Average minimum distance between $N$ points generate i.i.d. uniformly on the shell (sphere) of ball What is the expected minimum euclidean distance between $N$ points uniformly and independently chosen on the shell(sphere) of 3-D ball of radius $R$?   Note that the expected minimum distance might be difficult to compute, so a good lower bound is also fine. 
My approach  I take the approach similar to this   question. 
Suppose we put $N$ circles of radius $r$ uniformly on the surface of the ball.
Therefore we could compute the expected minimum distance by the following expression
\begin{align}
E[D]=\int_0^\infty P(D>t) dt.
\end{align}
So, it remains to compute  $P(D>t)$. Actually, since we are after a lower bound on  $E[D]$ it is enought to give a lower bound on $P(D>t)$.
Let $S_i$ be the event that the pair of circles does not intersect which should be given by
\begin{align}
P(S_i) =  \frac{ Surf(R)-Surf(r)}{Surf(R)}= 1-\frac{  \pi r^2}{ 4\pi R^2}=1-\frac{ r^2}{ 4 R^2}
\end{align}
Now the probability that euclidean minimum distance $D$ between $N$ balls is bigger than $t$ should be similar to 
\begin{align}
P[ D \ge t] \stackrel{?}{=}  P[ \cap S_i]
\end{align}
However, I think we somehow need to converte from the geodesic distance to euclidean distance but I am not very sure how to do that?

The asymptotics is a straightforward (or boring) generalization of my previous answer.
Sketch: Take a point over the unit sphere ($R=1$). Then the "excluded area" $a(D)$ (points that lie at a -euclidean- distance less than $D$ from the point) is $$a(D)=\pi D^2$$
The probability that a "link is free" (distance between points pair $i$ isgreater  than $D$ is 
$$p(S_i) =1-\frac{1}{4}D^2 \hspace{1cm} 0\le D \le 2 \tag{1}$$
Assuming independence as asympotical approximation:
$$p(\cap S_i) \approx (1-\frac{1}{4}D^2 )^M, \hspace{1cm} M=\frac{N(N-1)}{2} \tag{2}$$ 
Then, letting $t=$ minimum distance between points
$$E(t) = \int P(t >D) \, dD = \int p(\cap S_i)\, dD \approx \int_{0}^2 (1-\frac{1}{4}D^2 )^M \, dD \tag{3}$$
For large $M$ this tends to
$$E(t) \approx \frac{\sqrt{\pi}}{\sqrt{M}}\approx \frac{\sqrt{2 \pi}}{N} \tag{4}$$
For general ball radius $R$:
$$E(t) \approx R \frac{\sqrt{\pi}}{\sqrt{M}}\approx R \frac{\sqrt{2 \pi}}{N} $$
