# 質心

$r_m = {\sum {m_i} {r_i} \over \sum m_i}$

• $r$表示某一坐標軸
• $m_i$表示物質系統中，某i質點質量
• $r_i$表示物質系統中，某i質點的座標

## 動畫展演

 兩顆星體質量差不多，例如休神星。 兩顆星體質量不同，例如冥王星與冥衛一。 兩顆星體質量有很大的不同，例如地球與月球。 兩顆星體質量有極大的不同，例如太陽與地球。 兩顆星體以橢圓軌道互繞，此狀況通常稱為聯星。

## 重心

### 均勻重力場

$\mathbf{f}(\mathbf{r}) = -dm\, g\vec{k}= -\rho(\mathbf{r})dV\,g\vec{k},$

$\mathbf{F} = \int_V \mathbf{f}(\mathbf{r}) = \int_V\rho(\mathbf{r})dV( -g\vec{k}) = -Mg\vec{k},$

$\mathbf{T} = \int_V (\mathbf{r}-\mathbf{R})\times \mathbf{f}(\mathbf{r}) = \int_V (\mathbf{r}-\mathbf{R})\times (-g\rho(\mathbf{r})dV\vec{k} )= \left(\int_V \rho(\mathbf{r}) (\mathbf{r}-\mathbf{R})dV \right)\times (-g\vec{k}) .$

$\int_V \rho(\mathbf{r}) (\mathbf{r}-\mathbf{R})dV =0,$

### 非均勻重力場

Some of the inhomogeneity in a gravitational field may be modeled by a variable but parallel field: g(r) = g(r)n, where n is some constant unit vector. Although a non-uniform gravitational field cannot be exactly parallel, this approximation can be valid if the body is sufficiently small.[1] The center of gravity may then be defined as a certain weighted average of the locations of the particles composing the body. Whereas the center of mass averages over the mass of each particle, the center of gravity averages over the weight of each particle:

$\mathbf{r}_\mathrm{cg} = \frac{1}{W} \sum_i w_i \mathbf{r}_i,$

where $\mathbf{w}_\mathrm{i}$ is the (scalar) weight of the ith particle and W is the (scalar) total weight of all the particles.[2] This equation always has a unique solution, and in the parallel-field approximation, it is compatible with the torque requirement.[3]

A common illustration concerns the Moon in the field of the Earth. Using the weighted-average definition, the Moon has a center of gravity that is lower (closer to the Earth) than its center of mass, because its lower portion is more strongly influenced by the Earth's gravity.[4]

If the external gravitational field is spherically symmetric, then it is equivalent to the field of a point mass M at the center of symmetry r. In this case, the center of gravity can be defined as the point at which the total force on the body is given by Newton's Law:

$\frac {GmM (\mathbf{r}_\mathrm{cg} - \mathbf{r})} {|\mathbf{r}_\mathrm{cg} - \mathbf{r}|^3} = \mathbf{F},$

where G is the gravitational constant and m is the mass of the body. As long as the total force is nonzero, this equation has a unique solution, and it satisfies the torque requirement.[5] A convenient feature of this definition is that if the body is itself spherically symmetric, then rcg lies at its center of mass. In general, as the distance between r and the body increases, the center of gravity approaches the center of mass.[6]

Another way to view this definition is to consider the gravitational field of the body; then rcg is the apparent source of gravitational attraction for an observer located at r. For this reason, rcg is sometimes referred to as the center of gravity of M relative to the point r.[7]

## 参考资料

1. ^ Beatty 2006, pp. 45.
2. ^ Beatty 2006，第48页; Jong & Rogers 1995，第213页.
3. ^ Beatty 2006, pp. 47–48.
4. ^ Asimov 1988，第77页; Frautschi等 1986，第269页.
5. ^ Symon 1964，第259–260页; Goodman & Warner 2001，第117页; Hamill 2009，第494–496页.
6. ^ Symon 1964, pp. 260, 263–264.
7. ^ Symon 1964, p. 260.