An extended body's weight acts on all of its mass elements. For many mechanical problems, those distributed gravitational forces can be replaced by one equivalent resultant force acting through a particular point called the center of gravity.

Let mass elements $m_i$ be at positions $\mathbf r_i$ in a uniform gravitational field $\mathbf g$. Their individual weights are

$$\mathbf W_i=m_i\mathbf g.$$

The total weight is

$$\mathbf W=\sum_i\mathbf W_i =\left(\sum_i m_i\right)\mathbf g =M\mathbf g.$$

The total gravitational torque about an origin is

$$\boldsymbol\tau_g =\sum_i \mathbf r_i\times m_i\mathbf g.$$

Because $\mathbf g$ is the same for every mass element,

$$\boldsymbol\tau_g =\left(\frac{1}{M}\sum_i m_i\mathbf r_i\right)\times M\mathbf g.$$

The term in parentheses is the center-of-mass position $\mathbf R_{\rm CM}$. Therefore, in a uniform gravitational field, the distributed weight is mechanically equivalent to one force

$$\boxed{\mathbf W=M\mathbf g}$$

acting through the center of mass.

Thus

$$\boxed{\text{center of gravity} = \text{center of mass}}$$

when the gravitational field is effectively uniform across the body.

Line of action

A force applied anywhere along the same straight line produces the same torque about any point. The vertical line through the center of gravity in a uniform near-Earth field is therefore the line of action of the weight.

For equilibrium and tipping problems, the important question is often not only where the center of gravity lies, but where this line of action intersects the supporting surface.

Example: nonuniform mass distribution

A light horizontal bar of negligible mass carries two point masses:

$$m_1=2.0,\mathrm{kg}$$

at $x_1=0$ and

$$m_2=6.0,\mathrm{kg}$$

at $x_2=1.0,\mathrm m$.

The center of mass, and therefore the center of gravity in a uniform field, is

$$x_G =\frac{m_1x_1+m_2x_2}{m_1+m_2} =\frac{(2.0)(0)+(6.0)(1.0)}{8.0} =0.75,\mathrm m.$$

The entire weight

$$W=(8.0)g$$

can be represented as acting downward through $x=0.75,\mathrm m$ when computing external force and torque balance.

When center of gravity and center of mass differ

If the gravitational field changes appreciably in magnitude or direction across a large body, different mass elements experience different $\mathbf g$. Then the simple factorization above no longer works, and the effective center of gravity need not coincide exactly with the center of mass.

For ordinary laboratory objects near Earth's surface, the field variation across the object is usually negligible, so the center of mass is an excellent center-of-gravity approximation.

The center of gravity is therefore a force-and-torque representation of distributed weight, while the center of mass is a property of the mass distribution itself.