The center of mass

To simplify we first consider only two bodies. On each body there is an external force

Fand an internal force_{e}F. The internal force acts between the bodies, and for celestial bodies, this is a gravitational force._{i}Fwould then be the sum of gravitational forces from bodies outside the system. We assume that the bodies are on the_{e}x-axis, and that the forces act along this axis.

Choose a reference system which is an inertial system. Newton's second law gives:

According to Newton's third law the two internal forces are equal and opposite. Adding the equations then gives

We imagine that the two bodies are combined to one body with mass

m_{1 }+m_{2}, and that this body is acted upon by the same external forces as our two bodies. This imaginary body then gets an acceleration which we call the acceleration of the center of massaand given by:_{CM}

As the acceleration is the second derivative of the position, we have:

We may then define the position of a point called the center of mass in this way:

If the sum of the external forces is zero, the acceleration of the center of mass will be zero, and the velocity

vof the center of mass will be constant. Usually we choose a reference system where_{CM}v= 0. The center of mass will then be at rest._{CM}From the equations above, we see that the momentum of the system is described by the equation

Consequently,

if the center of mass is at rest, the total momentum is zero.If body 1 is the Earth and body 2 the Moon, and we put

x_{1}= 0, we getx_{CM}= 4660 km. The center of mass of the system is inside the Earth.

By considering components of forces, accelerations, velocities and positions, it is easy to see that the equations also are valid in three dimensions. If the bodies don't lie along a straight line, we must calculate thex-, y-andz-coordinate of the center of mass.It is also easy to extend the calculation to many bodies. With three bodies they-coordinate is calculated from:

If we include sufficiently many bodies when calculating the center of mass, the center of mass will be at rest or have a constant velocity with respect to an inertal system (see the two-body problem). As Newton's second law is exactly valid only in inertial systems, the natural choice is to give all velocities and accelerations with respect to the center of mass.

The center of mass is very useful when describing and calculating the orbits of celestial bodies.

Before running a simulation, Orbit Xplorer computes the position of the center of mass. If necessary, the velocities in

Parametersare adjusted so that the velocity of the center of mass will be the one given inParameters.