28

How difficult would it be? Unfortunately, your timeline is too tight for any of the easy options to work. The mass in the asteroid belt is highly concentrated. Between them, Ceres (~30%), Vesta (~10%), Pallas (~8%), and Hygiea (~5%) make up about half the mass of the asteroid belt. Combining these into a single body is (relatively) easy, but slow and ...


20

Just shrink the moon by 6%, and there will never be another total solar eclipse, but new moons will still happen as normal. The moon will be too small to fully cover the sun, so there will be partial and annular eclipses, but no total ones. If you still want the occasional total eclipse, make the moon slightly bigger again, so that eclipses are total if and ...


13

Short answer, no. The asteroid belt is incredibly sparse, where your chance of seeing another asteroid from the surface of one with the naked eye is pretty low. There's no simple way to overcome the momentum of every asteroid. They would all require manual adjustments to their orbits, many of them requiring multiple burns to reposition them and collide ...


13

Not easily. An orbiting object has the feature of being in free fall, which means that objects on it are not subject to gravitational pull from the object that is being orbited, no more than a person inside an orbiting space station feel no gravitational pull from the Earth. There is however such a thing as tidal forces. Since all parts of a planet orbits ...


13

Could a star exist with this cloud similar to earth atmosphere around it where a person could breathe without much effort? Nope. The solar wind and radiation pressure and a combination of planetary and stellar gravity would either blow away or hoover up all the gas cloud in relatively short order. That's why there's a fairly hard vacuum between planets in ...


10

In terms of planet formation, there are esssentially three ways to get a moon. It can form with the planet, by the same kind of particle accretion process, but with enough relative velocity not to get incorporated (this is how the larger "true" moons of the gas giants were formed -- Ganymede, Titan, etc.). It can form by capture, where an already formed ...


10

We have plenty of examples where stars have been hidden by nebulae - and not just newborn stars. Typically, the gas and dust comes from mass loss from one of the stars in the system. Examples include LL Pegasi, a binary system containing a carbon star that is sloughing off large amounts of dust as it nears the end of its life. This makes the star visible ...


10

I do not believe that any form of EMP would prevent electrical technology, as the EMP would induce currents in metals which might attract attention and at some point metal screens would be discovered. But there is one possible way that does not use EMP. If all of the common metals accessible by smelting such as iron, copper, zinc, lead, nickel etc were ...


9

Keep sun and moon sizes, make the orbit of the Earth around the sun more excentric and with a smaller mean distance. The new moon is unnafected, but full solar eclipses will only happen if the eclipse happens together with the Earth's apoapsis, or close to it. And that will only happen during a few specific days of the year. Any solar eclipse far from the ...


9

It already happens. Look at the schematic drawing below, where you see a planet and its Sun (not to scale), with two different places on the planet: one facing the Sun, the other on the opposite side from it. Gravitational pull can be schematized with a vector, and the resulting pull is the result of the summation of all the pulls in a given spot. In the ...


9

Off the top of my head, I feel like the easiest way to handle this is with a binary star system. Put your habitable planet in close orbit around a dwarf star. Planets close enough to a red dwarf to be habitable at all would almost certainly be tidally locked, which solves for the 'only one hemisphere is normally habitable' criteria in your question. Then ...


8

Interesting question, there should be a calculator around to check a scenario like that, only I could find none. There is several other calculators and tables which can help answering this question: Spectral type characteristics Orbital period of a planet Calculation of Habitable Zones From a quick estimate, this (1-day orbit) is not possible with a ...


8

Makre your own. The asteroid belt has about 3x1021kg of mass. 10% of the asteroid belt is made up of metallic-stype asteroids. Quite how much actual metal there is in there is hard to say, but lets go for 50%. So that's 1.5x1020kg. You need about .3% carbon in structural steel. About 75% of the belt are carbonaceous-type asteroids. These have a bulk carbon ...


7

Probably not all in one go, but along the way why not have some fun, save your job and give the Emperor something to smile about. Say you have a sixteen space tugs to work for you - first thing first, order 12 of them to pop out to the Kuiper belt and drag in a dozen matching 100 Km diameter KBO's (essentially comets), whilst this is happening, get the ...


7

Assumption: There's enough mass in the target asteroid belt to make a planet the desired size. As amazingly powerful as gravity is, it's quite a bit weaker than kinetic energy unless there's a whomping lot of mass nearby. Consider, for example, the impact of a baseball on a bat. Eventually it falls back to Earth — but for a moment, it's free. ...


