# Is it possible to focus sunlight from space on a tiny spot on Earth?

Recently I have read an article about orbital mirrors, benefits and future applications of these devices. There was a figure which basically explains how divergence angle affects spot size on Earth. For example if you take the diameter of the mirror 100 meters and put it on 400 km LEO it would create a spot about 4 km diameter.

This example was based on an orbital flat mirror. The author also included two formulas for determining the spot area.

for a non-focusing reflector:$$A_{e} = A_{r}+\frac{\pi}{4}(\alpha h)^2$$ for a focusing reflector: $$A_{e} = \frac{\pi}{4}(\alpha h)^2$$

In the latter equation we still can not get rid of the divergence angle which is approximately 9.3 mrad. Is there a way to further narrow the spot size on Earth? I also looked at Gauissian beam and beam waist equations but could not totaly figure it out. Can you also include some calculations with your answers?

• The key problem here is that you do not have a Gaussian beam, but the divergent light from the sun. The best you can do is to make a spot on the ground see the reflected sunlight across the entire sky. See the addendum to this blog post, aleph.se/andart2/space/a-sustainable-orbital-death-ray Jun 1 '18 at 12:50
• The mirror will, at best, form an image of the sun on the surface of the Earth. Calculate the size of that image (based on the relative distances between the Sun, Earth, and mirror. Magnification/demagnification $R$ is Jun 1 '18 at 14:24
• I don't hear anyone talking about off-axis parabolic reflectors/mirrors that is something you could look into ;) Jun 1 '18 at 15:34
• Thank you @BobvandeVoort . I found a related question and answer by using your suggestion. Here is the link. It says increasing the beam diameter reduces the divergence angle. So, I think a suitable solution might be using a small mirror for collecting sunlight and a larger mirror for focusing it on Earth. Jun 1 '18 at 16:32

The mirror will, at best, form an image of the sun on the surface of the Earth. Calculate the size of that image (based on the relative distances between the Sun, Earth, and mirror. Demagnification $R$ is $R = De/Dm$, where $De$ is distance to the Sun from the mirror and $De$ is distance from the Earth to the mirror. The Sun has a radius of about 432,000 miles and is about 93 million miles away. An orbiting mirror could be, say, 100 miles above the Earth's surface. So the image of the Sun would have a radius of about $432000 x 100/93000000 = 0.46$ miles wide. 2450 feet. If the mirror were, say, 10 miles wide, I would not want to be standing there (~ 100 x normal sunlight).