# Concept of reduced mass

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1. Mar 29, 2015

According to the concept: when a planet revolves around a star, and when both the bodies move in circular orbits due to the interaction between each other, both the bodies can be replaced by a single body of mass $\mu$ revolving in a circular orbit of radius equal to the distance between both the bodies.
so the total energy of the system becomes $\frac{1}{2}\mu v^2$
But it is positive. The total energy has to be be negetive
$-\frac{GMm}{r}+E_1+E_2$ where e1, e2 are kinetic energies of the bodies comes out to be negetive after solving.
Why do I get positive sign using concept of reduced mass?

2. Mar 29, 2015

### Orodruin

Staff Emeritus
Please provide references for this statement:
Did you try actually writing down the energy of the system in the CoM system?

3. Mar 29, 2015

Only the statement is given.
The energy of the body in circular motion should be ½μν2 right? Since both the bodies can be replaced by a single body of reduced mass $\mu$
I got the total energy using $-\frac{GMm}{r}+\frac{1}{2}I_1\omega^2+\frac{1}{2}I_2\omega^2=-\frac{GMm}{2r}$ which is also given in my textbook.
On using ½μν2, i get $\frac{GMm}{2r}$.

4. Mar 29, 2015

I understand now. I tried using the relation $$mv^2/r=GMm/R^2$$ which is wrong since there is no force of gravitation when you just have a single body.

5. Mar 29, 2015

### Orodruin

Staff Emeritus
No no no, you replace the two bodies with a single body of mass $\mu$ moving in a central gravitational potential. You still need to provide the reference, what book are you using?

6. Mar 29, 2015

$$m_1r_1=m_2r_2$$
$$m_1r_1\omega^2=m_2r_2\omega^2=\frac{Gm_1m_2}{r^2}$$
$$L=(I_1+I_2)\omega^2=\mu r^2\omega^2$$
Kinetic Energy of the system, $K=\frac{1}{2}\mu r^2\omega^2$