How many caps do you need to cover the intersection of an $n-1$-sphere and a $k$-plane? Consider a unit sphere in $\mathbb{R}^n$ centered at the origin and a $k$-plane passing through the origin with $1 \leq k < n$. If I cover this intersection by spherical caps of radius $r$ (with $r \ll 1$), my question is how many spherical caps do you need?
As an example, when $n = 3$ and $k = 2$, then the intersection is a great circle and so you need roughly a constant multiple of $r^{-1}$ since the great circle has circumference a constant multiple of 1 and each spherical cap of radius $r$ takes up a constant multiple of $r$ of that circumference.
What is this in the general $n$ and $k$ case? Is it a constant multiple of $r^{-(n - k)}$? If so why?
 A: The intersection is the unit sphere in the $k$-plane, so the question is how many spherical caps of radius $r$ do you need to cover a unit $(k-1)$-sphere. 
It's a difficult problem for which only lower and upper bounds are known. The thing is, in dimensions higher than $1$, you're going to have a lot of overlap between your caps, and this wasted volume is hard to evaluate precisely. For instance, the very recent Covering spheres with spheres by Ilya Dumer (2018) contains such results:
Call $\partial$ the density of a covering. That is, the ratio of the total volume of the caps divided by the volume of the sphere (volume means Lebesgue measure). Then any $n$-sphere of any radius $r$ has a covering by unit caps of density $$\partial \leq \left(\frac 12+\frac{3\ln\ln n}{\ln n}+\frac{3}{\ln n}\right)n\ln(n)$$
The author considers coverings of large spheres by unit caps, you will need to do the conversion to get coverings of unit spheres by small caps. You will then need to compare the volume of one small cap with the volume of the sphere to turn this density into an actual number of caps. And in the end, replace $n$ by $k-1$.
