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Q&A

How would lighthouses work in space?

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In my world spaceships are guided by lighthouses floating in space instead of electronic navigational systems. A lighthouse in space has the shape of a huge sphere emitting intense red light. Like the ordinary lighthouse a space lighthouse mark dangerous things, such as : black holes, supernovas "¦ They also mark space stations, docking bays, fuel stations and other things.

My questions are:

  1. Is this system viable?

  2. Is this scientifically possible?

  3. How much surface should a sphere (lighthouse) have, in order to emit enough red light?

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This post was sourced from https://worldbuilding.stackexchange.com/q/37842. It is licensed under CC BY-SA 3.0.

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1 answer

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1 - Is this system viable ?

I think the buck stops here, really. No, it's not viable.

What you are proposing is perhaps "possible" in some limited sense (but as already pointed out by others, you are up against some very fierce competition in terms of light sources). It isn't however viable.

This isn't for reasons of the amount of light you would need to put out. Throw sufficient amounts of handwavium at that, and you could explain it away, or just lampshade (pun only half intended) the whole problem.

The reason is spelled orbital mechanics combined with the finite speed of both light and space travel.

If we assume space travel by Newtonian or relativistic mechanics as currently understood (which would be an implied requirement, since you are asking for answers based in known science), then we are limited by the laws of orbital mechanics. Basically, spacecraft coast for all but a tiny fraction of their travel time. Practical spacecraft have very limited delta-v budget (ability to alter their velocity and vector) due to the tyranny of the rocket equation. In order to reduce the delta-v required for a particular position change after some amount of time, you need to increase the time between the maneuver and the time at which the position change needs to be completed. The earlier you can perform a maneuver, the less fuel you need to get the result you want. Compare the fact that in order to land, from an orbital velocity of on the order of 7-8 km/s, the space shuttle only needed to reduce its velocity by about 100 m/s under power before gravity and drag did the rest.

Objects in space generally refuse to stay put. If you take your fancy spacecraft into a low Earth orbit, park and lock the doors when you go for an EVA to grab a lunch, and look for the spacecraft about 45 minutes later, you will find it on the other side of the world. (Thankfully, you will also be on the other side of the world, which somewhat reduces the practical impact of this.) Geosynchronous orbits don't help, because you are still moving at orbital velocities; you just happen to have an orbital velocity that matches the angular velocity of the rotation of the planet beneath you (the point directly to your nadir). Lagrangian points don't help either, because as the relevant objects move those points move as well, which means you (in this case, the light source) is moving with them.

Let's say you can somehow engineer a light source that is bright enough to be possible to make out at the distance between Earth and Pluto when the two are at opposition (points farthest from each other), and we put it roughly where Pluto is in our universe. Pluto's orbit reaches out to about 49 AU from the Sun, and the Earth orbits at about 1 AU from the Sun. So we want something that is visible at 50 AU. (Pluto's orbit is in a different plane than the rest of the solar system, but for an example, this works anyway.) 50 AU really isn't far at all in terms of interstellar travel, which I take it you are concerned with because of the dangerous objects you mention in your question, but it works nicely once you approach a solar system.

Now, let's say your ships travel at 1% of the speed of light, or 3,000 km/s, relative to the light source. (This is far, far faster than anything we can accomplish with chemical rockets, but it is still somewhat within the realm of possibility with science and technology as we know them.) 50 AU is about 7500 Gm, so this distance will take your spacecraft about 2.5 million seconds to travel. (Back-of-the-envelope plausibility check: speed of light time delay from the Sun to Pluto, on the order of 7 hours. Speed of travel, 1/100 of the speed of light. Expected travel time, 700 hours. 700 hours is 2.52 million seconds. Check.)

Pluto's orbital speed averages about 4.67 km/s. In those 2.5 million seconds that the light needs to reach our intrepid spacecraft 50 AU away, Pluto (or our light source) moves almost 11.7 million km along its orbit. This is before the crew of the spacecraft even sees the light.

Spacecraft generally travel along elliptical transfer orbits selected to get it to some particular point in space at some particular time (usually at a time when an object of interest is going to, in its orbit, intersect that point, or a point near that one) within some given constraints (time, delta-v, payload mass, ...). Any time a spacecraft is going anywhere, it is doing so by assuming a transfer orbit (very often a Hohmann transfer orbit, which in many cases is the lowest-energy way we know of to get from point A to point B in space; another alternative sometimes considered is a bi-elliptic transfer orbit). In Hohmann transfer orbits, you basically trade time for delta-v; the more delta-v you can afford, the more direct a route you can take and the quicker you can get to your destination. Since as we saw above that we want to minimize the delta-v expenditure in order to reduce our spacecraft's mass ratio, this means that we need more time to get to where we are going than if we were travelling in a straight line.

So not only has the light source already moved over ten million km along its orbit by the time the light reaches the spacecraft at 50 AU away from the light source's original position, but you also have to consider how long the spacecraft will need to get to where the light source (or point of interest) is now. (And that's "now" in which reference frame?) And by the time you get there, where is the point of interest going to be then? It's a variation of the classic math trick question of halving something repeatedly:

$$ x + \frac{x}{2} + \frac{x}{4} + \cdots + \frac{x}{n} $$

This is exactly the type of problem humans are horrible at solving, and computers are excellent at solving.

Which is why anyone who wants something even remotely like this would be far more likely to use radio beacons and electronic computers than navigation by eyesight and estimation.

The type of light (or even EM) source you are using has no effect on this, because the problem is related not to the type of EM source but rather to the relative speeds of the objects involved, and orbital mechanics.

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