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So I have been doing a lot of reading on space elevators/tethers for a SciFi book I am writing and unfortunately there doesn't seem to be a whole lot of info on building one on Mars rather than Earth or the moon. Despite some handwavium for the necessary material science, I imagine it would be quite a bit easier to build one since Mars has around .38 Earth's gravity.

My book would place the elevator on top of Pavonis Mons which lies nearly on top of the equator. This should give me a head start of 14 km and negate most of the tidal effects from Mars rotation. The design would be more along the lines of Halo (elevator car inside tube rather than around cable).

Although I have a long, complicated construction method that supposedly explains the structure is able to continuously transfer fuel and power through hoses that run the length of the elevator along with the 3-6 (undecided) elevator cars that it lifts magnetically like a vertical maglev train. The only thing I haven't really worked out as far as the design is concerned is how to deal with Phobos as it regularly crosses the equator at around only 9,300 km above Mars.

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From what I understand, a cable from Earth would need to be around 35,786 km to reach the appropriate height. Would it be as simple as multiplying by Martian gravity, giving me the simple answer of 13,599 km above Mars?

Also, would there be a better location to place the elevator other than near the equator so that Phobos isn't an issue? This of course would require me to come up with something for the rotational effects.

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    $\begingroup$ Where exactly did you get that 100000 km from? And how much of the material science do you want to handwave? If it is simply indestructible, doesn't bend etc, all you need is to go to geostationary. If there are limits to its strength, you may need counterweights etc $\endgroup$ – Mołot Feb 13 '18 at 9:17
  • $\begingroup$ Reading over what I had read before again I realise I misunderstood something. I guess 100,000 was for using the excess length of the cable as its own counterweight. I will edit the question with corrections. $\endgroup$ – TitaniumTurtle Feb 13 '18 at 11:25
  • $\begingroup$ In Robinson's book, the elevator oscillates to miss Phobos. $\endgroup$ – Anton Sherwood Aug 26 at 22:03
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Depending on the exact design of your space elevator you'll be aiming for a station at or above geostationary (technically Areostationary) orbit, which is 20,428 km above Mars' surface. There must be at least enough cable and possibly counterweighting above areostationary orbit to ensure that the cable remains stable, but in theory if you're willing to put a heavy enough counterweight in place you can cap it off at geostationary orbit.

That's an engineering nightmare though, and you'll almost certainly need to extend the counterweight out further as a heavier counterweight means more stresses and more difficulty in getting material out there in the first place. If you put a construction station in geostationary orbit and then start feeding cable out of both sides to make sure the station stays in place as you build (as proposed by Jerome Pearson) then you need more cable on the outside, but again, if you add a heavier counterweight then you need less cable.

If we assume that it's 1:1 then you end up with a height of ~40000 km.

Phobos orbits inside the sweep of this structure at 9000 km, Deimos within at 23000 km. Sadly there's not much you can do about the latitude of the elevator: It requires the rotation of the planet in order to function (in fact you can show that ascending cars are really stealing orbital momentum from the planet and descending ones are giving it back). That isn't a killer though: You can design the elevator such that it doesn't directly hit either moon (they are very small compared to the area of space you can feasibly build in), but you will have to account for their gravitational effects, which is a lot more complex and will constantly drain energy from the system for corrections. If you're willing to spend enough energy they you could have the entire elevator oscillating back and forth, feeding off the gravitational energy of the two moons to ensure the cable safely weaves it's way between it's dangerous neighbours.

This may be of interest to you. Technically it's a discussion on using contemporary materials to build a martian elevator, but one of the answers proposes a different system that uses space elevators on Deimos and Phobos to achieve escape velocity but only taking a rocket as far as the innermost moon.

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  • $\begingroup$ What about putting the cable factory on Deimos (which is slightly above synchronous orbit)? Extrude it with enough velocity to overcome Deimou feeble gravity; when it reaches the appropriate length, catch it on Pavonis. (I do not mean to trivialize the problem of catching it.) If you do it right, the tension should put Deimos higher but, of course, stationary. $\endgroup$ – Anton Sherwood Aug 26 at 22:10
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A space elevator works by connecting a satellite in geostationary (well, on mars it's areostationary^^) orbit with a ground station. The satellite needs to be geo(areo)stationary so it's always above the same point of the planet - otherwise it would pull on the connecting cable and wrap it around the planet over time, or require the ground to hold it in position against its orbital tendencies (which would quickly destroy the cable, deorbit the satellite, or rip the ground station out... not good^^)

So for your appropriate height, we need to know how high an areostationary orbit is - luckily, wikipedia can help us there: it's 20,428 km.

