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# Smaller Solar Reflectors in Space

A variation on space mirrors to combat Global Warming
 (0) [vote for, against]

One of the proposed methods to deal with Global Warming is to reduce the amount of sunlight that reaches the Earth, and the main idea has been to simply put a lot of small mirrors in orbit around the Earth. If roughly 2% of sunlight can be blocked/reflected away from the Earth, that would suffice. See 1st link. However, details matter....

The diameter of the Earth is roughly 12,730km, which means its radius is half that, and the area of the circle that gets illuminated by sunlight is pi times the square of that radius, or ABOUT 12,730 tens-of-thousands of square kilometers. (That's kind of a unique calculation; try it for yourself!)

2 percent of that value is about 2.55 million square kilometers, and THAT is the amount of mirrored surface area we would need to put into space near the Earth, to reflect 2% of the sunlight that normally reaches us. It should be obvious that that is a rather large and possibly impractical magnitude of mirrored surface. This Idea is about a way to shrink it.

If we were to make a single large mirrored square with 2.55 million sq km of surface area, its dimensions would be almost 1600 km on each side. Few man-made things on Earth, besides the Great Wall of China, encompass enough materiel to give us a proper perspective of the magnitude of making a mirror that size. On the other hand, sunlight in space follows the Inverse-Square Law, and that is the key to this Idea.

The Earth normally orbits about 150 million km from the Sun. At half that distance, 75 million km, sunlight is 4 times the intensity it is at the Earth's distance. This means we could put a mirror there that is 1/4 the size of a mirror near the Earth, and still reflect about 2% of all the sunlight that normally reaches the Earth. So cut that square mentioned in the previous paragraph into quarters; each side is now 800 km, half the original length.

We can do that again! At about 37.5 million km from the Sun (significantly inside Mercury's orbit), sunlight intensity is quadrupled again, and so we could again shrink our square space mirror, down to 400 km on each side. The thing is, mirrors of roughly (yet larger than) THAT size have already been considered for another purpose! See 2nd link.

Logically, of course, if we could make a mirror tough enough for higher temperatures. we could put an even-smaller one even closer to the Sun, and so long as its sail-action was used to always keep it in-between the Sun and the Earth, about 2% of sunlight would still get blocked. Which is the goal of this method of combating Global Warming.

 — Vernon, Nov 10 2015

Space mirrors https://en.wikipedi...or_(geoengineering)
As mentioned in the main text. [Vernon, Nov 10 2015]

Solar Sail http://www.space.co...ght-solar-sail.html
As mentioned in the main text. [Vernon, Nov 10 2015]

Nifty Calculators http://orbitsimulator.com/formulas/
Orbital velocity and solar sail calculations. [MechE, Nov 10 2015]

Solar Sail Calculator http://www.georgedi...lar/Calculator.html
Plug the acceleration at launch at your desired orbit, and see if you can get enough to keep station. [MechE, Nov 10 2015]

Venus 2.0 Venus_202_2e0
As mentioned in an annotation. [Vernon, Nov 11 2015]

How about an infinitessimally small mirror, at the centre of the sun?
 — pocmloc, Nov 10 2015

It might be easier to cover an area of 2.5 million sq km by releasing into low Earth orbit 10^16 little flakes of aluminium foil, each one with a reflective surface of 2.5cm
 — hippo, Nov 10 2015

 Unfortunately, there's really only one position where you can actually place the mirror in a practical sense. If you go anywhere other than the Earth/Sun L1 point, you need to build a mirror ring rather than a single mirror, since orbital mechanics kills you. If you try to use the mirror as a solar sail to keep station, you're going to have two problems. First, your mirror is going to have to be absurdly flimsy, so it's light enough to maintain position. Second, any attempt to keep station is going to be a pain to balance. Because you're trying to slow the orbit (to track with Earth in a much wider orbit) significantly, you're going to tend to fall inwards, and the amount of thrust you'd need to balance that is going to be huge. It might be possible in a fairly wide orbit, where the speed difference is minimal, but the closer you get, the worse it's going to get, and it's going to scale up much faster than light pressure.

 Even Earth orbit requires two or more somewhat over sized mirrors instead of one (the oversized to offset the time their orbit is at the poles rather than between the earth and sun, the two so one is always on the day side).

