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N-body problem

  1. Apr 16, 2005 #1
    i am attempting to find a solution for the n-body problem, but i don't know the equations for gravity in three dimensions. if someone could post them for me, i would be most appreciative.

    thank you

    also, any advice as to how to approach this problem would be appreciated as well
  2. jcsd
  3. Apr 16, 2005 #2
    At the risk of sounding unintentionally arrogant, im going to assume that I need to be more specific when i say 'n-body problem'

    The n-body problem is the supposedly unsolvable method of calculating the orbit of more than two bodies influencing each other through gravity. It is currently believed that this is not possible, and that a three-body system is unpredictable, not because of a lack of proficiency in our current math, but because math itself is unable to solve it. I believe this to be wrong, and am interested in attempting to solve the n-body problem. However, in order to start this, i need the equations for calculating three dimensional gravity stuff. you know, like F=G*m1*m2/r^2, except for three dimensions, with x, y and z axes. also any tips that could help me are welcomed.

    thank you.
  4. Apr 16, 2005 #3
    Don't worry, you are not looking arrogant here, you are looking like pure mathematician.
    The 3D law of gravity is

    [tex]\vec{F_2}=-G\frac{m_1 m_2}{R_{12}^2}\frac{\vec R_{12}}{R_{12}}[/tex]

    The Force is directed along the line connecting the bodies and it is attractive.


    [tex]\vec{F_2}=-G\frac{m_1 m_2}{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}\frac{\vec R_{12}}{R_{12}}[/tex]
    Last edited: Apr 16, 2005
  5. Apr 16, 2005 #4
    Thank you very much, Shyboy.

    Just to clarify, what do the arrows above F and R represent?
    Last edited: Apr 16, 2005
  6. Apr 17, 2005 #5


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    It means they are vector quantities.

    I find shyboy has written the law in a peculiar manner. I would have written.. consider two point-masses m1 and m2 respectively, and respectively located by the position vectors [itex]\vec{r_{1}}[/itex] and [itex]\vec{r_{2}}[/tex]

    [tex]\vec{F}_{1\rightarrow 2} = -\frac{Gm_1m_2 \vec{r_{12}}}{|\vec{r_{12}}^3|} = -\frac{Gm_1m_2 \hat{r_{12}}}{|\vec{r_{12}}^2|}[/tex]


    [tex]\vec{r_{12}} = \vec{r_{2}} - \vec{r_{1}}[/tex]

    Or, in ugly cartesian coordinates...

    [tex]\vec{F}_{1\rightarrow 2} = \frac{-Gm_1m_2}{\Left[(x_1-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2\Right]^{3/2}} \left( (x_2-x_1)\hat{x} + (y_2-y_1)\hat{y} + (z_2-z_1)\hat{z} \right) [/tex]
  7. Apr 17, 2005 #6
    Haha I feel very ignorant here, but given my formal background in physics, i guess I am ignorant. What does the ^ above the R represent, as opposed to the vector arrows?
  8. Apr 17, 2005 #7


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    It stands for "unit vector"
  9. Apr 17, 2005 #8
    so in that equation, the unit vectors represent direction only, since (X2-X1) and so on would represent magnitude. I think Ive got it now. thank you to everyone for your help.
  10. Apr 19, 2005 #9
    Woudl this problem be similar to the Coulomb Force Law's inability to handle multiple moving charges?
  11. Apr 19, 2005 #10
    LizardKing23, what are you smoking? I want some too! Or is this some kind of a joke? Do you even know the solution to the two body problem? Hint: it is reducible to a problem of one body moving in a central potential, and the solution of the one body problem involves calculus.
    Last edited: Apr 19, 2005

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