7

Why not make the moon's orbit precess? If the precession chases the sun (from the perspective of earth) while staying a bit off from it, you could keep new moons as common as always and eliminate solar eclipses altogether. If you made the presession not quite at the same rate as the sun, you could also cause large periods without a single solar eclipse, ...


7

The simple answer is an infinite amount. Since you defined moon as "just a natural satellite" anything natural and in orbit counts. Be that a double planets partner (There are many definitions, yet I prefer the one with calculating the barycenter and if it is outside the bigger planet, it's a double planet. This makes Pluto and Charon double planets and Luna ...


7

Use self-replicating machines to disassemble Mercury, which is made of 30% silicates and 70% metal, so most of it could be used to build the panels that make up the swarm, and the mass is about right to build a swarm at the same distance that Mercury used to orbit. This paper on the Fermi paradox has some rough calculations about the time needed to do this ...


7

No to the star, Yes to the travel A star cannot exist in that type of atmosphere for many reasons. Not the least of which is heat. Temperatures range into the millions of degrees, which destroys any chance of a stable atmosphere. Furthermore, any particles in the air would result in continual dust explosions. This of course does not mention issues with ...


7

In the mare magnum of the Internet, one can find anything, literally anything. Well, it looks like someone created also a habitable zone calculator*! The calculator takes as input the luminosity of the star, which you don't give. However you give the estimated masses, thus we can estimate the luminosity of the star using the mass-luminosity relation $L\...


6

There is nothing that a meteorite could be made of that would do that. But it's not necessary. You see, all it really takes is for the cold to last long enough that the ice has time to significantly spread. Once that happens with more and more of the world coloured white, more sunlight will be reflected back into space. This increase in "albedo" will ...


6

The rock that did them dinos only managed to block sunlight for way less time than that: Recent computer simulations and atmospheric models indicate that within a few weeks to months temperatures and light levels would have begun to rebounded [sic] due to the release of heat stored in the oceans and the coagulation and fall of the dust and soot. The major ...


6

Yes, the habitable zone would be extended. And widened as well. This is because planets would gain some heat from infra-red radiation of the sphere and it would be uniform inside the sphere instead of centered on the sun. How big the difference is depends on the inner surface temperature of the sphere, which in turn depends on its ability to radiate heat ...


6

Atmosphere loss As you've suggested in your question, once a Sun-like star leaves the main sequence, it begins losing mass through a strong stellar wind, a stream of charged particles driven by photons. For a few hundred million years, it's a true red giant, expanding a bit and reaching luminosities of a few thousand solar luminosities. After spending some ...


6

In lieu of any other answers being written, the answer to your planetary configuration is primarily a matter of water. Some basics. The reason why living on the coast is so temperate (warmer nights, cooler days) to living in deserts and inland environments is that it is close to a massive body of water called the Ocean. Water is an ideal thermal mass; that ...


6

Your planet is a rogue planet near the center of an active galaxy. Active galactic nucleii, AKA quasars, produce enough light to create a habitable zone tens to hundreds of light years wide, rather than a measly fraction of an AU. Orbits in that habitable zone can take thousands or millions of years--i.e., timescales that are relevant to ecology, and much ...


6

Not in normal cases, but you might find some on the extremes. Consider a pulsar for example, it might have a habitable zone, but it wouldn't be overly friendly to humans. Although pulsars regularly blast out deadly gamma rays and X-rays, alien planets in orbit around them could theoretically be habitable, a new study finds. The researchers noted the ...


6

TL;DR: maybe. Orbital stability is pretty borderline, and some fairly unlikely circumstances have to arise to produce something that looks like maybe it will fit your needs. Tidal effects and orbital resonances will mess with the figures below, so they're only approximate Lets start with a star the size of the sun, putting the orbit of the planet at 1AU. We ...


6

At 20 AU, Sun-like star will produce 20*20 = 400 times smaller illumination. Based on this data: Lux, illumination will be in the range of 200-300 lux (with clear skies), which is more than on Earth on a "Very dark overcast day", but less than "Sunrise or sunset on a clear day" - more in line with well-lit building interiors.


5

It sounds like you're asking about the habitable zone of a star, and for that we have a relatively good idea of where that is for a star like our Sun- a relatively up to date 0.95-1.67 AU from 2013. Wikipedia has some nice reading on the subject But in case you were looking for a bit more variety in your stars, I direct you to the Hertzsprung-Russell ...


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