Also, regarding your second question: The only location to reasonably place a space elevator is ON (not even "near") the equator. Because only over the equator you can have a geo(areo)stationary orbit. I'm not sure how far away that mountain is, maybe it could be doable - but you definitely can not put the elevator at some random spot on the planet to avoid phobos.

I'm not sure whether there's a spot on the equator that phobos does not cross over, if there is - that would be the best place to put your elevator :)

Otherwise... Well, phobos is going to have to be out of the way. Maybe in your future it was hit by something else and already removed? Or maybe, since you want a counterweight on the upper end of the cable anyway, your engineers pushed phobos into areostationary orbit and are using it for that? (quite a feat, but you didn't specify how high-tech we're talking here^^)

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    $\begingroup$ Pavonis Mons is about a degree and a half off the equator (less then 100 km) so not sure if that's too far, but sounds about as close as I could get without starting from flat ground. However, I do believe that a counterweight is specifically there to stop the issue with wrapping around the planet. $\endgroup$ – TitaniumTurtle Feb 13 '18 at 11:57
  • $\begingroup$ @TitaniumTurtle With a counterweight, you can make the cable a bit shorter (and thus have the counterweight/satellite fly a bit lower) than needed for true geo(areo)stationary orbit. If you move north or south of the equator, you face a pull that's not straight up/down but slightly angled - which would, IF your material was truly indestructable and your ground station immovable, probably cause some kind of oscillating path for the counterweight. How quickly that would oscillate might be an interesting math problem - but to make it work at all, indestructable and immovable respectively ;) $\endgroup$ – Syndic Feb 13 '18 at 12:14
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    $\begingroup$ Just noticed I wrote "shorter" and "lower" in my comment... this is exactly the wrong way around. "longer" and "higher" are what is needed. $\endgroup$ – Syndic Feb 13 '18 at 15:08
  • $\begingroup$ @syndic Net acceleration for every part of the elevator below Mars stationary would be downward. Net acceleration at Mars stationary orbit is zero. Counter weight needs to be above aerostationery. Without a counterweight a Mars elevator would need to extend to an altitude of 66,000 kilometers. $\endgroup$ – HopDavid Jun 10 at 18:41
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    $\begingroup$ @Syndic Centrifugal force, ω^2 • r, scales with distance from body. Gravity, GM/r^2, falls with inverse square. So acceleration gradient isn't symmetric about Mars synchronous orbit. I used Pearson's equations to get upper length. See my Beanstalks, Elevators & Clarke Towers $\endgroup$ – HopDavid Jun 13 at 23:54
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You could give up on the beanstalk and try a skyhook dangled down from Phobos. There could be some sort of launch system to launch ships from the surface to rendezvous with the skyhook when it passes, and then be pulled up the skyhook to Phobos and beyond (or the payload transfered to elevator cars to go up the skyhook). Thus the ships will use a lot less fuel getting into Mars orbit than otherwise.

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Instead of a space elevator that is anchored to the surface of Mars, how about a modification of the Fulton surface to air recovery system? https://en.wikipedia.org/wiki/Fulton_surface-to-air_recovery_system

Hang a cable from Phobos and grab the end as it goes by. It's not a true space elevator in that the cargo has to get to the edge of the Martian atmosphere before it can catch a ride.

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Space elevator

Just to clarify the other answers, the requirement is that the counterweight has to be past the aerostationary orbit level. That makes it attempt to pull the rest of the cable up and away from Mars. There are basically two approaches, although a combination is certainly possible.

  1. Put the counterweight just outside the areostationary level. This involves a large, station-like counterweight. The counterweight would be located close to the areostationary level of 20,428 km (20,414 km above Pavonis Mons).

  2. Run the cable about twice as far as necessary. This turns half the cable into counterweight. So the cable would be longer than 40,000 km. People would leave the cable at the areostationary level though. The rest of the cable is just counterweight.

Either way, the counterweight has to be outside the areostationary level. But it doesn't need to be far outside. It can be pulled into one lump or it can be extended out. The math works either way.

It would probably be easier to move Phobos than to move the elevator. Moving Phobos is an engineering problem. We aren't really sure of the theory behind moving the base of a space elevator.

I'm not sure that Pavonis Mons helps you enough to be worth it. Really, if you can manage a 20,415 km cable, you should be able to manage 20,430 km.

