Visibility in a growing orchard This is a variant on Polya's orchard problem.1,2
Suppose trees are planted randomly in the plane.
The question is: How many trees are visible from the origin as
their radii grow? 
More precisely, confine attention to the first quadrant,
and distribute points uniform-randomly in that quadrant so that
each $1 \times 1$ lattice square has a mean density of $d$ points.
So, for example, if $d=4$ then a $5 \times 5$ portion of the quadrant
should receive $100$ points.
(Another model is to insist on $d$ random points per unit lattice square.)
Initially the radii of each tree is $0$ and all are visible from the
origin.3
(A tree is visible if at least one point of its bounding circle
has a clear line-of-sight to the origin.)
As the radii $r$ of the trees grows, fewer and fewer trees are visible
from the origin, and eventually some tree engulfs the origin and none
are visible.

          


          

$r=0.12$, $19$ (white) disks visible.


The function $V(r)$, the number of trees visible as a function of
$r$, decreases from $\infty$ at $r=0$ to eventually $0$.
A typical plot of $V(r)$ is shown below (for a different random example).

          


My first question is:


Q1. What is the expected form of the function $V(r)$?
  Can some heuristic reasoning suggest its rate of decrease with $r$?
  Perhaps $V(r) \sim 1/r^2$?

Added (27Jun2019). Here is the above example fit to @fedja's constant:
$V(r) = \pi /(4 d r^2)$:

          


And another example (as each run varies quite a bit):

          



Now imagine that you get to choose the planting of the trees, with the
stipulation that each unit lattice square must receive $d$ points,
and all points are distinct.
(Define each lattice square to be open on its bottom and left edges,
closed on its top and right edges, so their union covers the quadrant.)
The goal is to plant the trees to 
maximize visibility from the origin as the trees grow.
Ideally I'd like $V(r)$ for the ideal arrangement to dominate the
visibility functions for all other 
arrangements of the same number of points. I am not sure there even is such
an optimal arrangement.

Q2. What is an arrangement of trees, $d$ per unit lattice square,
  that maximizes the visibility function $V(r)$?


1
Thomas T. Allen, "Polya's orchard problem."
The American Mathematical Monthly
93(2): 98-104 (1986).
(Jstor link.)

2
Efficient visibility blockers in Polya's orchard problem.

3
In contrast to lattice points:
What fraction of the integer lattice can be seen from the origin?

 A: 
When the trees are of radius $r$, a tree at distance $s$ is visible iff there’s at least one sight line from the origin to the tree, at some angle $-{\sin^{-1} \frac sr} \le θ ≤ \sin^{-1} \frac sr$ from center, such that no other tree is centered within the oval of radius $r$ surrounding that sight line. Classify the potential obstructions within these ovals as

*

*center field, if they are within all such ovals;

*left field, if they are within only the ovals with $θ ≤ α$ for some $α$;

*right field, if they are within only the ovals with $θ ≥ β$ for some $β$.

To first order for small $r$, we can ignore the round ends and approximate these fields as three triangles of area $rs$; we’ll also use the small angle approximation to skip writing $\sin$ and $\sin^{-1}$.
The tree is visible iff: there are no center field obstructions, and the left field obstruction with largest $α$ (if any) and the right field obstruction with smallest $β$ (if any) satisfy $α < β$.

*

*The probability that there are no center field obstructions is $e^{-rsd}$.


*The probability that there are no left field obstructions is $e^{-rsd}$. But if there is one, the largest $α$ has probability density $\frac12 s^2de^{-\frac12(r - sa)sd}\,\mathrm dα$. Given that $α$, the probability that no right field obstructions satisfy $β ≤ α$ is $e^{-\frac12(r + sα)sd}$.
Therefore, the tree is visible with probability
$$e^{-rsd}\left(e^{-rsd} + \int_{-\frac rs}^{\frac rs} \frac12 s^2de^{-\frac12(r - sa)sd} e^{-\frac12(r + sα)sd}\,\mathrm dα\right) = (1 + rsd)e^{-2rsd}.$$
At each distance $s$, we expect $\frac π2 sd\,\mathrm ds$ trees, so the expected number of visible trees should be about
$$\int_0^\infty \frac π2 sd(1 + rsd)e^{-2rsd}\,\mathrm ds = \frac{\pi}{4r^2d}$$
for small $r$. (This agrees with the result suggested without proof in fedja’s comment.)
