# Trying to understand reduced mass

1. Mar 13, 2013

### DiracPool

I understand that the reduced mass is used to simplify a two-body problem into a one body problem, so that instead of calculating two masses rotating around a common center as in the earth and moon, you can treat the earth as stationary, and then used a "reduced" mass for the moon in the calculation.

My question is how is m1m2/m1+m2 calculated to derive the reduced mass. I'm not able to follow the arguments in my readings. Can someone give me a simple, straightforward derivation along with a conceptual "intuition" as to why this works?

2. Mar 14, 2013

### Staff: Mentor

The part in bold is incorrect. What is get is the relative motion of the two bodies, which follows the same equation as is you had a single body of mass "reduced mass" rotating around a point which is the center of mass of the system.

The trick is to think of center of mass motion (the two bodies moving as one system) and relative motion.

3. Mar 14, 2013

### Staff: Mentor

Can you point us to a derivation that you don't understand, and tell us where you get stuck? Then someone may be able to "unstick" you.

If someone simply posts a derivation, they run the risk of repeating something that you've seen already and don't understand.

4. Mar 15, 2013

### DiracPool

@DrClaude

Here is the reference I used to make the statement you said was incorrect, Dr...

http://www.ehow.com/how_5968910_calculate-reduced-mass.html

@jtbell

I'm trying to follow the derivation on the wiki page under the "Newtonian mechanics" heading. I get all the math but...why does a-rel = a1-a2 instead of say, a2-a1? Also, why is the result F12/Mred the result we are looking for? I get the derivation, I just don't understand how it works to turn the two body problem into a single body problem. So when they say at the end...

http://en.wikipedia.org/wiki/Reduced_mass

I'm not catching their drift. Any help would be appreciated.

5. Mar 15, 2013

### Staff: Mentor

The key word in that quote is "pretend." This is not rigourous. But indeed, if one of the objects is much more massive than the other, then the center of mass of the system will be located close to the middle of the massive object, and therefore you can imagine that it is as if you had an object of mass $m_\mathrm{red}$ orbiting in a closer orbit around the massive object.

I've just looked at the Wikipedia page, and I understand completely why you are confused. Maybe this derivation, taken from V. D. Barger and M. G. Olsson, Classical Mechanics: A Modern Perspective, 2nd ed., McGraw-Hill (New York), 1995, will be more helpful.

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6. Mar 15, 2013

### DiracPool

Thank you for that derivation, DrClaude

7. Mar 15, 2013

### Staff: Mentor

I expect that it doesn't make any difference in the end. That is, you could just as well start with arel = a2 - a1, and along the way you might have 1 and 2 switched around in some places, or maybe some + and - signs switched, but the differences will cancel out eventually. Try it and see what happens!

8. Mar 15, 2013

### Staff: Mentor

That just corresponds to a 180º rotation. So if in the inertial frame one object is at (-1,0,0) and the other object is at (3,0,0) then you can either adopt the convention that their relative position is (4,0,0) or (-4,0,0). Either way works fine, but your accelerations will all be 180º "out of phase".