Celestia Forums

I can't for the life of me figure out how to find the lagrange points mathematically. I know you must analyze things in an inertial frame, but then what? I can add three forces. How do i calculate that lagrange points though?? This is just a curiosity of mine.

For L4 and L5, I just tend to add/subtract 60 degrees from the Arg of Peri.

For L1:

We have the two masses M1, M2 orbiting their centre of mass C with orbital radius a, and the Lagrangian point L at a distance x from the centre-of-mass, as represented in the following ASCII-art diagram (note the points M1, M2, C and L should all be in a straight line).

Defining the mass ratio [tex]q=\frac{M_2}{M_1}[/tex], we have:

[tex]a_1=\frac{q}{1+q}a \mbox{ and } a_2=\frac{1}{1+q}a[/tex]

And the orbital angular frequency ? is given by:

[tex]\omega^2 = \frac{G(M_1+M_2)}{a^3}[/tex]

The gravitational forces on the object at the L1 point must sum to give a centripetal force for a circular orbit around the centre-of-mass, which leads to the equation:

[tex]\frac{GM_1}{(a_1+d)^2}-\frac{GM_2}{(a_2-d)^2}=\omega^2d[/tex]

Which when you write it out in terms of dimensionless quantities: the mass ratio q and the distance [tex]x=\frac{d}{a}[/tex] gives:

[tex]f(x)=\frac{1}{\left(\frac{q}{1+q}+x\right)^2} - \frac{q}{\left(\frac{1}{1+q}-x\right)^2}-(1+q)x=0[/tex]

You then can solve this iteratively using the Newton-Raphson method. A suitable first guess might be [tex]x_0=0.5[/tex] for q<1, and then the following sequence should converge to the true value.

[tex]x_{n+1}=x_n-\frac{f(x_n)}{f^{\prime}(x_n)}=x_n+\frac{\left(\frac{q}{1+q}+x_n\right)^{-2}-q\left(\frac{1}{1+q}-x_n\right)^{-2}-(1+q)x_n}{2\left(\frac{q}{1+q}+x_n\right)^{-3}+2q\left(\frac{1}{1+q}-x_n\right)^{-3}+(1+q)}[/tex]

This then gives you the distance from the centre-of-mass to the Lagrange point in units of the orbital radius of the secondary around the primary.

The distance from the primary to the L1 point (again in units of the orbital radius of the secondary around the primary) is [tex]x+\frac{q}{1+q}[/tex], where x represents the value to which the iteration converges.

The method to calculate L2 and L3 is very similar, I leave it as an exercise for the interested reader

I made a javascript calculator to find the Lagrange points:
http://orbitsimulator.com/formulas/Lagr ... inder.html

http://orbiter.dansteph.com/forum/read. ... 15&t=59315

click: Solar System Lagrange Points ADDON

Instead of copying, quoting the source of your post,

http://en.wikipedia.org/wiki/Lagrangian_point
(a reference which most people know anyway)

would have saved space.

Fridger

Simplified version of above.

Mass must be greater then 25 to 1 for Lagrange Points to exist
[tex]a = {\frac {m}{M+m}}[/tex]
Where
[tex]M =[/tex]Mass of Primary
[tex]m =[/tex] Mass of Child
[tex]R =[/tex]Distance Between Objects

L1 = [tex]r \approx R \times (1 - \sqrt[3] {\frac {a}{3}})[/tex] ; 0° Between Primary and Child
L2 = [tex]r \approx R \times (1 + \sqrt[3] {\frac {a}{3}})[/tex] ; 0° Past Child
L3 = [tex]r \approx - R \times (1 + {\frac {5} {12} \times a)[/tex] ; 180° Far Side of Primary (away from Child)
L4 = [tex]r \approx R[/tex] ; 60° Preceding Child
L5 = [tex]r \approx R[/tex] ; -60° Following Child

Mass of Primary (M) 1.99E+30 Sol
Mass of Child (m) 5.97E+24 Earth
Distance Between Objects (R) 149,597,870,691 m

Mass must be greater then 25 to 1 for Lagrange Points to exist
[tex]a = {\frac {m}{M+m}}[/tex] = 0.00000300150

L1 = [tex]r \approx R \times (1 - \sqrt[3] {\frac {a}{3}})[/tex] = 148,101,642,814 m; 0° Between Primary and Child
L2 = [tex]r \approx R \times (1 + \sqrt[3] {\frac {a}{3}})[/tex] = 151,094,098,568 m; 0° Past Child
L3 = [tex]r \approx - R \times (1 + {\frac {5} {12} \times a)[/tex] =-149,598,057,782 m; 180° Far Side of Primary (away from Child)
L4 = [tex]r \approx R[/tex] = 149,597,870,691 m; 60° Preceding Child
L5 = [tex]r \approx R[/tex] = 149,597,870,691 m; -60° Following Child

Google for Hill Sphere. Wikipedia has a good site. The Hill Sphere will give you the distance to the L1 and L2 points.

L3 is in Earth's orbit, on the opposite side of the Sun, so it is exactly 2 AU away.

L4 and L5 are 60 degrees ahead of and behind the Earth. The Earth, Sun, and L4 form an equilateral triangle, as do the Earth, Sun, and L5. So the Earth L5 distance is 1 AU. The Earth L4 distance is 1 AU. The Sun is also 1 AU from both these points.

Actually, one mass doesn't need to be smaller than the other two, at least for the L4 & L5. And I suspect the other 3 L points as well. The two orbiting masses combined need to be at most ~1/25 the mass of the large object. It's possible for an Earth-mass planet to be in Earth's L4 or L5 point.