# How big would a satellite need to be to block out the sun?

I have a story idea that involves a human attempt to halt and reverse global climate change. The idea is a simple one - a satellite that partially blocks out the sun. The satellite is located at the L1 lagrange point and is stabilised to always cast the most efficient shadow on earth.

The satellite would be similar to a solar sail with a surface area many square kilometers.

Ideally, I would aim to block 1% (or less) of solar energy. It would only need to cool the earth by a fraction (say 0.1) of a degree per year.

This is where I need clarity.

• Would 1% be too much cooling?
• What would the size of the satellite need to be to block that much sunlight? 100 km square?
• Is there anything else I haven't thought of?

Edit: This question isn't about being visible from Earth. In the world I'm working on, it's not even noticeable from earth without high tech equipment.

TLDR;

The equations:

$$r_{B} = \sqrt{\frac{F}{2.46\times 10^{-14}}}$$ or rearranging for $$F = r_{B}^{2} \times 2.46\times 10^{-14}$$

Where $F$ is the fraction of light blocked ($F=0.01$ gives your $1\%$) and $r_{B}$ is the radius of your satellite in meters which will achieve this.

For one percent reduction, using the equations above, we need a satellite of radius $6.376 \times 10^{5}$ m , or $637.6$ km - pretty big to say the least! (roughly the size of Alaska).

# The Maths

Initially you added a 'mathematics' tag onto this question - I'm assuming you wanted something more along the lines of a hard science tag (rather than asking about building a mathematical system as the tag is intended).

Distance to $L_1$

The wiki for Lagrangian points gives this equation:

$$d_{E} \approx D \sqrt[3]{\frac{M_{E}}{3M_{S}}}$$

Where $d_E$ is the distance $L_{1}$ is from Earth, $D$ is the distance between the Sun and Earth and $M_S$ and $M_{E}$ are the masses of the sun and earth respectively. Using:

$$D = 149597870700 \text{ m}$$ (This is 1 Au, the average distance, so will change but the equation is already approximate) $$M_S = 1.9885 \times 10^{30} \text{ kg}$$ $$M_E = 5.9724 \times 10^{24} \text{ kg}$$ As given by the NASA factsheet.

Giving us $d_{E} \approx 1.49656 \times 10^{9} \text{ m}$ or $1.5$ million kilometers.

Now lets look at what this means for how large a satellite you'll need.

The radius, $r_B$, of the Blocker projected onto the Earth gives a shadow with size $r_B^{'} = \frac{D}{d_{S}}r_B$ where $d_{S}$ is the distance the satellite is from the sun ($d_{S} = D - d_{E}$).

If we want to know the fraction, $F$, of light the satellite will block we can compare the areas of circles presented (the earth is actually a sphere so this won't be exact).

$$F = \frac{\pi r_{B}^{'2}}{\pi r_{E}^{2}} = \frac{(r_{B} \frac{D}{D-d_{E}})^{2}}{r_{E}^{2}} = r_{B}^{2} \times 2.46 \times 10^{-14}$$

Which you can use to calculate how much light you would block out for a satellite of a particular radius or rearrange to get the radius needed for a particular fraction ($r_{B}^{2} = \sqrt{\frac{F}{2.46\times 10^{-14}}}$ ).

• @sanchises The distance between L1 and Earth is less than the distance between the perihelion and aphelion of Earth. That is to say, the change in relative angular size of the sun at L1 is less than the relative change in angular size of the sun between January and July here on Earth. That is to say, not much. I think point source is reasonable. – kingledion Jan 17 '18 at 13:50
• @RonJohn, well, you may wish to do some calculations or you may join the failed group ) 0.5 degrees in astronomy isn't something you can easily handwave. – ZuOverture Jan 17 '18 at 19:43
• The story about column shadow was to show that all calculations in this answer after L1 distance, alas, aren't right, @Sanchises questioned them absolutely correctly. If the blocker size is less than ~14000km, there will be no umbra at all. – ZuOverture Jan 17 '18 at 20:18
• This doesn't seem right. L1 is approx 4 times farther away than the moon. During a total solar eclipse the moon doesn't even cut out 1% of total sunlight when you consider the entirety of the sunlit side of the Earth, yet you are suggesting an object much smaller than the moon would be adequate. – Octopus Jan 17 '18 at 21:30
• @ZuOverture Not having a dark shadow (umbra) visible on Earth would be a benefit, as that would have a lower impact on day-to-day life. What's at issue is: Does interposing something like this reduce the overall radiation delivered to the Earth? It does do that, by directing elsewhere the desired portion of the radiation which would have hit the Earth. Yes, to have an engineering-accurate number you definitely need to account for the Sun not being a point-source, not to mention needing to account for how far Sun-ward it needs to be from L1 to balance the force from the sunlight it redirects. – Makyen Jan 17 '18 at 21:48

The energy needed to produce such satellite and put it on orbit would deny the idea behind fighting with climate change.

