Center of Gravity vs Center of Mass

Question

1. To calculate the center of gravity of a distribution of mass, can I use gravitational potential (scalar) or must I use gravitational field (vector)?
2. And why, generally, the center of gravity goes to center of mass when the point where we calculate the potential goes to far away from mass distribution?

Answer 1

I am somehow puzzled about your question, but center of mass is vector and is calculated as a sum of vectors.

$$\vec{r}_\text{CM} = \frac{\sum_i m_i \vec{r}_i}{\sum_i m_i} = \frac{\sum_i m_i \vec{r}_i}{M},$$

where $m_i$ is mass and $\vec{r}_i$ is position of the individual (point) body.

If and only if you put the rigid body or system of masses into homogenous gravitational field, i.e. $\vec{F} = m \vec{g}$, then center of mass corresponds exactly to the center of gravity. This is important when calculating torque of gravity:

$$\vec{\tau}_{g} = \sum_i \vec{\tau}_{g,i}= \sum_i \vec{r}_i \times \vec{F}_{g,i} = \sum_i \vec{r}_i \times (m_i \vec{g}) = (\sum_i m_i \vec{r}_i) \times \vec{g} = M \vec{r}_\text{CM} \times \vec{g} = \vec{r}_\text{CM} \times (M \vec{g}) = \vec{r}_\text{CM} \times \vec{F}_g$$

Question 1: Center of gravity is always the vector, so whatever method you use, you must in the end get the vector. It is difficult to obtain center of gravity, as it is very difficult to solve the equation:

$$\vec{\tau}_{g} = \sum_i \vec{\tau}_{g,i} \equiv \vec{r}_\text{CG} \times \vec{F}_g$$

and obtain $\vec{r}_\text{CG}$ for non-homogenous gravitational fields. Sometimes it is even impossible.

Considering text in Wikipedia, I think the author said, that gravitational field can be expanded into several contributions, each of which corresponds to one specific and simpler force. Then you can find "CG" for each of these forces, i.e. instead of one point in which force acts on the rigid body, you have many such points with many forces.

Question 2: When you go far away from some mass distribution, its gravitational field becomes practically homogenous (same direction and same size), so CM and CG become one point.