Flying islands and shadows

Question

Let's say you have a "planet" of floating islands, and you still have a star (In similar size and distance to our Sun) around which this collection of floating islands resolve (Or maybe the sun actually orbits all the floating islands).

For a given size of floating island, how far away from each other do the islands have to be, to avoid casting shadows on each other? (Ignoring the fact that it might be unlikely, due to whatever errant movement and positions they may have)

Or, more directly, if on our world, and in our solar system, how far are shadows thrown, for a given size?

Aside from probably being quite notably visible, would a disk of 1 square kilometer, throw a shadow of the same size (Let's just go with the sun being directly above), and would you have sunlight enough to see in the "covered" area, if the disk was 1km away? 5km? 20?

Answer 1

The Sun is an extended light source, and thus the length of the cone of shadow cast by the Moon (or a floating island) is finite; in the simple diagram below, the cone of shadow is the grayed out area.

^{The Sun, the Moon (or a floating island, if you wish), and an observer. $D_S$ is the diameter of the Sun; $D_M$ is the diameter of the Moon (or of the floating island); $R_M$ is the distance from the Sun to the Moon (or to the floating island); $R_S$ is the length of the shadow; and $R_O$ is the distance between the Moon (or the floating island) and the observer. Own work, available on Flickr under the Creative Commons Attribution license.}

Let's assume that the Sun, the Moon (or the floating island) and the observer are all placed neatly in a straight line.

From considerations of triangle similarity, we see that the shadow cone cast by the Moon (or by the floating island) is

$$R_S = \frac {D_M R_M}{D_S - D_M}$$

Plugging in the numbers for our own Sun, Moon, and distance from the Sun to the Moon we have

$$\frac {3470 \times 149{,}000{,}000}{1{,}392{,}000} \approx 371{,}000\,\text{km}$$

which is inline with the observation that the length of the cone of shadow of the Moon is a bit larger than the smallest distance between the Moon and the Earth (and thus total solar eclipses and possible), but a bit smaller than the longest distance between the Moon and the Earth (and thus sometimes when the Moon passes in front of the Sun we see an annular eclipse instead of a total eclipse).

Now, and observer placed farther away from the Moon (or the floating island) will see the dark Moon (or floating island) blocking part of the Sun. The angular diameter of the Sun seen by the observer is $D_S / (R_M + R_O)$; the angular diameter of the Moon (or floating island) seen by the observer is $D_M / R_O$.

^{Note 1: Small angle approximation applies.
Note 2: The angles are in radians. Multiply by $180 / \pi$ to get degrees.}

When the Moon (or the floating island) passes in front of the Sun at a distance from the observer greater than the length of the cone of shadow, it blocks part of the light coming from the Sun; the relative decrease in illumination is proportional to the square of the ratio between the angular diameters. For practical purposes, if the relative decrease in illumination is less than 1/4 or so nobody will notice, not even photographers.

Answer 2

This is not a question of distance, it's a question of geometry.

If the islands form a sphere about the sun such that none of them would form a line with one island between another and the sun, you could put them close enough that a small child could step from one to the other without having one cast a shadow on another.

Once you have the line, you stuck, although if it's a large enough distance, you could say that the shadow is too small to consider.