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L2 Lagragian point?

  1. Oct 30, 2008 #1
    Hi this is my first time posting on this forum. I have an question about Lagragian points.

    I was trying to find L2 lagrangian point, a point that lies on the line defined by the two large masses, beyond the smaller of the two. Here, the gravitational forces of the two large masses balance the centrifugal force on the smaller mass. Let's say the larger object has mass M and the smaller object has mass m, and the distance between the two objects is R and the distance from the smaller object to L2 is r, then how do you guys derive a formula to calculate r? I was given a hint that I need to use approximation method meaning R is exceedingly greater than r such that (r/R)^2 is equal to zero. Again, Unlike L1 which is located between the two objects, L2 lies behind the smaller object.

    At least if you can give me some hints, I would really appreciate. Thanks.
  2. jcsd
  3. Oct 30, 2008 #2

    D H

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    Work in the frame with origin at the center of mass and rotating with the objects so that the location of the two massive objects in this frame is constant. The five Lagrangian points will also have a fixed location in this frame. The accelerations on a fixed point in a rotating frame are the gravitational accelerations toward the two bodies and the centrifugal force. The accelerations must sum to zero at some point if it is to remain fixed. The first goal is to find this polynomial. Once you have this polynomial, you can solve it using numerical methods, or you can find an approximate solution.
  4. Oct 31, 2008 #3


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    DH, i just wanted to be a little more anal about this, to include a couple of points that i needed to remember when i did this (for recreational purposes, i don't think i ever needed to do this for a class).

    of the two massive objects. ignoring any masses of anything that might be living at the L2 point, we can easily model the two planets as point masses revolving around their common center of mass. and we don't have to think about them being affected by the test mass at L2.

    the other thing i needed, but had to remember to something i only used in freshman physics or in my Statics and Dynamics class required for engineering was D'Alembert's principle. this centrifugal force is a fictitious force that we put in because of this.

    i think you can get an closed form solution, no? the order of the polynomial doesn't exceed 3, does it?

    i think the L4 and L5 points are the cool ones. those are the ones that are "pits" or "depressions" at a local minimum of potential energy
  5. Oct 31, 2008 #4


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  6. Oct 31, 2008 #5

    D H

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    The equation for the L2 point is

    [tex]\frac{G(M+m)}{R^3}\left(R\frac{M}{M+m}+r\right)\,-\,\frac{GM}{(R+r)^2}\,-\,\frac {Gm}{r^2}= 0[/tex]

    where M and m are the masses of the two massive objects (M > m), R is the distance between the two massive objects, and r, the unknown quantity, is the distance between the smaller object and the L2 point. Denoting α=m/M, x=r/R, this becomes

    [tex](1+\alpha)x+1-\frac{1}{(1+x)^2}-\frac {\alpha}{x^2}= 0[/tex]

    Clearing the fraction leads to

    [tex]((1+\alpha)x+1)x^2(1+x)^2-x^2-\alpha(1+x)^2 = 0[/tex]

    which is a fifth order polynomial.
  7. Oct 31, 2008 #6
    If this is the case, the above polynomial becomes r = -R/2 if you use (r/R)^2 = approximately zero. But the solution to this question is, I believe, r = R(m/3M)^(1/3). Am I missing something here?
  8. Oct 31, 2008 #7

    D H

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    [itex]r=R\,(m/3M)^{1/3}[/itex] is an approximate solution. How to get there from the fifth degree polynomial:

    First rewrite the polynomial this way:

    [tex]x^2((1+x)^3-1) -\alpha(1+x)^2(1-x^3)= 0[/tex]

    Expanding the first term,

    [tex]x^3(3+3x+x^2) -\alpha(1+x)^2(1-x^3)= 0[/tex]

    Assuming [itex]x\ll 1[/itex], [itex]3+3x+x^2\approx 3 [/itex] and [itex](1+x)^2(1-x^3)\approx 1[/tex], so

    [tex]3x^3 \approx \alpha[/tex]


  9. Nov 3, 2008 #8
    Thanks for your help. But I still not comprehend the last approximation method part. How come x^3 still survives when the other terms vanish?
  10. Nov 3, 2008 #9

    D H

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    Dropping the x2 and higher power terms from the original polynomial leads to [itex]\alpha(1+2x) \approx 0[/itex], which means [itex]x \approx -1/2[/itex] or [itex]\alpha \approx 0[/itex] (or both). The former is nonsense and the latter is just a reconfirmation of the assumption [itex]\alpha \ll 1[/itex]. In other words, the naive first-order approximation doesn't work. Refactoring the polynomial as a polynomial in [itex]alpha[/itex] and then eliminating higher order terms in x in each term does work.
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