Reduced mass and energy representation? If we have a gravitational system where $m<<M$ then we can write down the energy of $m$ as follows:
$$E=\frac{1}{2}m(\frac{dr}{dt})^2+\frac{L^2}{2mr^2}-\frac{GMm}{r}$$
but if $m$ is not small compared to $M$ then we have to use reduced mass to get:
$$E=\frac{1}{2}\mu(\frac{dr}{dt})^2+\frac{L^2}{2\mu r^2}-\frac{GMm}{r}$$
But whose mass is this? Do both masses individually have this same energy or is it the total energy of the whole system? or is it just the energy of the mass, $m$, if so what is the energy of the mass $M$, since we would get the a very similar expression? The same goes for $L$ the angular momentum, which mass does this belong to??
 A: The reduced mass is a mathematical trick.
Consider the rotation of a diatomic molecule with two atoms of mass $m_1$ and $m_2$, which is similar to your example. (bear with me on this)
The molecule rotates about its centre of mass and to work out the kinetic energy of the rotation we should use 
$K.E. = {1 \over 2} I \omega^2$
where $I$ is the moment of inertia given by 
$I = \sum m_i r_i^2$
where $r_i$ is equal to the distance from the centre of mass to mass $m_i$
It turns out that we can also calculate $I$ with
$I = \mu r^2$
where $r$ is the total bond length of the molecule and $\mu$ is the reduced mass $\mu = {m_1 m_2 \over m_1+m_2}$. Note that $r = r_1 + r_2$
Now generally it is much easier to use the actual bond length of the molecule than the individual distances from the centre of mass to each mass. 
Similarly in your problem above it is much easier to use the separation between the two masses, $r$, rather than worry about the individual distances to the combined centre of mass of the system.
In the case that one mass is much smaller than the other we still need to use the reduced mass in principle, but because 
$\mu = {m_1 m_2 \over m_1+m_2}$ 
if $m1 \lt\lt\lt m2$ then
$\mu \sim {m_1 m_2 \over m_2} = m_1$ 
so we can just use the lighter mass instead.
$L$ the angular momentum is the angular momentum of the whole system around the centre of mass - so it is the angular momentum of both bodies. In the case that one body is very light, nearly all the angular moment will be with the light body, so we could assume that the light body has all the angular moment. The is a bit like regarding the sun as fixed in the solar system and not moving as it is so heavy compared to the planets, but in reality it will move around the centre of mass of the 'solar system'.
