Abstract:
The N-body problem looks into solving the gravity effect on all
bodies in a system most effectively.

My problem:
To get to the solution, I have to calculate the center of gravity
in a given section of the problem area that will represent all
bodies as one. This will be solved in two dimensions. All
bodies have a mass, and a position with x, y coordinates.
Could someone help me with the formula to find this point? Also
the gravity force for this point would help (and I would imagine
it is necessary to find the solution.)

"Torstein Raanes" <[Only registered users see links. ].uit.no> wrote in message
news:bqega8$2p5f$[Only registered users see links. ].no...

In a Cartesian coordinate system, the coordinates of the center of mass of a
group of point masses is the *weighted* average of the coordinates of the
individual masses. In two dimensions:

X-bar = Sum[m(i)*x(i)]/Sum[m(i)]

and

Y-bar = Sum[m(i)*y(i)]/Sum[m(i)]

Again, outside the circumscribing sphere, the collection can be treated as a
single body with

m(tot) = Sum[m(i)]

The difficulty with the N-body problem is that there are more momenta to be
conserved than there are degrees of geometric freedom, so separation of
variables into orthonormal coordinates is not possible in general. In the
two-body problem we get a solution by transforming the x, y, and z
coordinates of two bodies into six *new* coordinates with orthogonal
momenta: the x, y, and z momenta of the center of mass and the r, theta, and
z momenta (cylindrical coordinates) of the rotating binary. With only three
bodies you get nine momenta to be conserved, and it is not easy to find
places in 3 dimensions to put 9 orthogonal momenta.

Sometimes a three body problem can "rearrange" itself into two two-body
problems by "ejecting" one body leaving the other two as a binary. Then the
ejected body and the binary become an effectively independent two-body
problem.