For a separable metric space $(X,d)$, we say it satisfies the global $N$ covering property if any ball $B_r(a)$ can be covered by $N$ balls of radius $r/2$. Let's see an example, let $(\mathbb{R}^2,d)$ be 2-dimensional Euclidean space with standard metric. Then it is not hard to see that we can cover $B_r(a)$ by $4^2$ balls with radius of $B_{r/2}$ by slicing a tangent square (edge length is 2r) around the ball into $4^2$ smaller squares and covering them by half balls (of course this covering number is not optimal, but at least it is global), so naturally we can extend this covering to higher dimensions $\mathbb{R}^n$, concluding covering number $4^n$.
Now I want to find the covering number on hypersphere $(\mathbb{S}^n,d)$ where $d$ is the distance induced by the standard metric. Obviously above slicing dosen't work in this case since it is not flat. But I saw in one paper that the covering number can be chosen to be $9^n$, I tried to do it on $\mathbb{S}^2$, calculating the volume of geodesic disc $B_{r}$: $2\pi(1-\cos r)$, but I think volume ratio has nothing to do with the covering number. Does anyone know how $9^n$ is obtained? Or is there any idea how to approach this?