For Which Countries Does PayPal Require a State?

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 $\vec{r}$.