Practically, the L1 point isn't stable, and requires station keeping operations, but much less so than anywhere else.
 — MechE, Nov 10 2015

Why not just reduce the rate of fusion reaction in your primary ? They do a kit now, not expensive ...
 — 8th of 7, Nov 10 2015

 I haven't done the math, but see the link for calculators for both orbital velocity and solar sail max velocity. (Unfortunately not solar sail max acceleration, which is what you really need).

ETA see second calculator for acceleration, for real world materials. Plug the acceleration into the orbital velocity equations and see if station keeping is actually possible.
 — MechE, Nov 10 2015

hmmm - 2.5 million sq km of aluminium foil (0.005mm thickness) would weigh 33 million tonnes. The energy usage in getting that into orbit might itself have a significant impact on global warming.
 — hippo, Nov 10 2015

Seems like blimps for high altitude that would make ozone to block uv rays would be better as it would reduce skin cancer and since uv light has more power should reduce global temperatures by .5 degrees or so.
 — travbm, Nov 10 2015

Mylar is your best bet if you're producing it from terrestrial materials. It also has the advantage of maximizing available acceleration. (Roughly, it looks station keeping is possible for a mylar sail as long as the weight of the required structure is half or less the weight of the mylar itself, assuming I didn't drop a zero somewhere).
 — MechE, Nov 10 2015

 There's something not quite right with your geometry. If you put a 12,730km disc right above us, it would block the entire sun. Move the disc halfway to the sun, and it will be light again.

Also, move the moon to L1. It's about the right size.
 — the porpoise, Nov 10 2015

 //12,730km disc right above us, it would block the entire sun//

True, I'd missed that. He's treating the sun as a point source. Since it's not, radiation would slip past the disk. It would still block out a significant chunk of direct sunlight, however, so it probably just adds somewhat to the required diameter as you get closer to the sun, so it's not a strict geometric progression.
 — MechE, Nov 10 2015

I dunno, move an Earth-sized disc to the orbit of Mercury and it's not going to block much. I think the original idea is backwards. The mirror can be made smaller if it's closer to Earth.
 — the porpoise, Nov 10 2015

//treating the sun as a point source// see anno no.1
 — pocmloc, Nov 10 2015

 It is true that my calculation can simply be taken as considering the Sun to be a point-source, but 2% of the total light reaching the Earth, whether from a point or from a disc, is still 2%. ALL sunlight has quadruple intensity at half the distance. Tiny mirrors in Earth orbit won't be casting a detectable shadow on the surface of the Earth, yet if 2.55 million sq km of them were in-between the Sun and the Earth at any one moment, 2% of the total sunlight would still be reflected away. So, I don't see any big problem with a single mirror reflecting 2%.

Regarding solar sails, they have to have a large enough surface-to-weight ratio to be pushed by sunlight against the Sun's gravitation. We would use rockets to make them escape Earth's gravitation before unrolling/unfolding them. The thing is, both sunlight pressure and gravitation obey the Inverse Square Law, so any sail that can accelerate away from the Sun at Earth orbit can accelerate away from the Sun at any other distance from the Sun. Tilting it, making it less- perpendicular to sunlight, would change sunlight pressure without changing gravitation, and so that is how any specific location can be controlled/maintained. (consider tilting parts of it in different directions, like a parasol)
 — Vernon, Nov 10 2015

 If you want to block 2% of the Sun's light with a mirror next to the Sun, then you need a mirror that is 30 billion sq. km. in area, using the same math as your did. (2% * Sun's radius^2 * pi). Yes, the light the near-Sun mirror blocks is much more intense, but that's because most of it wasn't going to hit the Earth anyway.

 You can block 2% of the light at the Earth with a much smaller mirror, as you say, 2.55 million sq. km.

Or look at it this way. Model the Sun as a wall of point sources each with an area of 2.55 m. sq. km. You need 30b/2.55m or 11,764 of those point sources. When the 2.55 m. sq. km. mirror is beside the Earth, it blocks 2% of incident light from each of those point sources. When the 2.55 m. sq. km. mirror is beside the Sun, it blocks only a single point source (because hey it's exactly the same size as one of them). Blocking 1 out of 11,764 point sources is the same as blocking 0.009% of the light, which is much less than 2%.
 — the porpoise, Nov 10 2015

 Another simple way to look at it: when a small mirror is relatively close to earth, all the light it is reflecting would have hit the earth. Since the sun is ~10x the diameter of the earth, if the mirror is more than 1/10th the distance to the sun, some of the light from one edge of the sun that hits the mirror would not have hit the earth anyway.