I ignore the Lagrange point and just assume you want to block 100 km square (nothing much really, clouds are blocking much more). So you want to create a shadow that is 100 km by 100 km large. ISS is 400 km above the earth. Mir was 350 km. So let's say 300 km from earth.

$$\text{Umbral size} = 2*(b S - B s)/(S - B)$$

Where: $b =$ blocker size
$B =$ blocker distance (from Earth's surface)
$s =$ sun radius
$S =$ sun distance (from Earth's surface)
$\text{Umbral size}=100$

From this we have
$$(50-b)S=B(50-s)$$ so $$(50-b)149600000=300(50-695700)$$ Which give us around 51.39 km. To put that into perspective – ISS is 109 metres wide. Titanic was 269 metres long.

And to go further. You know why NASA use gold foil? Because it's the best in blocking radiation. But use the cheapest tinfoil on the market (also the lightest). One sheet of 3 square meters of 0.3 mm thick foil weight around 2,63 kg (assuming density 2.80 g/cm³, to switch to gold just use 19.32). 51 square kilometres would require to put 44710000 kg of just foil (so no mechanism to unfold it, hold it together, counter engines, extra fuel etc.)
Next perspective, the ISS weight 419455 kg. 106 times less than you would want to put up there. ISS in numbers

Also this equation tells us that the 300 km from earth is not enough as you would need a lot of (wild guess, more than 100 times the ISS need) energy to counter gravitation and stay on the orbit.

Also you know how much aluminium we produced in 2016 in the world? 57600000 kg according to www.world-aluminium.org. So 70 % of all the world production would be needed for your project.

With this you can see that the Lagrange point would need to be much further away from Earth. Which would, of course, require larger size and result in larger mass.

So to summarize – in your story, around fifth calculation your humans would realize they can't produce enough tinfoil or producing enough tinfoil and fuel to put the satellite would spike up the energy production of the whole world.

Which could be a nice story on itself. Humans try to stop global warming in the worst way possible. By producing least effective solution that use so much energy they speed up climate change to one year.

• Consider using MathJax, as far as I know, it works here as well. math.meta.stackexchange.com/q/5020/121236 – Mołot Jan 17 '18 at 10:54
• A few links wouldn't go amiss either. Not that I doubt the validity of your claims but I'm sure you had to look up the 2016 world aluminum production so including the link to show where you got it from wouldn't be too hard. – Lio Elbammalf Jan 17 '18 at 11:52
• @SZCZERZOKŁY Mining asteroids is not comparable to magic – Callum Bradbury Jan 17 '18 at 12:54
• 0,3 mm thick foil That's not foil, that's sheet metal. Aluminium household foil is about 1/20th that thickness. That leaves you with only 5 ISS masses and 3.5% of a year's aluminium production, which doesn't seem to bad (considering it saves the planet instead of just letting astronauts play in space). Delta-V required for low earth orbit is about 9.4km/s; only an additional Delta-V of 3.77km/s is required to get to L1. – Sanchises Jan 17 '18 at 13:55
• @SZCZERZOKŁY The James Webb telescope uses 100nm aluminium coating on 25µm kapton films. Gold is mainly used for its high emissivity (so that it doesn't heat up the spacecraft through conduction); it blocks radiation just fine if conduction is not an issue. – Sanchises Jan 17 '18 at 15:00

A simple way to think about this without all the math is to look at the projected path of any total solar eclipse. Only a small swath of land is generally affected, which is why during the latest total eclipse visible from the USA, people were traveling to other states to be able to get better views.

That said, it will also depend on a combination of how big it is and how far away it is. Your thumb can block out the sun... at least from your point of view.

• Welcome to worldbuilding! This is true, but doesn't fully answer the question, so would you be able to edit your question to answer how big the satellite would need to be? – Mithrandir24601 Jan 17 '18 at 17:59

No reason to build a giant anything. Simply spread a plate shaped cloud of minute particles at L1 that would then block the appropriate amount of earth bound radiation. I visualize a multi-nozzeled spinning emitter to cancel its own thrust. I trust it could be determined what the resulting expanding donut shaped (or a different shape?) cloud's duration and hence effect would be. Of course it may take multiple releases over time to do the trick.

• I have edited the post to fix spelling and clarify the question mark. It was confusing, at least to me, as to why the sentence seemed to stop and then have a fragment. – John Locke Oct 24 '18 at 22:49

Echo I was an early satellite made from aluminum coated mylar. 100 foot diameter for 159 pounds. It was inflated with 33 pounds of powders that sublimed in vacuum. This was done with 50's tech. The film was 1/2 mil (12 microns) and the aluminum coating was .2 microns.

