# Centre of gravity for a very tall building

Imagine a very tall building. Centre of gravity is defined as the point of application of force of gravity, right? Now, is the centre of gravity the same as centre of mass of this building? I am including the change in g with height, so force of gravity varys with height

You are right to be wary. When the gravitational field is nonuniform, the centre of mass and the centre of gravity are in general different.

The centre of gravity is the point around which the nett torque from the gravitational forces is nought. In your problem, where gravity varies with position $\vec{r}$, we seek the position $\vec{r}_0$ where the nett torque of the gravitational forces vanishes, to wit:

$$\int \rho(\vec{r})\, (\vec{r}-\vec{r}_0)\wedge \vec{g}(\vec{r}) {\rm d}V = \vec{0}$$

To illustrate, let's assume the simplified case where $\vec{g}$ is in one direction $\vec{z}$, but varies slightly with position, so we can write $\vec{g} = g(\vec{r}) \vec{z}$. Then the above equation becomes:

$$\left(\int \rho(\vec{r}) g(\vec{r}) (\vec{r}-\vec{r}_0) {\rm d}V \right)\wedge \vec{z} = \vec{0}$$

Now, if $g$ is constant and $\vec{r}_0$ is chosen to be the centre of mass, i.e. the point where:

$$\int (\vec{r}-\vec{r}_0)\,\rho(\vec{r}) {\rm d}V=\vec{0}$$

then indeed the torque will vanish. But in general, we must choose the point $\vec{r}_0$ where:

$$\int (\vec{r}-\vec{r}_0)\,\rho(\vec{r}) \,g(\vec{r}) \, {\rm d}V=\vec{0}$$

which is different if $g$ varies with $\vec{r}$.

• Thank u for this wonderful answer. U understood my question perfectly, sir. – user34304 Dec 3 '13 at 14:26

Center of gravity is the same as center of mass. We usually put mass point (a whole mass of the body in one point) into a center of mass. This is why gravity is acting from the same point.

• This is only true for uniform gravitational fields. See for example: tides. – lionelbrits Dec 3 '13 at 15:00