# Why objects at lagrangian points 4 and 5 are stable?

1. Jun 9, 2007

### magnetar

why objects at lagrangian points 4 and 5 are stable?

2. Jun 9, 2007

### cesiumfrog

Is the http://en.wikipedia.org/wiki/Lagrangian_point" [Broken]explanation ("[w]hen a body at these points is perturbed, it moves away from the point, but the Coriolis effect bends the object's path into a stable, kidney bean‐shaped orbit around the point") insufficient for a particular reason?

Last edited by a moderator: May 2, 2017
3. Jun 9, 2007

### pervect

Staff Emeritus
Small masses at the L4 and L5 points are stable if and only if the mass ratio between the primary and secondary is greater than around 22:1. (I don't recall the exact figure offhand).

Demonstrating this using calculus is a rather lengthly calculation, involving finding the equations of motion of the small mass, linearizing the resulting differential equations, and showing that these linearized differential equations don't have exponentially growing solutions.

If you have the necessary background in calculus and the interest, I could provide a few links.

4. Jun 9, 2007

### tony873004

Big masses are stable there too. To get the exact figure you would need to specify how the secondary masses are divided. For example, you could have a star, a planet, and a massless test particle in the Planet's L4 or 5 point. Or you could have a star, and 2 equal-massed planets spaced 60 degrees apart, or any combination inbetween. It doesn't make a huge difference, changing by something like (my guesses) 22:1 to 23:1 between these two extreme conditions.

Here's the results of a Gravity Simulator simulation I performed trying to find out the mass ratio between the star and two combined equal-mass planets. The first column is is the mass ratio ((Planet1 mass + Planet2 mass) / star mass). The 2nd column is how many years this trojan configuration is stable. In this example, the equally massive planets are 1 AU away from a 1 solar mass star.

The interesting part is that it is unstable for all values ratios greater than or equal to 0.041, then it becomes stable for a small range of ratios from 0.04 to 0.031, then becomes unstable again for a small range of ratios from 0.03 to 0.0286, then becomes stable again for all ratios less than 0.0285.

Code (Text):

2m/M     years stable
0.05        21
0.041      430
0.04    10000+
0.0388  10000+
0.0386  10000+
0.0385  10000+
0.031   10000+
0.03       800
0.299      658
0.0295     400
0.0294     450
0.0293     400
0.029      400
0.0289     440
0.0288     490
0.0287     536
0.0286     990
0.0285  10000+
0.028   10000+
0.02    10000+

A book called "The Three-Body Problem" by Christian Marchal verifies this strange island of instability around 0.029 in Chapter 8, although he computes it with math rather than simulating it like I did, and concludes that there are 2 stable regions: 0 <= R < 0.02860...and 0.02860... < R <= 0.03852.

In these examples, the mass ratio between the planet and the trojan object is 1:1. I believe that the above results (both numerical and analytic) would change, but not by much, if you altered this 1:1 ratio.

5. Jun 9, 2007

### pervect

Staff Emeritus
I've taken a more detailed look at your results:

Interestingly enough, there does seem to be something going on here that I'm not familiar with. The equation that I give below predicts that instability starts when m/M = .019819, or 2m/M $\approx$ .0396

The stability analysis I did was a purely linear one - what sort did the author do in your textbook?

The results that I know about are as follows:

In order for three massive bodies to be stable (as defined by a linear analysis) one must have the following relationship between their masses:

$$27 \, (m_1*m_2 + m_2*m_3 + m_3*m_1) < (m_1 + m_2 + m_3)^2$$

The best literature reference I have for this is Volume 5 of of "What's Happening in the Mathematical Sciences" by Barry Cipra.

If m_3 is zero, this reduces to the usual relationship (see for instance http://www.physics.montana.edu/faculty/cornish/lagrange.pdf [Broken] for the usual 2-body case).

Once upon I time I took it upon myself to confirm this result via a rather involved computer analysis based on the Hamiltonian approach. One can find a summary of what's involved in the following post.