# How much energy from the Sun could we get if we use Jupiter or the Earth as an atmospheric lens?

How much energy (watts) from sunlight could arrive to the focal point if we use Jupiter or the Earth as an atmospheric lens by using refraction?

How far the focal point would have to be placed for each case?

Thanks!

• How would you plan on using all this power? You would be zipping by so fast that there would be little chance to do anything with it. Oct 18, 2020 at 15:43

Let's try to make some quick and dirty estimation for the upper limit in the case of Earth. (And in the end figure out that we have an actual measurement of this...)

First, the bending power of the atmospheric lens. It's not uniform as @uhoh mentioned in his answer. But, for rays passing just above surface we can get a accurate number. During sunset atmospheric refraction makes the Sun appear about 0.5° higher than it actually is. That is, a ray that barely makes it past Earth is bent by 1°. If we treat the incoming rays as parallel, the focal point is about 360,000 km behind Earth.

Now assume that the whole Earth is a perfect lens projecting an image of the Sun at this distance. We can easily calculate the apparent size of this lens and compare it to that of the Sun. The lens is about 30 times larger. Optics conveniently tells us that power per area scales with the apparent size, hence is thirty times higher than the ordinary solar constant.

Unfortunately most of the Earth is opaque. If we assume the useful thickness of air at 10 km, 99.7% of the light is blocked, dropping the power to 10% of the solar constant or about 100 W/m².

Air has a far from perfect transmittance either. At noon, 70% of the Suns energy reaches to the surface, but (making some generous assumptions) this drops to about 1% around sunset. The light passing through the lens passes through this atmosphere twice and hence only 0.01% of intensity remains, leaving us with a mere 10 mW/m². Compared to the solar constant, this is a factor of 0.00001. Given the assumptions, this is an upper limit of the actual brightness due to the lensing alone.

Did you note the peculiar distance of 360,000 km? How nice would it be to have a canvas there to measure its brightness from your backyard? Something the size of, say, the Moon? Passing right through Earth's shadow? That's exactly what happens during a Lunar eclipse. From several photography blogs I got numbers of $$2^{15}$$ to $$2^{18}$$ for the change of the brightness of the Moon as it enters into totality. This corresponds to a attenuation between 0.00002 to 0.000004, right around the value we got above, although this contains not only the effect of the atmospheric lens, but also all other kinds of refraction and reflection.

• thanks for your answer! perfect, 10 mw/m2, but how many watts would reach the focal point?
– John
Oct 25, 2020 at 13:54
• @Albert - the Sun is an extended object, so there is no focal point. There would be a projected image of the Sun with a diameter of 3000km. But in reality, even this doesn't exist, because the atmosphere is not a lens. Oct 25, 2020 at 14:08
• thanks for your answer. I see. I thought that maybe there would be a focal point, since Yu Wang proposed a Very High Resolution Space Telescope Using the Earth Atmosphere as the Objective Lens: trs.jpl.nasa.gov/handle/2014/22381
– John
Oct 25, 2020 at 18:45

Partial answer, because this is a really hard question!

We can learn how lenses work using the thin lens approximation where the amount that a ray is bent (the angle, or the tangent of the angle) is proportional to the distance from the center of the lens.

For example if my lens bends rays 1 cm away from the axis by 0.57 degrees (0.01 radians) then it's going to intersect the axis 1 meter downstream. If rays at 2 and 3 cm away from the axis are bent 2 and 3 times more strongly, then they'll also intercept the axis at 1 meter. This is the main ingredient for a focussing lens, the strength of the bending is proportional to the distance from the axis.

Some lenses don't focus well, or not at all. In the simplest case of a gravitational lens, they do tend to concentrate light somewhat because for a thin ring a given distance from the object, those rays will intercept the axis some distance later. Rays passing through another ring farther or closer will be bent differently, but here's the problem. While a thin lens' strength is proportional to r, a gravitational lens' strength is proportional to 1/r. It's only in very special weird cases where some degree of imaging is possible, and it's still quite distorted.

Now we have a planetary atmosphere which is definitely a thin ring, and we have the unfortunate problem that the strength gets weaker, not stronger as we move away from the center of the Earth, so while we might get some concentrating behavior, we might not because the change in strength is extremely rapid. The ring we are talking about is around 6400 kilometers in radius, and increases in density by a factor of 2 every 5 or 6 kilometers (log(2) times 8 km scale height).

I thought about calculating this but you really have to do ray tracing because each ray passes through every altitude down to some limit before going back up again. Hopefully someone will find a paper about atmospheric refraction and cite it; I am sure there are many!

• Original comment didn't work. Delete, I will try again. Oct 18, 2020 at 16:00
• "atmosphere thinner with altitude" -- for those familiar with fiber optics, this is sort of like a GRIN lens. Raytracing through the spherical shell that is our atmosphere, plus 100% blockage of rays which meet up with the central rock, should be relatively simple to set up with a decent optics package, maybe OSLO or CODE V Oct 19, 2020 at 12:09
• @CarlWitthoft [Ball lenses]() are usually uniform index but something that can simulate Luneberg-like lenses would come in handy. But no approximations because with the index of refraction doubling every 1/1000 of a radius a planet's atmosphere is one crazy lens.
– uhoh
Oct 19, 2020 at 13:08