3
$\begingroup$

Think of a giant hemispherical shell, constructed of spaceships, sitting just far enough from Earth to be too far for Earth-based defences and close enough to effectively cast a shadow on the entire daylight side of the Earth.

Assume the ships are constructable with current physics; they can be mostly a simply blocking material (wing, or sail) with a small habitable/engine zone (think giant self-propelled tarps). They don't have to be strong, just sufficiently space-worthy to stay intact outside the range of Earth's defences (more or less at current capability) and stay between the Sun and the Earth.

Since these three variables are dependant, please include:

  • How big (area) would/could the ships be?
  • How many would you need to construct the shield?
  • How far from Earth would they be?
$\endgroup$

This question asks for hard science. All answers to this question should be backed up by equations, empirical evidence, scientific papers, other citations, etc. Answers that do not satisfy this requirement might be removed. See the tag description for more information.

  • $\begingroup$ The problem is that anything this flimsy isn't going to be out of range of Earth's current defences at any practically achievable point between the Sun and Earth. One nuke of decent size could wipe out a whole lot of them. This question should add the proviso that any defence must be undertaken before Earth turns into an ice-ball, i.e. within a few days of the sunshades being deployed. $\endgroup$ – Monty Wild Mar 11 '16 at 1:43
  • $\begingroup$ I could figure out the math, but the short answer is a whole lot. $\endgroup$ – Caleb Hines Mar 11 '16 at 1:56
  • $\begingroup$ I don't think a nuclear explosive would "take out a bunch of them". With no shock wave, the damage radius will be small. And the area of the shade is huge. So a more accurate assessment is that a bumb would make a pinhole. $\endgroup$ – JDługosz Mar 11 '16 at 2:38
  • $\begingroup$ I +1 answers below; but it depend on if you want total darkness, or just "a shadow" as written. Sizes could be much smaller if you just want to freeze Earth, and don't mind partial lighting: For example, our North and South poles (when lit) have daylight but do not receive enough solar heating to be habitable (i.e. with farms, ranches, surface water, working outside without massive protection against frostbite, etc). Our avg temp is 16C, drop to 0C and we become Antarctica, oceans, crops, livestock, die. Light enough to see, no warmth to survive. Google Glaciation, Snowball Earth. $\endgroup$ – Amadeus Sep 26 '17 at 11:12
8
$\begingroup$

Location

You envision these ships "sitting" in space, but nothing truly sits in space. Everything is constantly moving along curved paths through gravitational fields. In order to not consume obscene amounts of fuel, these ships need to find a location that will let them naturally move in such a way that they are always between the Sun and the Earth. Fortunately for your attack fleet, such a point exists: it's called the Sun-Earth Lagrange point 1, or SEL-1. Objects placed at this point will require very little station-keeping to stay balanced between the Earth and the Sun. This point is outside of the Moon's orbit, approximately 1% of the distance from the Earth to the Sun. The wiki says that we currently have five active probes in the vicinity of SEL-1, including the solar observatory SOHO, and the climate observatory, DISCOVR, which took the following photograph:

The view from SEL-1 (via DISCOVR)

Size of Shield

Before we can determine the number of ships, we need to know the total size of the shield they need to form. Imagine a very long cone. At the large end is a giant ball for the Sun. Near the small end is a smaller marble for the Earth. Even though there is a huge size discrepancy between the two, the length of the cone (the distance from the earth to the sun) is so much larger than the Sun, that the cone is still relatively narrow (hopefully that picture makes sense).

In order to block all the sunlight, the shield needs to be a disc wide enough to fill the cone at its location. As a reminder, this disc is located 1% of the way from the Earth to the Sun. This means that the disc of the shield needs to be larger than the Earth, but not by very much (because the cone is narrow, and the shield is relatively close to earth). In fact, rather than bother with the exact trigonometry, I'll just do a rough estimate and assume the shield has approximately the same radius as the Earth.

This leads to an area of:

Ashield = π * Rearth2 ~ 128,000,000 km2

Size and Number of Ships

You want a ship "constructable with current physics" which I assume means something roughly on par with today's engineering capabilities.

Solar sail technology is still in its infancy, but what was originally planned to be the world's largest solar sail (granted, not a hard bar to surpass), was the Sunjammer. Interestingly, before it got cancelled, it was meant to launch in 2015 with DISCOVR towards SEL-1. It was to have an area of 1200 m2 (which is only 0.0012 km2).

The ISS is the largest man-made object in space, and its solar panels have an area of about twice that, at 2500 m2 (0.0025 km2).

