# Could a space colony 1g from the sun work?

Let me break down my thoughts and I have no science background so let me know if this could even be possible.

1. Have the colony on an object like a big asteroid or a similar man made construct.

2. Have the object orbit around the sun at a distance where the sun's gravity is 1g

3. Have the object tidally locked into its orbit around the sun

4. This way we would get the 1g gravity from the sun on the opposite side of the asteroid

5. Solar power should give us most of our power supply

6. Reflecting a small portion of the light could give plants enough to grow and simulate night and day

So the questions are:

1. Could we survive at the distance of 1g from the sun?

2. And if so, how thick would the object need to be in order to protect us from the heat and any radiation from the sun at that distances?

3. Could we keep an object orbiting the sun at that distance for a long period of time?

4. Any other input you think would be nice

Thanks and I hope I explained my crazy thoughts thoroughly enough.

• Point 2 is an Oxymoron. Orbits are Free-Fall trajectories, i.e. They must feel 0g. The Moon is constantly falling, it just keeps missing the Earth. – Aron May 15 at 8:15
• Hi Matthew, you might consider moving this to Worldbuilding.se - there's a community who likes to use speculative science with hard science and can help you work through options for this. – Mikey May 15 at 15:09
• Yeah...I think what the asker is actually imagining is a fixed (not orbiting) platform at that point, which would have to be constantly thrusting to keep from falling into the sun. I think that would produce the result he's looking for. – Chris B. Behrens May 15 at 16:15
• Yeah, "1g from the sun" is not really a thing. Sure, there is a distance from the sun at which its gravitational attraction has that strength (3.7 million km or 12 light seconds, as per google.com/search?q=sqrt+(mass+of+sun+*+gravitational+constant+%2F+(9.81+meters+per+(second+squared))) ). But the only way to "feel" that gravity would be to stand on a platform somehow fixed relative to the sun's surface; otherwise you will simply be in free fall. – Christoph Burschka May 16 at 14:13

Interesting but no, it wouldn't work for the same reason that astronauts in the International Space Station, other space stations, or orbiting shuttles or capsules do not "feel" gravity with respect to their station or capsule.

When you are inside an object which is in orbit, you are in orbit too! The Earth pulls on the station with nearly 1 g and it pulls on you the same amount, but you are both orbiting so you move in the same circles. This gives you the experience of weightlessness within the object, be it an asteroid, shuttle, capsule, or space station.

Watch ISS Expedition 22 Commander Jeffry Williams in the video Demonstration of Acceleration Inside the International Space Station During a Reboost. As soon as the space station starts to accelerate due to a reboost burn, you can see the camera appear to move backwards (towards us).

What's really happening is that the camera is simply remaining in its original orbit and the ISS is accelerating forwards (in the same direction we are looking). Both feel nearly 1 g downwards from Earth, and that's what keeps both in a circular orbit rather than shooting straight out into space.

• You could build a Dyson sphere at the correct distance from the sun, but that would be rather uncomfortably close to the sun (2.5 million km if my maths is correct) – hdhondt May 14 at 5:58
• +1. In short: When you're in orbit, the centrifugal acceleration cancels out gravity. You have to stop orbiting to feel the gravitational pull. – TooTea May 14 at 7:13
• How about a statite, maybe using an electric solar sail? – JollyJoker May 14 at 8:16
• At exactly the time he said "now what happens if I let go of myself?" I thought that exact same thing hah! – Magic Octopus Urn May 14 at 11:50
• @JdeBP: Centrifugal force only exists in non-inertial reference frames, such as a reference frame that is accelerating in a circular path around the earth. In that reference frame, yes, centrifugal force does cancel gravity, leaving a net zero force relative to that moving reference frame. – Peter Cordes May 14 at 15:55
• Have the object tidally locked into its orbit around the sun

• This way we would get the 1g gravity from the sun on the opposite side of the asteroid

Interesting idea, but you missed something in your math.

