The center of mass

To simplify we first consider only two bodies. On each body there is an external force Fe and an internal force Fi. The internal force acts between the bodies, and for celestial bodies, this is a gravitational force.  Fe would then be the sum of gravitational forces from bodies outside the system. We assume that the bodies are on the 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  m1 + m2, 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 mass  aCM and given by: 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 vCM  of the center of mass will be constant. Usually we choose a reference system where  vCM = 0. The center of mass will then be at rest.

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  x1= 0, we get xCM= 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 the  x-, y- and z-coordinate of the center of mass.It is also easy to extend the calculation to many bodies. With three bodies the y-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 Parameters are adjusted so that the velocity of the center of mass will be the one given in Parameters. 