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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.
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]
[link]
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How about an infinitessimally small mirror, at the centre of the sun? |
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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 |
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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. |
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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). |
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Practically, the L1 point isn't stable, and requires station
keeping operations, but much less so than anywhere else. |
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Why not just reduce the rate of fusion reaction in your primary ? They do a kit now, not expensive ... |
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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). |
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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. |
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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. |
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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. |
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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). |
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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. |
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Also, move the moon to L1. It's about the right size. |
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//12,730km disc right above us, it would block the entire
sun// |
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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. |
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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. |
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//treating the sun as a point source// see anno no.1 |
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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%. |
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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) |
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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. |
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You can block 2% of the light at the Earth with a much smaller mirror, as you say, 2.55 million sq. km. |
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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%. |
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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. |
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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 |
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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. |
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While you're busy cooling the earth, perhaps you
can redirect some of that light towards Mars? |
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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.) |
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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. |
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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. |
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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. |
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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. |
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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. |
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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. |
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Thanks to all who participated; I'm not deleting this
Idea because it may be educational to someone in the
future. |
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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. |
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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. |
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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. |
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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**. |
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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. |
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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! |
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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! |
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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. |
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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. |
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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. |
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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! |
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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. |
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[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. |
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//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. |
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// 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. |
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//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.// |
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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. |
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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). |
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Insulating the back so that more energy radiates off the
front works, but it adds mass and volume. |
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