# Calculate Reduced Mass

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### Calculate Reduced Mass

If we use the center-of-mass frame of reference we can show easily that for a two-body system the total kinetic energy (K.E.) in any other frame is equal to the K.E. calculated from the reduced mass and the reltative velocity, $v$, between the two masses, $m_{1}$ and $m_{2}$. To calculate reduced mass, $\mu$, we take one mass and weight it by the other:

\[

\mu \ \equiv \ \frac{m_{1}m_{2}}{m_{1}+m_{2}}

\]

and if the speeds of $m_{1}$ and $m_{2}$ are $v_{1}$ and $v_{2}$ respectively, in some arbitrary frame of reference, the total K.E. is

\[

{\rm K.E.} \ = \ \frac{1}{2} m_{1}v_{1}^{2} +

\frac{1}{2} m_{2}v_{2}^{2}

\]

but since $m_{1}v_{1} + m_{2}v_{2}=0$ and $v_{2}-v_{1} \equiv v$,

\[

v_{1}^{2} = \frac{m_{2}^{2}}{(m_{1}+m_{2})^{2}} \\

\\

v_{2}^{2} = \frac{m_{1}^{2}}{(m_{1}+m_{2})^{2}}

\]

so

\[

{\rm K.E.} \ = \ \frac{1}{2} m_{1}m_{2} v^{2} \left(\frac{m_{2}}{(m_{1}+m_{2})^{2}} \ + \

\frac{m_{1}}{(m_{1}+m_{2})^{2}}\right) \\

\ = \ \frac{1}{2} m_{1}m_{2}\frac{(m_{1}+m_{2})}{(m_{1}+m_{2})^{2}} .

\]

Hence

\[

{\rm K.E.} \ = \ \frac{1}{2}\frac{m_{1}m_{2}}{(m_{1}+m_{2})} \ v^{2}

\ = \ \frac{1}{2}\mu v^{2}

\]

Thus, the two-body problem with gravity is equivalent to the one-body problem in which one of the masses orbits the other, and the reduced mass is used in place of either of the individual masses.