Space catapult

You also should know that a space elevator doesn't give much benefit on a world without an atmosphere. You might be better off with a space catapult. Of course, you also may be planning on giving Mars an atmosphere, which would then make a space catapult non-feasible. The basic issue is that adding atmospheric resistance (which increases with speed) to gravity makes the speeds required at ground untenable. But without an atmosphere, you simply have to crank up the distance to make the acceleration work.

Another problem with an atmosphere is that it dilutes the amount of sunlight reaching the ground, so you have less power. Not so big a deal with an elevator, which can get its energy from space. May be a problem with a catapult, which is entirely on the ground.

A space catapult is helped significantly by increasing the distance above the ground from which it releases. So you might terminate it in Pavonis Mons. Probably not at the top, but above the normal ground level.

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  • $\begingroup$ So catapults are actually something pretty prevalent in my universe. I don't intend to give Mars an atmosphere, but the space elevator has more of a cultural significance in the backstory. Also, I expect enough space traffic around the attached station that the elevator is much faster than surface based shuttles. $\endgroup$ – TitaniumTurtle Aug 27 at 1:27
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Construct the elevator by extending cables down from Deimos and Phobos during the construction period, adding eccentricity to their orbit during the construction process... You don't have to have the final anchor-point directly over equator if there is a secondary anchor-point equal and opposite of the primary one on the other side of the equator attached to Phobos as well (which has presumably been lifted to areostationary orbit while lowering the initial tether.)

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  • $\begingroup$ I am not sure I understand what you mean with "You don't have to have the final anchor-point directly over equator if there is a secondary an'chor-point equal and opposite of the primary one on the other side of the equator $\endgroup$ – L.Dutch Aug 26 at 3:04
  • $\begingroup$ @L.Dutch, the idea there is a Y-shaped cable symmetric about the equator. $\endgroup$ – Anton Sherwood Aug 26 at 22:52
  • $\begingroup$ This might be a good idea actually considering the amount of weight the cable would have to support. It would also serve to allow multiple places from which to reach the station. $\endgroup$ – TitaniumTurtle Aug 27 at 15:40
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Several possible solutions. First, as others said, you need the cable to run from the planet to aerostationary orbit, ~20.000km, and a counterweight beyond. This counterweight can be simply cable (not exactly the same mass as the main cable, because what you want is to cancel the force, and net gravity is not constant, but about the same order of magnitude) or it can be an asteroid. Deimos is near the aerostationary orbit (~23.000km), but it is too heavy for a counterweight. Maybe you can nuke it to have a big chunk of it to use as weight ;)

As for Phobos: it orbits mars at ~9000km, about 3 mars radii. One possible solution is to synchronize the cars up and down the cable to deform it (like the strings of a guitar), so that whenever phobos is going to crash, the cable is away from it. This needs a carefully controlled choreography, but it is possibly feasible. This is what A. C. Clarke did in The Fountains of Paradise. Another solution is not to use an equatorial cable, but an Y shaped one. It is harder to build, but I have seen proposals for this. One of the sides of the Y is anchored north (maybe latitude 45-60º) and the other, south. Then Phobos can pass through the Y. I am not completely sure this would work, because the Y wouldn't be straight, but more like a ≻- (curved). The two lower cables have to join beyond Phobos, but far below the aerostationary orbit. And lastly... if you have already nuked Deimos, why not move Phobos to another orbit?

And if you use cable as a counterweight, you have another advantage: as the whole cable is moving at the angular speed of the aerostationary orbit, when you are below that orbit you are at suborbital speed, and if you detach from the cable you fall to mars. But if you are above that orbit you are faster than aerostationary: if far enough, you have speed to leave the martian system or even the solar system without using any fuel (think of it as a sling).

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This may be a minor point but it seems too important to leave in a comment:

If anchored away from the equator, the cable will approach the equatorial plane as an asymptote.

There are three forces to consider (in the rotating frame):

  • centrifugal force, away from the axis of rotation;
  • gravity, toward the center of the planet;
  • tension, toward the anchor.

If the anchor is not on the equator, these last two each have a component parallel to the axis of rotation, opposing each other.

This makes the cable longer and more costly, but it is a way to avoid Phobos.

Such a cable is not straight: it sags. There is a maximum latitude for the anchor, at which the cable lies flat on the ground; I do not know how to compute this.

(How massive can the cable and counterweight be without causing a measurable wobble in the planet's rotation?)

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