There is one advantage to moving this reflector away from the earth a bit. If your group of mirrors is too dense it could create nonuniform shading, the extreme being local eclipses. If you move the mirrors out almost 1/10th the distance to the sun, the shadow will be softened and spread evenly over the earth. Though I must say that
 — scad mientist, Nov 10 2015

Could we drag an asteroid to Earth/Sun L1, and grind it up into a dust cloud? Perhaps the gravity is low enough to allow a large cloud, but without too much dust drifting off (it could also very slowly orbit a small central mass). We'd dirty up the place, which is not good news for new telescopes, but perhaps we could trap some light and heat there to radiate in a non-Earth direction.
 — TIB, Nov 10 2015

While you're busy cooling the earth, perhaps you can redirect some of that light towards Mars?
 — RayfordSteele, Nov 11 2015

 OK, this appears to be a classic case of reaching a conclusion without taking all the available relevant data into account. With some of the data, one conclusion is reached; with more data, another conclusion can be reached. (I say this is a "classic" thing because certain long-lasting debates have been following that pattern for decades. More specifically, those who oppose abortion are most certainly not taking into account all the available relevant data.)

 In the case of the present Idea, we can imagine a "frustum", which is basically a cone that has had its pointy end lopped off. The base of this frustum is the diameter of the Sun, about 1.4 million km, and the top of the frustum is the diameter of the Earth. This frustum fully defines the Earth's default/unmodified illumination by sunlight.

 I'm pretty sure I DID take a different frustum into account in another Idea ("Venus 2.0", linked), but that was because that Idea needed to block all the sunlight, not just a particular fraction of it.

 It should be intuitively obvious that 2% of the surface area of the top of the Sun-Earth frustum is smaller than 2% of any circular area defined by a geometric plane that cuts through the frustum anywhere between the Sun and the Earth (and parallel to its base). That is the foundation of the arguments made in other annotations here. I accept that.

 However, it remains to be explained why what SEEMED to initially make sense failed in the end. To answer that, lets consider a shorter frustum that is 75 million km tall, with a 2.55 million sqr km circle on top (about 900km radius). That's 2% of Earth's disc-area, considered before thinking about the Inverse-Square Law.

 So, let us think of all the sunlight that can reach this area from the surface of the Sun --if it wasn't blocked, where would it go? Well, we simply ray-trace the sides of THIS frustum to make a complete cone, and extend beyond, creating another cone. This second, expanding-light-cone, will be quite wide by the time the cone's length reaches Earth's distance from the Sun, 75 million km from the top of that shorter frustum.

 Since the base of this cone will be far wider than the diameter of the Earth, THAT is why that 2.55 million- sqr-km area can't actually block 2% of sunlight reaching the Earth, when relocated halfway toward the Sun.

Thanks to all who participated; I'm not deleting this Idea because it may be educational to someone in the future.
 — Vernon, Nov 11 2015

 My preceding anno hints that there MIGHT be some points where a smaller disc or square, than 2.55 million sq km, can successfully block 2% of sunlight reaching the Earth, in accordance with the effects of the Inverse Square Law.

 On paper draw a line representing the diameter of the Sun, and another parallel line representing the diameter of the Earth. The second line should be located such that if the first was bisected by a perpendicular line, the second would be bisected, too.

 Now add two more lines to the sketch, to make two isosceles triangles. The base of each triangle is one of the first two lines (the Sun-line and the Earth-line), and the apex of both triangles is the same point, located in- between the first two lines. These triangles represent two cones SIMILAR to those described in my previous anno. The main difference is that in the previous anno, the apex of the two cones is located far closer to the half-way mark, in-between the Sun-line and the Earth- line, than the apex of the two triangles is located here. Here that apex is located rather close to the Earth-line.

 In the much-larger of the two triangles, which has the Sun-line as its base, and near the apex of this triangle, we can imagine cutting across the triangle to make a trapezoid (equivalent to the second frustum described in my previous anno). The length of this cut would represent the 1800-km-diameter of the disc that can DIRECTLY block 2% of sunlight, **when** **that** **width** **is** **considered** **to** **be** **part** **of** **the** **Earth-line**.

