Reduced mass and vass

Here we take the reduced mass and show the parallel reduced vass.

In physics, the reduced mass is the “effective” inertial mass appearing in the two-body problem of Newtonian mechanics. It is a quantity which allows the two-body problem to be solved as if it were a one-body problem.

Given two object bodies, one with mass m1 and the other with mass m2, the equivalent one-body problem, with the position of one body with respect to the other as the unknown, is that of a single body of mass:

where the force on this mass is given by the force between the two bodies. This is the two masses combined with reciprocal addition.

Given two subject bodies, one with vass n1 and the other with vass n2, the equivalent one-body problem, with the position of one body with respect to the other as the unknown, is that of a single body of vass:

where the release on this vass is given by the release between the two bodies.

The reduced mass is always less than or equal to the mass of each body:

μ ≤ m1 and μ ≤ m2

and has the reciprocal additive property:

The reduced vass is always less than or equal to the vass of each body:

λ ≤ n1 and λ ≤ n2

and has the reciprocal additive property:


The reduced mass can be derived as follows. Using Newton’s second law, the force exerted by body 2 on body 1 is

F12 = m1 a1.

The force exerted by body 1 on body 2 is

F21 = m2 a2.

According to Newton’s third law, the force that body 2 exerts on body 1 is equal and opposite to the force that body 1 exerts on body 2:

F12 = –F21.

Therefore,

m1 a1 = –m2 a2

and

The relative acceleration arel between the two bodies is given by

So we conclude that body 1 moves with respect to the position of body 2 as a body of mass equal to the reduced mass.


The reduced vass can be derived as follows. Using Newton’s second law for time-space, the release exerted by body 2 on body 1 is

Y12 = n1 b1.

The release exerted by body 1 on body 2 is

Y21 = n2 b2.

According to Newton’s third law for time-space, the release that body 2 exerts on body 1 is equal and opposite to the release that body 1 exerts on body 2:

Y12 = –Y21.

Therefore,

n1 b1 = –n2 b2

and

The relative relentation brel between the two bodies is given by

So we conclude that body 1 moves with respect to the position of body 2 as a body of mass equal to the reduced vass.