May 28th, 2015, 10:38 AM  #1 
Senior Member Joined: May 2012 Posts: 203 Thanks: 5  reduced mass
I am not sure what the reduced mass between two bodies represents... How does one use it to figure out the relative motion between two bodies. Is the reduced mass, hypothetical mass located at the centre of mass? but then the centre of mass will change depending on the distance between the two bodies....So in the end we get at hypothitcal 'reduced mass' oscillating between the two bodies as the distance between them varies? Last edited by Kinroh; May 28th, 2015 at 10:48 AM. 
May 28th, 2015, 11:42 AM  #2  
Math Team Joined: May 2013 From: The Astral plane Posts: 1,807 Thanks: 722 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
As an entity the reduced mass can get a little crazy. For the Sun and Earth the reduced mass is essentially just the mass of the Earth, and for an electronpositron pair the reduced mass is actually only half the mass of the electron. Dan  
May 31st, 2015, 05:49 PM  #3 
Senior Member Joined: May 2012 Posts: 203 Thanks: 5 
I am not sure I follow : 'The variable that designates the position of the remaining mass does not affect the position or size of the reduced mass.' If I could bother you to elaborate? 
May 31st, 2015, 06:12 PM  #4  
Math Team Joined: May 2013 From: The Astral plane Posts: 1,807 Thanks: 722 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
For the two body problem we have two masses m1 and m2 with a force F12 between them. This gives the system of equations: $\displaystyle m_1 \frac{d^2x_1}{dx_1^2} = F_{12}(x_1, x_2)$ $\displaystyle m_2 \frac{d^2x_2}{dx_2^2} = F_{21}(x_1, x_2)$ where $\displaystyle F_{21}(x_1, x_2) = F_{12}(x_1, x_2)$ by Newton's 3rd Law. We can reduce this to a single body equation by the substitutions: $\displaystyle R = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$ and $\displaystyle \mu = \frac{m_1 m_2}{m_1 + m_2}$ This gives the "equivalent" single body problem $\displaystyle \mu \frac{dR^2}{dt^2} = F(R)$ which only depends on R as a variable. Now to answer your question. Yes, the reduced mass is changing in the original problem but you need to consider which part of the problem you are trying to solve. When we convert the two body problem into the one body problem we are changing the variables. You can look at it via the two body equations of motion or you can look at it via the one body equations. In the one body equation the reduced mass is considered to be a constant for the solution. You are going to drive yourself crazy trying to keep track of both at once. Dan  

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