probability
How can I find an average distance between two points lying on surface of a sphere of a certain radius?
More importantly : can knowing the average distance between two points on surface of a disk ( this question has already an answer on MSE) be useful to answer the question about average distance between two points on surface of sphere? Or there are no immediate obvious relationship/generalization between the two question?
I don't know how you arrived at your formula for "$P_2(2)$", whatever is meant by this.
At any rate you may assume your circle of radius $1$, the first point $z_1$ as $(1,0)$ and the second point $z_2$ as $(\cos\phi,\sin\phi)$ with $\phi$ equidistributed on $[0,\pi]$. Then $|z_2-z_1|=2\sin{\phi\over2}$ and therefore $${\mathbb E}\bigl[|z_1-z_2|\bigr]={1\over\pi}\int_0^\pi 2\sin{\phi\over2}\ d\phi={4\over\pi}\doteq1.273\ .$$
Let $(X_i, Y_i)$ for $i=1,2$ be i.i.d. and has uniform distribution on $[0, 1]^2$. Then $|X_1 - X_2|$ has PDF
$$ f(x) = \begin{cases} 2(1-x), & 0 \leq x \leq 1 \\ 0, & \text{otherwise} \end{cases}. $$
This shows that the average distance is
\begin{align*} \int_{0}^{1}\int_{0}^{1} 4(1-x)(1-y)(x^2 + y^2)^{1/2} \, dxdy &= \frac{1}{15}(2+\sqrt{2}+5\log(1+\sqrt{2})) \\ &\approx 0.52140543316472067833 \cdots. \end{align*}
If $l_n$ denotes the average distance between two uniformly chosen points in $[0, 1]^n$, then the following formula may help us estimate the decay of $l_n$ as $n \to \infty$:
$$ l_n = \frac{1}{\sqrt{\pi}} \int_{0}^{\infty} \left \{1 - \left( \frac{\sqrt{\pi} \operatorname{erf}(u)}{u} - \frac{1 - e^{-u^2}}{u^2} \right)^{n} \right\} \frac{du}{u^2}. $$
Using an estimate (which I believe to be true but was unable to prove)
$$ e^{-u^2 / 6} \leq \frac{\sqrt{\pi} \operatorname{erf}(u)}{u} - \frac{1 - e^{-u^2}}{u^2} \leq 1 $$
and hence we get
$$ l_n \leq \frac{1}{\sqrt{\pi}} \int_{0}^{\infty} \frac{1 - e^{-nu^2/6}}{u^2} \, du = \sqrt{\frac{n}{6}}. $$
For example, for $2 \leq n \leq 20$ we have
Best Answer
Without loss of generality, assume the first point is at the "north pole"; also without loss of generality, assume the second point is along the "prime meridian." Then the probability of being at "latitude" $x$ degrees north is equal to the probability of being at "latitude" $x$ degrees south (and is proportional to $\cos x$). Therefore, the average latitude is at the "equator," and the average distance is $\pi r/2$, as stated by Henry and achille hui.