Using this tech a 59 km diameter sphere would take 270 million kg. This ignores the extra gas required to inflate a larger sphere. (Volume of gas goes up with the cube, while the area goes up with the square of diameter.

I suspect an easier way would be to build induction catapults on the surface of the moon. and fire bags of rocks into L1, similarly to what G. Harry Stine proposed for this book "The Third Industrial Revolution. at that point you want to process them into a powder and give them a very slight electrical charge to keep them from clumping.

Unfortunately L1 isn't stable, so you will need to continuously replenish it. It may be better to give Earth rings, like Saturn. This is a short sighted solution, as it puts a lot of crud in orbit. Eventually we will want to industrialize the solar system, and having large quantities of orbital sand is a significant traffic hazard.

Edit: Such a satellite has to be 1/10 the diameter of the earth to block 1/100 of the sunlight. So instead of 60 km in diameter it has to be ~1200 km in diameter. My answer is off by a factor of 20^2. This is a non-trivial project.

Partial answer to the 3rd question "Is there anything else I haven't thought of?":

Politics. Such a sunshade made of reflexive material makes an enormous weapon. Earth receives about $1.7\cdot 10^{17} W$ from the Sun, 1% of that is $1.7\cdot 10^{15}W$. If you can focus the reflexive area (say to an area of a big city), you'll get an equivalent of 20 Hiroshima explosions per second. In short, Die Another Day on (enormous) steroids.

Now, who is going to control the sunshade? The UN?

• @F1Krazy, a mirror flipped around remains a mirror... – L.Dutch Jan 17 '18 at 9:08
• Nowhere in the question it says anything about reflective. And on the image it is shiny, but flat. Seems like you are solving problem that's not even there, and it feels more like a comment than an answer. – Mołot Jan 17 '18 at 9:16
• @ZuOverture, a mirror is a mirror. A lens is a lens. They are not the same. – L.Dutch Jan 17 '18 at 10:08
• @Mołot, excuse me, I tend to forget that people here prefer to perceive everything literally. Read "reflective surface" instead of "mirror" if you prefer it that way. And forget about using foil surfaces for photovoltaics, that's definitely not possible. Also you can't use them as solar sails and rotate them to change the direction of thrust. Definitely no way. – ZuOverture Jan 17 '18 at 10:59
• @Mołot Very good point. Satellite DXers are very well aware of the danger from white dishes, melted LNB's are not uncommon: nzdtv.com/forum/showthread.php?4944-Melted-LNB . And that's just around a m², not million times more. – Radovan Garabík Jan 17 '18 at 13:21

Use the moon

Why build an giant satellite to orbit the earth at all? Extend from the moon itself! The same side is always facing the earth. Just make static structures that extend from the edges of the moon (as it is seen from Earth)

This reduces the amount of materials you need to transfer up into space to build such a structure, plus you can mine the moon itself for materials.

As the moon casts a bigger shadow on earth, global temperatures will drop on average. This causes minimal distruption to existing systems. The shadow is not being cast over the same spot on earth all the time. This merely extends the area covered during solar eclipses that are alraedy happening.

• I'm not sure you read the question... OP's looking to block 1% of solar energy. Moving the Moon to a point where it would block only 1% of solar energy would be far more difficult than building a satellite. – Azuaron Jan 18 '18 at 20:33
• Moon already already exists as satellite that casts a shadow on the earth. No need to move it from its existing orbit at all. Only need to extend it. – NatraxYN Jan 18 '18 at 20:50
• You would need to lock the orbit so that it was always between the Earth and the Sun, and you would need to push it out farther from the Earth so that it blocked only 1% of the solar energy. – Azuaron Jan 19 '18 at 15:21
• No need to lock the orbit. Extending a sail light structure from the moon would simply extend the duration and coverage of existing solar eclipses. So that on average, over an year, less sunlight reaches the earth. – NatraxYN Jan 19 '18 at 20:03
• @NatraxYN The moon passes between (but not necessarily blocking) the sun and earth every 28 days and at most blocks the sun for 100 minutes. Even if you could expand the moon to total eclipse every lunar month, you would only block 1/4 of one percent of the time. You'd need to expand the lunar radius x4 at least. This is not an insignificant project. – Coomie Oct 25 '18 at 2:49

Although you've edited to remove the 3rd question, "Is there anything else I haven't though of?", I feel duty-bound to suggest that, yes, there is. It's the moon. L1 is at about 1.5 million km, and the moon's orbital radius is about 380,000 km. In the long run the moon is going to perturb anything at L1, so your "stabilization" mechanism will need to be a lot more robust than you think.