You'd only get the tidal difference between the sun's gravity at the centre of mass vs. the sun's gravity 1 object-radius farther away. This varies with $$1/R^2 - 1/(R+r)^2$$, and is very small unless the radius of your object is an appreciable fraction of the radius of your orbit. See https://en.wikipedia.org/wiki/Tidal_force for more, including a picture if you're having trouble visualizing.

The sun doesn't produce 1g gravity gradients over any reasonably small distance. (Look for a much denser more compact object like a white dwarf star for that: an object squished by its own gravity into an electron-degenerate state, somewhat short of collapsing further into a neutron star).

And if you did have a gravity gradient strong enough, it would tear a natural asteroid apart! 1g "outward" force on both sides of the object is huge; you'd be way way inside the Roche limit, and would thus need a very strong station.

Or maybe connect your station to a counterweight (with a long tether) that orbits closer to the dwarf star, so the only large part is the part in gravity. This assembly of 2 masses + a tether would be tidally locked, held in place by the 1g difference between the radii of their positions.

This makes it possible to get a pretty large size so you can get 1g without too steep a gradient, i.e. not too close to the white dwarf. Still not practical for the Sun, though.

Maybe you'd also build a couple intermediate modules of your station so you could take advantage of lower gravities, and one in the middle for zero g.

Like a space elevator, the tether has to support its own weight. (Where weight is calculated as the integral of "felt" acceleration times delta-mass over the length of the tether.)

• Well written answer, +1! – uhoh May 14 at 12:09
• In the tether solution you cold have a station on either or both ends. The gravity would be close to identical, (although opposite direction). – Taemyr May 14 at 14:19
• @Taemyr: yeah that may make more sense than towing dead weight into place. I was thinking that you wouldn't want 2 habitats really far away on opposite ends of your tether, so you'd just use the side farther from the star. But if moving dead weight into position is costly, then you'd want to make both sides count. Or maybe one side for garden / storage / whatever. – Peter Cordes May 15 at 7:40
• Neat but rather huge failure surface and catastrophic failure mode... :-P – R.. May 15 at 13:12
• @R.. What do you mean? If the tether breaks, your station ends up in freefall in a slightly elliptical orbit. (Assuming it was in a circular orbit before). Or do you mean that if you screw up anything when you're close enough to a white dwarf for a 1g gradient over reasonable distance, you're hosed :P – Peter Cordes May 15 at 13:15

FWIW, since no one mentioned it, you could construct a shell around the Sun on which a stationary observer would experience a downward gravitational acceleration of 1g. This shell would have a radius of about five times that of the Sun (5.28 Rsun), and be well within the orbit of Mercury (0.0669 Lmerc). Of course, I am not saying that this would be a good idea, nor suggesting what materials could survive at this distance. Such a discussion would be more appropriate for the SF StackExchange.

• Note that in addition to being irradiated by the sun, the materials of the shell would need to withstand some humongous compressive forces in order to keep the shell from buckling under its own weight. – Henning Makholm May 14 at 22:02
• well, it was mentioned in a comment – JCRM May 14 at 22:09
• Probably better for Worldbuilding.SE than Science Fiction.SE. – CJ Dennis May 15 at 1:57
• Or have a platform supported by a column on the sun! The column would only need to be 5.28 Sun radii tall and withstand the heat of the Sun. – dotancohen May 15 at 10:00
• It might be illegal to write this answer without using the words "Dyson sphere". :) – David Richerby May 16 at 12:54

I'll keep this reasonably short and generic as you don't have a science background. If you want artificial gravity, it isn't based on distance, but on rotation. Then you just need to worry about the size of your habitat and the RPM. SpinCalc is a handy little calculator that can help you figure out what size of habitat, angular velocity, and tangential velocity you need to get whatever gravity you desire. The last numbers I saw for protection against any and all forms of radiation was shielding about six feet thick. Gerard O'Neill's 1974 Physics Today article goes into a fair amount of technical detail on space habitats in free space.

