Center of Gravity equals Center of Mass if it exists
Theorem
Let $B$ be a body in a gravitational field $\mathbf G$.
Let $B$ have a center of gravity $P$.
Then the center of mass of $B$ is also $P$.
Proof
Let $Q$ denote the center of mass of $B$
First suppose that $\mathbf G$ is uniform.
From Center of Gravity in Uniform Gravitational Field is Center of Mass:
- $P = Q$
$\Box$
Now suppose that $\mathbf G$ is non-uniform.
There are two possibilities:
- $(1): \quad$ $B$ is barycentric
- $(2): \quad$ $B$ is not barycentric.
If $(1)$, then from Center of Gravity of Barycentric Body is Center of Mass:
- $P = Q$
If $(2)$, then from Center of Gravity in Non-Uniform Gravitational Field, the forces on $B$ induced by $\mathbf G$ consist of:
- a single force $\mathbf F$
- a couple $C$ whose plane is perpendicular to the line of action of $\mathbf F$.
The line of action of $\mathbf F$ does not necessarily pass through some fixed point as $B$ rotates in $\mathbf G$.
Hence $B$ has no center of gravity.
All cases are covered, and the result follows.
$\blacksquare$
Sources
- 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): centre of mass (barycentre; CM; mass centre)
- 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): centre of mass (CM; barycentre, mass centre)