sun-Jupiter system, which is relatively robust against external disturbances because of the large mass of Jupiter, should easily be able to hold another Jupiter mass at either of its "trojan" points, but the combined mass of all asteroids that might accumulate there does not even match our Moon. The "negligible mass" assumption is therefore accurate for all systems of interest in astronomy and space expolration.

Reference

1. Kemp, Sean, "An Examination of the Mass Limit for Stability at the Triangular Lagrange Points for a Three-Body System and a Special Case of the Four-Body Problem" (2015). Master's Theses. 4546. https://doi.org/10.31979/etd.4tf8-hnqx https://scholarworks.sjsu.edu/etd_theses/4546

$\endgroup$
3
  • $\begingroup$ This is very helpful, thank you. However, when you rearrange the equation given by Kemp, your expansion of the brackets seems incomplete. The expansion of the brackets would be (changing the M-terms to a, b, and c for clarity), a^2 + b^2 + c^2 + 2ab + 2bc + 2ac. The equation then becomes quadratic and offers two solutions (one of which is so large Saturn could fit in it, and the other is 2.4, just barely larger than your estimate- given it's a less-than term, I take the latter, I imagine). Apart from that (which seems to come out to a very similar answer anyway), this is perfect. Thanks! $\endgroup$
    – Connie
    Oct 4, 2021 at 1:13
  • $\begingroup$ Not sure what you mean @connie. The apparently missing terms in the numerator are in fact subtracted off, decreasing the RHS from 27 to 25. My root 2.215..., which I specify as the one smaller than Earth's mass, does appear to agree numerically with the original equation drawn from Kemp. Thank you. $\endgroup$ Oct 4, 2021 at 1:31
  • $\begingroup$ Ah, I see. Not sure how I missed the change from 27 to 25, but that's very clever. When I was looking at the 2(81 +82x) term on the top, and the 81 + 82x term at the bottom, I was annoyed it didn't look like there was some quick and neat way to get rid of them since the squared terms were in the way, but apparently there was! Guess I get one point for Intuition Connie guessing right, and taking two for Lazy Connie not following up that hunch. I'll try to remember that for the future, your way was much easier. $\endgroup$
    – Connie
    Oct 4, 2021 at 1:37
-1
$\begingroup$

It's a great question!

Incomplete answer ending with a question:


Lagrange points are mathematical concepts that were derived assuming only two massive bodies (in the universe!) in circular orbits around their center of mass.

There will then be a gravitational potential field, and its gradient is a force field.

In the rotating frame those can be expressed as a pseudo-force field and pseudo-potential field. Looking at it in a rotating frame, the two massive bodies are fixed, and this makes solving some problems easier.

As you've mentioned already, the third body of the "circular restricted three body problem" or CR3BP is basically a "massless test particle" and you can put it anywhere to see what happens. The Lagrange points are mathematically derived places where a hypothetical test particle would sit still if placed exactly at the point. Just like a bowling ball would sit at the top of something. Depending on the shape, it might be stable against a tiny displacement (inside the caldera of a tall, inactive volcano), or it might be unstable (at the convex top of a pointy mountain).

The third body doesn't have to exist. It's an abstraction that we move around to answer "What if?"

...how big could that secondary moon be before it no longer fits the 'comparatively negligible mass' category?

There are of course more than two bodies in the universe pulling on each other, and even just in our solar system. And none of the orbits is exactly circular.

So we've already broken both assumptions necessary for Lagrange points to exist mathematically. If you put something at the Sun-Earth L1 point, gravitational effects from the Moon and Jupiter, and perturbing effects because the Earth's orbit is elliptical and not perfectly circular ruin any chances of putting something "exactly" at L1. It's not really there.

Instead there's sort of an "L1-ish area" where things will be somewhat more unstable than they would be in the pure CR3BP.

Adding mass to "the third" body in our actually many-body, elliptical orbit-based solar system won't make much of a difference to its trajectory until it starts changing the motion of the two bodies we're focusing on when pretending that the Lagrange points exist.

So if you removed Earth's Moon from its normal orbit and put it at Sun-Earth L1, That's about a 1% effect.

note: The best way to think of an object at the Sun-Earth Lagrange points is that they are orbiting the Sun as the Earth is, (about 1% closer to the Sun), but in a resonant orbit with the Earth.

If the Sun and Earth orbited in circular orbits and there were no other planets, I think that they could still be in 1:1 resonant heliocentric orbits, but the distance will now be a little different than the classical L1 distance.

To double check on this, I've just asked

in Astronomy SE, and will update here based on the outcome there.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.