# Reduced mass

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 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 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 The force exerted by body 1 on body 2 is According to Newton's third law, for every action there is an equal and opposite reaction: Therefore, and The relative acceleration 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. Alternatively, a Lagrangian description of the two-body problem gives a Lagrangian of 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 Then substituting above gives a new Lagrangian 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 $V(\vert \mathbf{r}_1 - \mathbf{r}_2 \vert ) = - m_1 m_2 / \vert \mathbf{r}_1 - \mathbf{r}_2 \vert\!\,$ 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 "Reduced mass" may also refer more generally to an algebraic term of the form that simplifies an equation of the form 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. Full article ▸
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