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?

1 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.

  • $\begingroup$ take a example: a solid hemisphere with mass M , Homogeneous density ... to calculate the center of gravity at a pont P in the symmetry axes, above the top.. at a distance Z from the center of the hemisphere.. why center of gravity is only equal to the center of mass when this point P go far away from the hemisphere? 2. and to calculate that, we need use the Potential i think, instead field, that is more complicated. $\endgroup$
    – AlexandreH
    Commented Apr 16, 2012 at 20:52
  • $\begingroup$ 1. Because the gravitational field becomes homogenous. That is its direction as well as size becomes practically constant. $\endgroup$
    – Pygmalion
    Commented Apr 16, 2012 at 20:54
  • 1
    $\begingroup$ But can i use Potential to search the center of gravity? or only works using the g , gravitational fiel (vector)? $\endgroup$
    – AlexandreH
    Commented Apr 16, 2012 at 20:57

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