Why COM and COG are different things?

Question

In the formula for Newton's law of gravitation i.e. $$F = G(m_1)(m_2)/r^2 $$ here $r$ is the distance between COM of the two objects , so we consider that the object is a single point located at it's COM and at this point gravitational force of attraction acts . Then why is sometimes COM and COG different for an object ? Please give some practical example .

Answer 1

While your question is essentially a duplicate of the one I linked to, none of the answers have used the equations, so I will supply them here in a comparison.

The COM is weighed based on the mass density $\rho(\mathbf r)$: $$\mathbf r_\text{COM}=\frac{\int\text dV\,\rho(\mathbf r)\cdot \mathbf r}{\int\text dV\,\rho(\mathbf r)}$$

Whereas the COG is weighed based on, well, the weight $$\mathbf r_\text{COG}=\frac{\int\text dV\,\rho(\mathbf r)\cdot g(\mathbf r)\cdot\mathbf r}{\int\text dV\,\rho(\mathbf r)\cdot g(\mathbf r)}$$

When $g(\mathbf r)=g$ is uniform over the object, then the two definitions become equal.

Answer 2

The COM for a rigid body is a fixed position that does not change. The center of gravity can depend on orientation, and can be in a different place on the rigid body, when in a non uniform gravitational field. In such a situation, the COG will always be closer to the main attractive body than the COM. For example, a satellite in orbit around Earth would have it's COG slightly lower than it's COM, because gravitational attraction decreases with distance, so the lower part of the satellite has slightly higher gravitational attraction to Earth than the upper part. see; https://en.wikipedia.org/wiki/Center_of_mass#Center_of_gravity for more information.