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I've come across many definitions. Is it

1) The point from which the weight of the body acts, i.e., the point at which if the entire mass of the body is assumed to be concentrated, the gravitational force acting on the body remains the same

2)The point at which if the entire mass of the body is assumed to be concentrated, the gravitational potential energy of the body remains the same

3)The point at which if the entire mass of the body is assumed to be concentrated, the gravitational torque acting on the body about the origin remains the same

4) All/some of the above?

I came across this while trying to answer this question.

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Meh! The definition is that is is the first moment of the mass distribution divided by the zero'th moment ($\sum \vec{x}_i m_i / \sum m_i$). It's affect is as all of the above. – dmckee Apr 17 '12 at 19:37
@dmckee: one can call them corollries of $\Sigma \vec{x_i} m_i / \Sigma m_i$ – Vineet Menon Apr 18 '12 at 6:38
Doesn't the center of gravity depend on the gravitational field? In a uniform gravitational field, I agree that it is (∑x⃗ imi/∑mi) but what about a non-uniform gravitational field? – Amu Apr 18 '12 at 6:43
Related: and links therein. – Qmechanic Dec 3 '13 at 13:42

I think that the second definition is wrong: It does not define a point, but rather a line.

For whatever potential energy the body has, there is a height such that putting all the mass in that height would give the same potential energy. The line that this height defines is the line I am talking about.

It seems to be true that the center of mass is in that line, though

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You are assuming that the gravitational field has the same magnitude at all points on a horizontal line. It need not be that way. Though I agree that the second definition alone won't suffice, it could be a number of different points. – Amu Apr 18 '12 at 6:47

All of those! The definitions are inter-connected. For example, 2nd definiton comes directly from 1st, as you get potential energy as the work of gravitational force:

$$U = - \int \vec{F} \cdot \text{d}\vec{s}$$

But the focal point here is the torque. As long as you don't consider rotation of the rigid body, but only translations, you do

$$\sum \vec{F} = m \vec{a}$$

and it is irrelevant where you put the force(s) that acts on the body. Note, $\vec{a}$ above is the acceleration of center of mass.

However, when you calculate rotations

$$\sum \vec{\tau} = I \vec{\alpha}$$

(the expression above only applies to constant axes), you must find the torque and you must know, where the force(s) acts. And it can be shown for gravitational field, that one and only one point exist for which you can say that if all the gravitational force act on it you will get the right torque and its name is center of gravity. For homogenous gravitational fields, this point equals to center of mass.

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This doesn't address the case of a nonuniform field, which is the only thing that makes the question nontrivial. – Ben Crowell Oct 26 '13 at 19:59

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