
#1
Apr1605, 06:46 PM

P: 13

i am attempting to find a solution for the nbody 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
Apr1605, 09:27 PM

P: 13

At the risk of sounding unintentionally arrogant, im going to assume that I need to be more specific when i say 'nbody problem'
The nbody 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 threebody 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 nbody 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. 



#3
Apr1605, 09:55 PM

P: 138

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. or [tex]\vec{F_2}=G\frac{m_1 m_2}{(x_2x_1)^2+(y_2y_1)^2+(z_2z_1)^2}\frac{\vec R_{12}}{R_{12}}[/tex] 



#4
Apr1605, 10:44 PM

P: 13

Nbody problem
Thank you very much, Shyboy.
Just to clarify, what do the arrows above F and R represent? 



#5
Apr1705, 12:14 AM

Sci Advisor
HW Helper
PF Gold
P: 4,768

It means they are vector quantities.
I find shyboy has written the law in a peculiar manner. I would have written.. consider two pointmasses 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] Where [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_1x_1)^2+(y_2y_1)^2+(z_2z_1)^2\Right]^{3/2}} \left( (x_2x_1)\hat{x} + (y_2y_1)\hat{y} + (z_2z_1)\hat{z} \right) [/tex] 



#6
Apr1705, 10:09 AM

P: 13

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
Apr1705, 10:58 AM

P: 13

so in that equation, the unit vectors represent direction only, since (X2X1) and so on would represent magnitude. I think Ive got it now. thank you to everyone for your help.




#9
Apr1905, 12:53 AM

P: 2,223

Woudl this problem be similar to the Coulomb Force Law's inability to handle multiple moving charges?




#10
Apr1905, 05:53 AM

P: 21

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.



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