Reduced Mass

#Physics

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

• Reduces two body problem into a one body problem
• Always smaller than or equal to $\displaystyle m_{1}$ or $\displaystyle m_{2}$ because $\displaystyle m_{1}=\frac{m_{1}}{1}\left( \frac{m_{1}+m_{2}}{m_{1}+m_{2}} \right)=\frac{m_{1}^{2}}{m_{1}+m_{2}}+\mu\geq \mu$

$\displaystyle \vec{F}{21}=\mu \vec{a}{\text{rel}}$

• This is the force object 2 exerts on object 1
• $\displaystyle \vec{a}{\text{rel}}$ is $\displaystyle \vec{a}{1}-\vec{a}_{2}$
• Here, we fix the position of object 2 and treat object 1 as the object moving
• See Wikipedia article for derivation