iSoul In the beginning is reality

# 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:

${\mu}=\frac{1}{\frac{1}{m_{1}}+\frac{1}{m_{2}}}=\frac{m_{1}m_{2}}{m_{1}+n_{2}}=\frac{1}{n_{1}+n_{2}}$

where the force on this mass is given by the force between the two bodies. This is half of the harmonic mean of the two masses.

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:

${\lambda}=\frac{1}{\frac{1}{n_{1}}+\frac{1}{n_{2}}}=\frac{n_{1}n_{2}}{n_{1}+n_{2}}=\frac{1}{m_{1}+m_{2}}$

where the force on this vass is given by the surge 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:

$\frac{1}{\mu&space;}=\frac{1}{m_{1}}+\frac{1}{m_{2}}=n_{1}+n_{2}$

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:

$\frac{1}{\lambda}=\frac{1}{n_{1}}+\frac{1}{n_{2}}=m_{1}+m_{2}$

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

$\textbf{a}_{2}=-\frac{m_{1}}{m_{2}}\textbf{a}_{1}.$

The relative acceleration arel between the two bodies is given by

$\textbf{a}_{rel}=\textbf{a}_{1}-\textbf{a}_{2}=\left&space;(1+\frac{m_{1}}{m_{2}}&space;\right&space;)\textbf{a}_{1}=\frac{m_{1}+m_{2}}{m_{1}m_{2}}=\frac{\textbf{F}_{12}}{\mu&space;}.$

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 surge exerted by body 2 on body 1 is

Γ12 = n1 b1.

The surge exerted by body 1 on body 2 is

Γ21 = n2 b2.

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

Γ12 = –Γ21.

Therefore,

n1 b1 = –n2 b2

and

$\textbf{b}_{2}=-\frac{n_{1}}{n_{2}}\textbf{b}_{1}.$

The relative expedition brel between the two bodies is given by

$\textbf{b}_{rel}=\textbf{b}_{1}-\textbf{b}_{2}=\left&space;(1+\frac{n_{1}}{n_{2}}&space;\right&space;)\textbf{b}_{1}=\frac{n_{1}+n_{2}}{n_{1}n_{2}}=\frac{\mathbf{\Gamma}_{12}}{\lambda}.$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.