Answer:
Reduced mass is the "effective" inertial mass appearing in the two-body problem of Newtonian mechanics. This is a quantity with the unit of mass, which allows the two-body problem to be solved as if it were a one-body problem. Note however that the mass determining the gravitational force is not reduced. In the computation one mass can be replaced by the reduced mass, if this is compensated by replacing the other mass by the sum of both masses.
Given two bodies, one with mass m_{1}\!\, and the other with mass m_{2}\!\,, they will orbit the barycenter of the two bodies. 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
m_\text{red} = \mu = \cfrac{1}{\cfrac{1}{m_1}+\cfrac{1}{m_2}} = \cfrac{m_1 m_2}{m_1 + m_2},\!\,
where the force on this mass is given by the gravitational force between the two bodies. The reduced mass is always less than or equal to the mass of each body and is half of the harmonic mean of the two masses.
This can be proven easily. Use Newton's second law, the force exerted by body 2 on body 1 is
F_{12} = m_1 a_1. \!\,
The force exerted by body 1 on body 2 is
F_{21} = m_2 a_2. \!\,
According to Newton's third law, for every action there is an equal and opposite reaction:
F_{12} = - F_{21}.\!\,
Therefore,
m_1 a_1 = - m_2 a_2. \!\,
and
a_2=-{m_1 \over m_2} a_1. \!\,
The relative acceleration between the two bodies is given by
a= a_1-a_2 = \left({1+{m_1 \over m_2}}\right) a_1 = {{m_2+m_1}\over{m_1 m_2}} m_1 a_1 = {F_{12} \over m_\text{red}}.
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.
Alternatively, a Lagrangian description of the two-body problem gives a Lagrangian of
L = {1 \over 2} m_1 \mathbf{\dot{r}}_1^2 + {1 \over 2} m_2 \mathbf{\dot{r}}_2^2 - V(\vert \mathbf{r}_1 - \mathbf{r}_2 \vert ) \!\,
where m_i, \mathbf{r}_i are the mass and position vector of the i th particle, respectively. The potential energy V takes this functional dependence as it is only dependent on the absolute distance between the particles. If we define \mathbf{r} \equiv \mathbf{r}_1 - \mathbf{r}_2 and let the centre of mass coincide with our origin in this reference frame, i.e. m_1 \mathbf{r}_1 + m_2 \mathbf{r}_2 = 0 , then
\mathbf{r}_1 = \frac{m_2 \mathbf{r}}{m_1 + m_2} , \mathbf{r}_2 = \frac{-m_1 \mathbf{r}}{m_1 + m_2}.
Then substituting above gives a new Lagrangian
L = {1 \over 2}m_\text{red} \mathbf{\dot{r}}^2 - V(r),
where m_\text{red} = \frac{m_1 m_2}{m_1 + m_2} , the reduced mass. Thus we have reduced the two-body problem to that of one body.
The reduced mass is frequently denoted by the Greek letter \mu\!\,; note however that the standard gravitational parameter is also denoted by \mu\!\,.
In the case of the gravitational potential energy V(\vert \mathbf{r}_1 - \mathbf{r}_2 \vert ) = - G m_1 m_2 / \vert \mathbf{r}_1 - \mathbf{r}_2 \vert\!\, we find that the position of the first body with respect to the second is governed by the same differential equation as the position of a body with the reduced mass orbiting a body with a mass equal to the sum of the two masses, because
m_1 m_2 = (m_1+m_2) m_\text{red}\!\,
"Reduced mass" may also refer more generally to an algebraic term of the form
x_\text{red} = {1 \over {1 \over x_1} + {1 \over x_2}} = {x_1 x_2 \over x_1 + x_2}\!\,
that simplifies an equation of the form
\ {1\over x_\text{eq}} = \sum_{i=1}^n {1\over x_i} = {1\over x_1} + {1\over x_2} + \cdots+ {1\over x_n}.\!\,
The reduced mass is typically used as a relationship between two system elements in parallel, such as resistors; whether these be in the electrical, thermal, hydraulic, or mechanical domains. This relationship is determined by the physical properties of the elements as well as the continuity equation linking them.