# Does total angular momentum of the Earth-Moon system include individual rotational angular momenta?

To calculate the angular momentum of a body we need to specify a point (or an axis?) from which to define the displacement vector $\vec{r}$, so that $\vec{L} = \vec{r} \times \vec{p}$.

For a rigid body, the formula becomes $\vec{L}=I\vec{\omega}$, assuming the moment of inertia is not a tensor. So in this case we need to specify an axis.

NOW: what if I wanted to calculate the total angular momentum of the Earth-Moon system? What is the most sensible point/axis to choose?

I would say, intuitively, the centre of mass of the system.

\begin{align}\vec{L}_\text{tot} &= \vec{L}\text{ of the Earth due to its orbit about centre of mass} \\ &+ \vec{L}\text{ of Moon due to its orbit about centre of mass} \\ &+ \text{angular momenta due to rotations of Earth and Moon around their own axes}?\end{align}

I am not sure if and how to include the rotations of Moon and Earth around their axes: I know they have to enter somehow because of the tidal friction effect on the orbit of the Moon, but I don't know how to reconcile this with the fact that I chose the centre of mass as the point of reference here, and none of the rotation axes have anything to do with the centre of mass of the Earth-Moon system.

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I changed your vectors to use arrows so that $\vec{\omega}$ would show up properly - I don't think it's possible to make a bold omega here. Hopefully that's all right. –  David Z Feb 16 '14 at 22:55
@DavidZ Well actually: $\boldsymbol\omega$. –  joshphysics Feb 17 '14 at 2:42
@joshphysics ah, I forgot about that. –  David Z Feb 17 '14 at 3:05
The parallel axis theorem lets you calculate the rotational angular momentum around each body's center and then translate it to the center of mass of the system. You still have to take into account that the axes aren't perpendicular the orbits; you'll probably need to assume the moment of inertia is a tensor for this. –  Javier Mar 29 '14 at 20:52

Consider two bodies A and B. With respect to an inertial coordinate system with origin at point O, the coords of the particles in A are vectors $x_{a}\in V_{3}$ with $a=1,2,\ldots, N_{A}$ and similarly the coords of the particles of B are $x_{b}\in V_{3}$ with $b=1,2,\ldots,N_{B}$. The momenta wrt the inertial frame with origin at point O of the particles of A are $p_{a}\in V_{3}$ and the momenta of the particles of B are $p_{b}\in V_{3}$. The total angular momentum of the system wrt point O is, $$J=\sum_{a}x_{a}\times p_{a}+\sum_{b}x_{b}\times p_{b} \ .$$ Let's introduce the centre of mass of body A as $X_{A}\in V_{3}$, $$X_{A}=\frac{\sum_{a}m_{a}x_{a}}{\sum_{a}m_{a}}=\frac{\sum_{a}m_{a}x_{a}}{M_{A}}$$ and the centre of mass of body B as, $$X_{B}=\frac{\sum_{b}m_{b}x_{b}}{\sum_{b}m_{b}}=\frac{\sum_{b}m_{b}x_{b}}{M_{B}}$$ Adding and subtracting the centre of mass coords, $$J=\sum_{a}(x_{a}-X_{A})\times p_{a}+\sum_{b}(x_{b}-X_{B})\times p_{b}+X_{A}\times \sum_{a}p_{a}+X_{B}\times \sum_{b}p_{b} \ .$$ The first two terms on the RHS are the angular momenta of the bodies about their respective centres of mass $X_{A}$ and $X_{B}$. Let's write these contributions as $J_{A}\in V_{3}$ and $J_{B}\in V_{3}$. The total angular momentum is now, $$J=J_{A}+J_{B}+X_{A}\times \sum_{a}p_{a}+X_{B}\times \sum_{b}p_{b} \ .$$ Let the linear momentum of the particles of body A be, $$P_{A}=\sum_{a}p_{a}$$ with a similar formula for the sum of the momenta of the particles of body B. Put these formulae into the equation for the total angular momentum, $$J=J_{A}+J_{B}+X_{A}\times P_{A}+X_{B}\times P_{B} \ .$$ This is precisely the splitting of the total angular momentum that Harold wrote in his question. $J_{A}$ and $J_{B}$ are the angular momenta of A and B about their own centres of mass. $X_{A}\times P_{A}$ is the orbital angular momentum of A about the origin O of the inertial coords and $X_{B}\times P_{B}$ is the orbital angular momentum of B about O. The point O could be taken as the centre of mass of the complete system, but it doesn't matter for the above result.