 According to the logic of my previous anno, that disc- line can STILL/INDIRECTLY block 2% of sunlight when located at that special place, closer to the Sun than the apex of the two triangles in this sketch. That's because the expanding cone of light would be exactly the Earth's diameter, and not vastly greater than that diameter, as mentioned in my prior anno. The disc is located closer to the Sun than the Earth, and the Inverse Square Law DOES apply, allowing it to affect a greater intensity of light than the intensity at the Earths surface.

 It's been so long since I used trigonometry for anything that I'm unsure of how to compute the exact distance from the Sun we could place that 1800-km-wide disc, to block 2% of sunlight, as per the two cones/triangles indicated in this description. But once we have located the apex of the two triangles, THIS NEXT THING is the thing to think about!

 Between the disc and the apex is a shrinking width where a smaller disc should still be able to block 2% of sunlight. AT the apex, a quite-small solar-sail/mirror, easily launched with just one of today's rockets, might suffice!

Am I deluding myself, or have I found the simplest Solution to the Global Warming problem?! If it is located too close to the Earth's gravitational field, then an actual solar sail wouldn't work, because it would fall to the Earth. We would need a solar-powered ion drive (manipulating the Solar Wind), or maybe one of NASA's experimental reactionless drives, to keep that small mirror in the right place over the long term.
 — Vernon, Nov 11 2015

 Some quick figurings with geometry, not so much trig, leads to the following: The two isosceles triangles previously described are proportional, so since the bases of the two triangles have a ratio of about 113:1 (Sun's diameter to Earth's diameter), their heights have that ratio, too.

 So we divide 150 million km by [113+1=114] to get the "unit" of about 1,316,000 km; the apex of the two triangles is that distance from the Earth; it is 113 times that distance from the joined apexes to the Sun.

 Now recall I described a hypothetical 1800-km-wide disc closer to the Sun than the location of the two triangle-apexes. We can think of ANOTHER proportionate triangle, having a base of 1800 km. The ratio of the Earth's diameter to 1800 km is about 6.87:1. So we divide 1,316,000 by [6.87+1=7.87] to get about 167,000 km --and we add that to 1,316,000 to get 1,473,000 km, the distance from the Earth to the 1800-km-wide disc that still manages to block 2% of sunlight. Believe it or not, that location is quite close to the L1 point!

 Anyway, it happens that those triangle-apexes are closer to the Earth than the L1 point (which is only 1/100 the Sun-Earth distance). Since we want a smaller disc diameter than 1800 km, located NEAR the apexes, we most certainly need an appropriate way to keep the disc from falling to the Earth.

[Voice], yes, I agree that simply to block sunlight is good enough; it doesn't really have to be reflected away. However, the block can't be in orbit, because only a tiny part of the time would it actually be directly in- between the Sun and the Earth, blocking sunlight. It has to be in that single location all the time, which means powered location-maintenance is critical.
 — Vernon, Nov 11 2015

 //mirrored surface. // If the goal is just to block sunlight you don't need a reflective, much less mirrored, surface. Just one that doesn't emit much radiation backwards. You only need a mirror if you're also generating power.

// we most certainly need an appropriate way to keep the disc from falling to the Earth.// Set it in orbit. You'll have to make it twice as large but you won't have to artificially stabilize it much.
 — Voice, Nov 11 2015

 //If the goal is just to block sunlight you don't need a reflective, much less mirrored, surface. Just one that doesn't emit much radiation backwards.//

 A mirrored surface is much cheaper than the sort of structure you'd need to keep the object from radiating a decent chunk of it's energy backwards. Aluminized Mylar rejects about 85% of the energy hitting it, and radiation off the front and back would be roughly equal, so about 7.5% headed in the general direction of earth.

 A blackened aluminum shield (for instance) would warm to a fixed temeprature, and radiate roughly equaly off the front and back, so you've got about 48% heading in the general direction of earth. (I'm assuming low enough earth orbit and ignoring edge effects that most of the backside radiation is hitting the earth. This isn't correct, but is probably within a factor of 2 or so).

Insulating the back so that more energy radiates off the front works, but it adds mass and volume.
 — MechE, Nov 11 2015

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