Velocity of earth due to the rotational motion of the earth moon system about the center of mass I calculated the location of the centre of mass of Earth-Moon system using $$\frac{\sum m_ix_i}{M} = \frac{mx}{M}$$ where m is the the mass of the Moon $= 7.35\times10^{22} kg$, x is the distance between centres $= 3.844\times10^8 m$ (taking the centre of the Earth as origin), $M$ is the sum of the Earth mass and the Moon mass $= 5.97\times10^{24} + 7.35\times10^{22} kg$.
Now I want to calculate the velocity of the Earth due to the rotational motion of the earth moon system about the center of mass that I evaluated from above to be $= 4675km$
I was wondering if I could use the orbital speed formula here? That is, 
$$\sqrt{\frac{GM}{R}}$$ where $M$ is the sum of masses of the Earth and the Moon and $R$ is the evaluated $4675km$.
When I do this, the answer is different to when, for example, I try using:
$$V=\omega R = \frac{2\pi R}{T}$$ where $T = T_{moon} = 30\times24\times3600$ 
(I understand that the moon rotation period is not exactly one month, but this is an approximation since I just want to understand the concept).
I was also thinking that I could possibly calculate it using energy conservation, but I can't get my head around it.
Any help/hint would be greatly appreciated!
 A: You asked for hints:
Your second approach is more appropriate here; you know it will provide a correct answer because you know the orbital period (~ 1 month) and the radius of the orbit (distance from Earth center to the Earth/Moon center of mass).
Your first approach needs to be modified.  The force of attraction between the Earth and Moon depends on the distance between them, not on the distance to the center of mass of the system.  The force is proportional to the product of the two masses divided by the square of their separation, so the gravitational acceleration of the Earth due to the Moon's gravity is proportional to the Moon's mass divided by the square of the separation (the gravitational acceleration of the Moon due to the Earth's mass is proportional to the Earth's mass divided by the square of the separation).  The centripetal acceleration of the Earth depends on the orbital velocity and the distance from the center of the Earth to the center of mass of the Earth/Moon system; and the centripetal acceleration must equal the gravitational acceleration. The rest is just algebra.
A: You can try to use the angular momentum conservation, or the fact that the torque on the system should be zero. 
Or I think another approach could be use the fact that center of mass and moon can be taken as inertial frames which, when the moon be taken as inertial frame the mass of the earth will be the reduced mass. 
We know that if two frames are inertial the velocity of the observed object can be written as, $V_{earth} = V_{CM}+V'_{earth}$ Where $V_{earth}$ is the velocity of earth respect to the moon and $V'_{earth}$ Velocity of earth respect to the center of mass.