Let's stretch the limits of what is currently possible, and assume for the sake of simplicity that, with some effort, we are able scale this up by about 3 orders of magnitude, and build and deploy a large solar sail with an area of 1 km2 (for comparison, the Golden Gate bridge is 2.7 km long). Then, given the required area we computed above, we would only need 128 million of them.

Earth's Defenses

Given enough lead time, we can get an object pretty much anywhere in the solar system... or in some cases, even outside of it. The length of time that it takes to get to a destination is very much determined by the trajectory taken and the type of propulsion used. DISCOVR took 100 days to reach SEL-1, but it was no doubt taking a fuel-efficient route. If we took a more energetic route, we could probably get an object there within a week or so (just spit-balling it, based on a 3-day lunar trajectory used for Apollo, and a need for urgency)

However, we do not currently have the weaponry to take out a literally-planet-sized flotilla of ships. Or do we? According to this list, it looks like if we pooled our international resources we could muster about 4,000 active nuclear weapons. Mark makes an excellent point about the damage of these weapons being minimal in space, since there is no gaseous medium to transmit a shock wave. Even if we hit 4,000 of the 128 million sails, that's less than 1 percent of 1 percent of the total fleet. But... maybe we might get lucky, and initiate a Kessler Syndrome-style chain reaction. After all, these sails are presumably flying in a rather tight formation in order to block the sun. Taking out one might be enough to take out its neighbors, which each take out their neighbors, and so on. Whether this actually happens would depend on a number of factors, including the toughness of the sails, the tightness of the formation they are flying, and the force of the nuclear weapons detonated.

Even still, it's not clear how helpful this would be, because you've still got a massive debris field sitting between Earth and the sun. Without active station keeping, the debris will eventually start to dissipate, but this can be expected to take on the order of at least several months. Probably longer.

So in short, our defenses would consist of building enough rockets to launch every single nuke on the planet to SEL-1, waiting at least a week for them to get there, then hoping that they can initiate a chain reaction that leads to a debris field that will eventually disperse after several months. I hope someone can come up with a better defense plan than that, because I don't think that would be sufficient.

On the bright side, if each of these sun-blocking ships only cost the projected amount of Sunjammer (\$20 million to build + \$10 mil. to launch) -- despite having 3 orders of magnitude more surface area -- then the total cost of the shield would be \$30 million per ship * 128 million ships ~ $3.8 quadrillion. That's about 200 times the US national debt, so we're probably safe for now.

$\endgroup$
  • 1
    $\begingroup$ The Planetary Society flew a solar sail in June. It was 32 square meters. $\endgroup$ – JDługosz Mar 11 '16 at 7:17
  • 1
    $\begingroup$ Nuclear weapons can be "enhanced" as detailed at the Atomic Rockets website. Using the nuclear weapons to drive "shotgun" pellets at 100km/s is perhaps the simplest of the various schemes.....projectrho.com/public_html/rocket/spacegunconvent.php $\endgroup$ – Thucydides Mar 12 '16 at 2:08
  • $\begingroup$ I think you need to address station keeping more. With these giant solat sails deployed the ships would be blown away from the sun $\endgroup$ – Ewan Mar 12 '16 at 11:25
  • 1
    $\begingroup$ Potentially, but not necessarily, depending on how much mass they have, or whether they can find a way to balance the radiation pressure against the forces of gravity (perhaps by being slightly sunward of SEL-1). But I think the whole idea that you need a planet-sized fleet in the first place makes any other detailed considerations seem rather ludicrous. $\endgroup$ – Caleb Hines Mar 12 '16 at 15:00
4
$\begingroup$

How big, and where to put it: the questions are interrelated. You should draw it out geometrically.

Get a roll of kraft paper, thin paper sold for masking and catching paint spills, or whatever. Or use chalk on a sidewalk! You can certainly get a large scale that way.

Draw the sun, earth, and distance between them to proper scale.

Now you need to model the umbra and penumbra shadows. Because the sun is larger than the earth, the shadow is complicated. Take some string held tight to form lines. Draw (lay down string, that is) a line tangent to both circles, on each side. Now now you can see that anything that spans between those lines will block the entire sun from the entire earth.

It's cone shaped, so the closer it is to the Earth's end of it, the smaller it needs to be. If it's right up against the Earth, it's the size of Earth. On the other extreme, if it's right up against the sun it must be the size of the sun.

Now wherever you choose, how do you keep it there? To let it stay put without constant thrust, look at the inner Langrange point. The radiation pressure will blow it outward, so consider that another outward force and move a bit inward from the Lagrange point so the pull of gravity is stronger toward the sun. If you want to be really accurate you can figure out the force involved and solve for the correct distance so the sum of forces in the direction of the sun, and a radial velocity to keep it from falling, match the 1-year period of the Earth.

