I have a question when I'm learning the electric quadrupole radiation in E&M. In the radiation zone, the vector potential $$\vec{A}(\vec{r},t)$$ generated by the current distribution $$\vec{J}(\vec{r}',t')$$ is given by \begin{align} \vec{A}(\vec{r},t)\approx\frac{\mu_0}{4\pi r}\int d^3r' \vec{J}\left(\vec{r}',t'=t-\frac{r}{c}+\frac{\hat{r}\cdot\vec{r}'}{c}\right). \end{align} One may then obtain the E and B fields using \begin{align} \vec{E}\approx\hat{r}\times\left(\hat{r}\times\frac{\partial\vec{A}}{\partial t}\right),\qquad \vec{B}\approx-\frac{\hat{r}}{c}\times\frac{\partial\vec{A}}{\partial t}. \end{align} These formulas are standard in textbooks of electrodynamics. It turns out the radial component of $$\vec{A}$$ along $$\hat{r}$$ contributes no E and B fields in the radiation zone.

Consider a uniformly charged sphere with a changing radius. The sphere will not generate any E and B fields in the radiation zone because all currents are along the radial direction, and thus $$\vec{A}$$ is always along $$\hat{r}$$ by symmetry. So the changing radius of the charged sphere will not radiate any E and B fields to the far field. This situation shows that the trace of the electric quadrupole will not contribute any radiation, which is why we define the electric quadrupole in its traceless form.

But then I seem to remember that the core collapse of a supernova does radiate gravitational waves for us to detect. I have very limited knowledge of the gravitational wave. I only know that Newton's law of gravitation and Coulomb's law are both inverse-square laws. Can someone explain to me 1) am I right to conclude that the uniformly charged sphere does not radiate as its radius changes, and 2) does a supernova radiate graviational waves as its core collapses and why or why not?