Unfortunately, that won't work. The 1 gee of force downwards would be canceled out by the 1 g of force upwards that you experience as centripetal force by being in orbit. Even if you let the station/colony/thing free-fall (which it is in orbit), you won't feel gravity until you get close enough the difference in gravity between your station and you is severe enough that you'll be able to feel it. I'm guessing that being that close wouldn't be a great idea. When you're free falling, which is what orbit is, you and the colony are accelerating at equal rates. Because of this, you don't feel any acceleration relative to the colony. If you've ever been in a plane, you feel this when there's turbulence. If the plane makes a big enough drop, you can feel yourself lose contact with the seat and "float" up a little bit. This usually doesn't last long (unless the plane is crashing, in which case you have better things to be doing than reading this). You can sometimes feel this in cars going down hills as well. Anyway, a better solution would be to spin the station. However, if it's spinning, then you need the station to be able to withstand those forces, and the size of station needed increases dramatically. Artificial gravity in space is difficult!

Like other people have said, an orbit will not induce local gravity from the sun, because an orbit is in freefall by definition. What you want is a statite, which uses solar radiation pressure to maintain a fixed distance from the sun. This relies on the sail's lightness number, which is the ratio of its maximum acceleration to the sun's gravity, and which normally does not vary with distance. The maximum available acceleration with current technology is 0.26, which is not very good. The theoretical maximum, with some kind of lattice sail, is 22, which is enough. However, that's only the beginning of the troubles you might have.

To get 1g of gravity from the sun, you need to be at about 5.29 solar radii, or 3.68 million kilometers, or 0.025 AUs. At that distance, you receive about 1600 times as much sunlight as you do on Earth. The article on solar sails says that carefully designed sails can maintain safe operating temperatures down to 0.25 AUs.

If you can get a sail working at that altitude, you'd have to protect the station from the sun's light. Using information from designs of probes like the Parker Space Probe, we could probably keep the temperature of both the sail and the station down by having the sail's incident surface be sharply angled from the sun's light, like a cone, and by keeping the station in the sail's shadow, like a broom balanced on someone's hand. Note that angling the surface reduces how much thrust you can get from it, and the total lightness number of the sail and the station together needs to be greater than 1 even in the worst contingency. Don't forget that the lightness number (which is normally independent of distance) is actually reduced while close to the sun, because it's no longer a point emitter; it takes up about 22° of the sky, so your solar cone will have to be large enough to cover that much from the station's perspective.

Complicating matters is the fact that the station will be in the sun's corona, which is much more active in terms of radiation than the solar wind is. It's also well within the distance that the sun forms coronal loops of magnetic and plasma energy, which are the structures that give rise to coronal mass ejections. Those ejections can cause blackouts and wreak havoc on Earth, millions of kilometers away; who knows what will happen to the station if a coronal loop forms on top of it? To protect against such a thing, you need radiation shielding, which is invariably heavy, and goes against the statite being as light as possible.

Then there's the problem of maintenance. How do you repair the solar sail if it tears? You certainly can't be on the sun side of it, or you'll explode like a watermelon in an incinerator. The hole in the screen will let massive amounts of light through, which will sear anything in their rays, and even if you're patching it from the shady side, the reflections from the piece of solar sail you're attaching will burn out the eyes of any astronaut unlucky enough to be working there. Also, if whatever supports hold the station in place relative to the sail fail, you don't have the luxury of it being in orbit; it will fall like an actual rock on actual Earth. If you're lucky and find out about it when it happens, you'll have just enough time for last rites before you fall through the sail and burn.

All in all, there are much easier ways of generating artificial gravity aboard space stations.

• I thought of this when I wrote my response above about a static shell and got similar numbers. However I am unsure if the solar wind, coronal mass ejections and the like might exert a significant force. – Kieran Mullen May 17 at 3:41