I'm simulating the restricted three body problem. But I don't know how to found the initial conditions of the object located at the Lagrange points analytically or computationally. I mean my problem is that the object located at the Lagrange points as initial position have to remain at the same distance relative to the second object.

I've already look the derivation of the Lagrange points, but when deriving it, people look at the system as if were stopped in a moment in time and find the saddle points when $\vec{F}_{\Omega}=\vec{0}$. My problem is the following:

Problem: Given three objects with masses, $m_1,m_2,m_3$, such that $m_3<<m_1,m_2$. Find the initial conditions of the third object, such that it's initial position is one of the Lagrange points and that it remains orbiting always in the Lagrange point for a long period of time.

I can use all the initial conditions of the first, second, and third object. I also know the solution to the first,second and third objects position velocity for any time. I just have to know the initial conditions such as the constraint above is satisfied.

I managed to find a initial velocity by putting numbers, in following picture(but that's not the idea):

Initial Conditions for this picture:

$x_1=1.0 $cm ,
$y_1=6.0 \cdot 10^3$ cm ,
$vy_1=1.0 \cdot 10^6$ cm/s
$m_1=8.0 \cdot 10^{22}$g

$x_2=4.0 \cdot 10^3 $cm ,
$y_2=6.0 \cdot 10^3$ cm ,
$vy_2=2.0 \cdot 10^6$ cm/s
$m_2=2.0 \cdot 10^{21}$g

$x_3=3.194 \cdot 10^3 $cm ,$L_1$ $x$ component
$y_3=6.0 \cdot 10^3$ cm , $L_1$ $y$ component
$vx_3=0$cm/s ???
$vy_3=1.080680314 \cdot 10^6$ cm/s ???
$m_3 =0$g


  • $\begingroup$ This is the derivation I'm mentioning: Derivation $\endgroup$
    – Keith
    Jun 18, 2014 at 15:05
  • $\begingroup$ Look at Figure 2 of the document and rotate it; note that the L-points remain fixed relative to $m_1$ & $m_2$. This seems like good reason to look at it as a moment in time. $\endgroup$
    – Kyle Kanos
    Jun 18, 2014 at 16:03
  • $\begingroup$ Also, your L1 in the image seems to be way too far away from the masses to be right. $\endgroup$
    – Kyle Kanos
    Jun 18, 2014 at 16:09
  • $\begingroup$ I computed it with this formula : $L_1$=$[R(1-(\frac{M_2}{3(M1+M2)})^{1/3}, y_1]$ , so I believe it's right $\endgroup$
    – Keith
    Jun 18, 2014 at 16:12
  • $\begingroup$ Oh, I see now. I just misread your graph. I thought the blue & purple lines were opposite of what they are (that your L1 point was sitting outside the two objects). $\endgroup$
    – Kyle Kanos
    Jun 18, 2014 at 16:20

1 Answer 1


The Lagrange points are fixed in the rotating coordinate frame where the two masses are fixed. You should be able to start with $m_3$ having zero velocity in that frame. Transform that zero velocity into inertial space (if that is where you are doing the simulation) and there you should be.

Looking at your data, you don't have the CM fixed at $(0,0)$, which would be convenient. The initial $x$ values should be opposite in sign, with the magnitudes a factor of $40=\frac {m_1}{m_2}$ different. Also the initial $y$ velocities should have opposite signs and the same ratio. Then the initial $y$ velocity for the small object should be to keep it on the line between the two other masses.

  • $\begingroup$ what is the initial $y$ velocity for the small object to keep it on the line between the two other masses ?. How can I find it ? $\endgroup$
    – Keith
    Jun 18, 2014 at 16:39
  • $\begingroup$ Since the CM is stationary, you want $vy_3=\frac {x_3}{x_2}vy_2$. That means it has the same angular velocity around the CM as mass $2$ $\endgroup$ Jun 18, 2014 at 16:40

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.