Does the 3-body problem have stable solutions?

  • Thread starter Thread starter kjknohw
  • Start date Start date
  • Tags Tags
    Stable
Click For Summary
The discussion centers on the stability of three-body systems with non-negligible mass under Newtonian physics. It is established that stable configurations can exist, particularly when the bodies form an equilateral triangle, provided a specific mathematical condition involving their masses is met. The stability can be analyzed through Hamilton's equations and eigenvalue problems, with negative real parts indicating stability. Additionally, practical calculations can be performed using modern computer software to explore these configurations. Overall, while stable solutions are possible, the conditions for their existence are mathematically complex.
kjknohw
Messages
6
Reaction score
0
Given 3- bodies of non-negleble mass, are there stable configurations or not.


for example,
Will a planet (with non-zero mass) orbiting 1000 au away from two stars 0.1 au apart eventually leave the system if you give it 10^9999999... years; or is the system stable for a truly infinete time?
assume: Newtonian physics with point masses. (no gravitational waves no time dialation, etc.)

Note: this is a more of a mathmatical question than physical.

If no stable system exists, is there any way to estimate the time it will take for the system to destabalize.
 
Astronomy news on Phys.org
kjknohw said:
Given 3- bodies of non-negleble mass, are there stable configurations or not.

Yes.


Take a look at

http://groups-beta.google.com/group/rec.arts.sf.science/msg/05738a5d682962cc

and

http://groups-beta.google.com/group/rec.arts.sf.science/msg/1fd8079db42c1137

Three bodies orbiting around their common center of mass in an equilateral triangle will be stable if

(m[1]+m[2]+m[3])^2 -27*(m[1]*m[3]+m[3]*m[2]+m[1]*m[2]) >= 0

Sources

Volume 5 of of "What's Happening in the Mathematical Sciences" by Barry Cipra.

and a personal (rather messy) computerized calculation (see the second link). Basically you start with Hamilton's equations, you linearize them, and you wind up with a 12x12 eigenvalue problem. If all of the eigenvalues have negative real parts, the (linearized) system is stable.

Home computers with the right software are good enough nowadays to solve this symbolically. (It helps a lot that the matrix is very sparse).
 
one way to put it is also to make 2 of the stars orbit around their comon center and then let those 2 orbig with the third star around the comon center, this solution is the one nature prefer cause the chance of 3 stars bieng created at the right spot with the right distance with the right mass are slim. if u have a planet then it will have to orbit in the hill sphere of one of the stars. The most stable one i can think of is to orbit the third and "lonly" star. its hills sphere will probebly be bigger.
hill sphere radius
r=a(m/(3M))^(1/3)
the hill sphere radius (r) for a body with mass (m) orbiting a heavier body with mass M at a distance of a is approximently to that.
 
I recently purchased a lunar clock intended as a Christmas present for a young family friend who has become interested in astronomy. The clock face shows twelve images of the Moon in all its main phases, with the Full Moon depicted at the "twelve o'clock" position. Now I purchased this clock especially early to allow time to check on its accuracy. After waiting for a full lunar orbit to occur, I have since discovered that the clock time when measured relative to its lunar images (that's to...

Similar threads

  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 21 ·
Replies
21
Views
5K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
8K
  • · Replies 26 ·
Replies
26
Views
3K
  • · Replies 9 ·
Replies
9
Views
5K