# On a frozen planet with a tower that generates an artificial sun, how tall would the tower have to be to illuminate an area with a 1000 mile diameter? [closed]

Yes this is a fictional worldbuilding brainstorming question. But I'm sure there's a mathematical equation to find the answer. I just don't know what it is. Anyone? Let's assume the planet is the size of Earth.

## closed as unclear what you're asking by nitsua60, fi12, Aify, Xandar The Zenon, 458Feb 16 '16 at 10:20

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• Too few data. It depends on what you mean by „frozen” and „warm”, what is chemical composition of the atmosphere and surface, the planet radius, how much energy does the artificial sun radiate and in what form. – jaboja Feb 16 '16 at 1:00
• If I know how hot the artificial sun is, I can tell you the answer. – fi12 Feb 16 '16 at 1:03
• We offer you cryptic welcomings Toby. We love new questions but this one may not be fit for this site in its current state... – Xandar The Zenon Feb 16 '16 at 1:03
• It can be interpreted as any heat though, for which a meaningful answer exists – Hohmannfan Feb 16 '16 at 1:04
• How large is the planet? Do you really want to know about "warming" the planet, or are you just looking at how to have the light from this "sun" hit the specified area. – nitsua60 Feb 16 '16 at 1:08

As @Hohmannfan points out, the curvature of the earth limits the region illuminated by a light source. If we want an area of diameter $d$ to be illuminated, the (planar) angle subtended by this diameter is $\theta=d/r$ (in radians), where $r$ is the radius of the earth. We can draw a four sided shape with vertices at the center of the planet, the top of the tower, and two opposed points on farthest reaches of the illuminated area. If the tower is the shortest it can possibly be, the angles at the vertices at the farthest reaches of the illuminated region are both right angles. Thus, with a little trig, we can calculate that the distance from the center of the earth to the top of the tower is $D=r\sec(\theta/2)=r\sec(d/2r)$. To get the height of the tower above the surface of the planet we simply subtract the planet's radius: $h=D-r=r(\sec(d/2r)-1)$. Plugging in $d=1000$ mi, and $r=3959$ mi, we get $h=35.1$ mi. This is the absolute minimum height of the tower.

Of course, it is quite a bit more complicated than that. The problem is that the light from the tower will hit the farthest reaches of its area of illumination at a very oblique angle, which has very little warming effect. This is why the far north and far south of earth are so cold. We could make up for this by making the artificial sun extremely intense, but this would bake the area directly beneath it so as to make it uninhabitable. The solution is to make the artificial sun more intense, and further above the surface. Since you do not elaborate on what you mean by making the illuminated zone inhabitable, let us suppose that you want the fringes of the region warmed like the latitude $\phi$ on earth. The geometry here is quite complicated, but I come up with $$h=r\sin(d/2r)\cot(\phi-d/2r)-r-r\cos(d/2r).$$ So if you want the area $500$ miles from the tower to be like $45^\circ$ north or south on earth, the tower needs to be $626$ miles high. This means that the light from the tower hits the ground at and angle greater than $45^\circ$ at points inside the $1000$ mile diameter region.

There are even more complications like atmospheric absorption, shadows from the tower itself, and probably other things I've not though of.

• With the 35 mile high tower you could terrace the planet's surface so that the terraces face the sun and the steps between the terraces are in darkness. It might make for some interesting interplay between habitable and frigid zones. The terrace steps would get progressively shorter as you approach the base of the sun, of course, so perhaps the outermost ring of habitability could be the boonies, separated by a wide band of frigid wasteland from the inner kingdom. – Jason Patterson Feb 16 '16 at 4:18

Yep, there is one:

The horizon is curved, so your area that receives light is limited.

$$h = \frac{r_e}{\sqrt{1-\left(\frac{r}{r_e}\right)^2}}-r_e$$

$r_e$ is the radius of the planet, and $r$ the radius of your illuminated area.

For Earth, that is a minimal height of $51.35km$ in order to be able to cover that area.

This yields the minimum required height of the tower to cover that area. "suitable for humans and animals" is very vague, and makes it impossible to get a more accurate result. • The OP refers to heat rather than light. In addition, where did you find this formula? Can you give me a source? – fi12 Feb 16 '16 at 1:16
• @fi12 The OP actually does mention illumination in the question. And he gave too few details to specify the heat. – Xandar The Zenon Feb 16 '16 at 2:18