Plot that distance on your graph. Measure the width. Actually, you might need a bit of a bowl shape rather than a flat disk, for practical reasons in building it and balancing it there.

Take a photo of your drawing and use it inside the cover of the book you write where this features.

$\endgroup$
2
$\begingroup$

How big? Slightly larger than the Earth.

How many? One

How far? A bit outside the Moon's orbit.

If you want to cast the Earth into eternal darkness, the easiest way to do so is with a solar sail. Between size, reflectivity, and flimsiness, it's essentially immune to anything that Earth-based defenses can do to it: in space, a nuke's blast effect is limited to the gas produced by the casing of the bomb itself, thermal effects are mostly reflected by the sail's material, and radiation just passes through.

Your main concern won't be the Earth, but the Moon: you'll need to maneuver parts of the sail to compensate for gravity changes as the Moon goes by, and perform station-keeping to keep the Earth fully shadowed over the course of each month.

$\endgroup$
  • $\begingroup$ How do you figure "a bit outside the Moon's orbit"? $\endgroup$ – JDługosz Mar 11 '16 at 3:00
  • $\begingroup$ Makes station-keeping easier if you only have gravity coming from one side. $\endgroup$ – Mark Mar 11 '16 at 3:02
  • $\begingroup$ He wants something "constructable with current physics". Assuming that includes engineering techniques, I very much doubt we are currently anywhere near capable of building a single Earth-sized structure! (but you are correct that this is about how large the shield would have to be. $\endgroup$ – Caleb Hines Mar 11 '16 at 3:09
  • 1
    $\begingroup$ With gravity from one side, wouldn't it just drop? How does that allow you to keep station? $\endgroup$ – JDługosz Mar 11 '16 at 3:11
  • 1
    $\begingroup$ But "just outside moon's orbit" means well within Earth's Hill Sphere. It will fall to Earth without some kind of thrust. How are you keeping it up? $\endgroup$ – JDługosz Mar 11 '16 at 3:22
1
$\begingroup$

A highly motivated alien civilization could block the sun's rays more efficiently by using a large cloud of gas located at the Lagrange point.

A sulfur(?) cloud capable of reflecting only 20% ​​of sunlight would reduce the Earth's average temperature by 2 ° C per year [Universe SandBox].

In 10 years the temperature would drop from 15ºC to -20ºC. With most of the planet under ice, more solar radiation would be reflected by the white surface, accelerating cooling [Universe Sandbox].

20 years after the beginning of the siege all ecosystems of the surface would be deactivated, leaving only viable life in the oceans, in the water below the icing.

Since nuclear attack on a cloud is useless, Earth's resources would be summed up to survival, perhaps creating heated geodetic habitats for a very small part of the population. More than seven billion people would die of hunger, cold or mayhem.

It would be ironic to see humanity succumb to a giant cosmic fart..

$\endgroup$
  • $\begingroup$ how can a cloud of gas be held in place in the vacuum of space? $\endgroup$ – L.Dutch Aug 28 '18 at 18:58
0
$\begingroup$

This is easy to determine with some math and physics related calculations.

Area Covered By The Ships

Radius of Earth: 6371 km

Surface area of Earth: $4\pi r^2$ = 40589641 $km^2$

Daylight side surface area = half of total surface area = 255032235 $km^2$

This is the area you want to shadow.

Now, the area of equator-slice (if you cut Earth at the equator, the surface area of that slice) is: $\pi r^2$ = 127516117 $km^2$

This is the maximum size of the plate you can ever need for your purpose. Practically, you can use a slightly smaller plate, considering the dispersion angle of sunlight (I haven't researched that topic as it is too complex for a person like me).

The area covered by the ships would be 127516117 $km^2$ and these ships would block out 127516117 $km^2$ area on Earth (the daylight side).

enter image description here

How Many Ships Would Be Needed?

Depends on what is the size of your space ships. For example, the area covered by International Space Station is ~2500 $m^2$

If you use spaceships as large as that, you will require some 1.5 billion or so spaceships. Sigh.

How Far From Earth Would They Be?

The area of spaceships I calculated is enough to block out all sunglight if you place them as a plate on the surface of Earth. Which means that wherever you place them, they would work fine. For the sake of your question, I would suggest them to be placed between Earth and Moon. These would have to be almost geostatic, accounting for Earth's tilt.

Considering that Moon is approx 384000 km from Earth, your spaceships would be approx 348000 km from Earth.

enter image description here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.