How fast would a Martian space elevator travel?

Question

READ EDITS

I am writing a passage about the main character of my book travelling up a space elevator to the spaceport that rests in geostationary orbit above Mars. This means that the elevator would have to travel roughly 17,000 km [previously wrote 20,400 km] from ground to station. I'm trying to figure out how fast to make the elevator car travel in order to determine how long it will take, but my math is concerning me so I thought I would turn to you my fellow worldbuilders for help. I am having trouble finding information on the maximum G-forces a person can stand comfortably. I need this so that I know my maximum acceleration. So first question:

What is the maximum G-forces someone can stand COMFORTABLY?

I use the term "comfortably" loosely, as I mainly mean tolerable for long periods of time.

What I did find is that maglev trains accelerate up to about 0.5 Gs (I am imagining a sort of maglev elevator). This means that if I am remember my kinematic equations correctly, (and not fudging it up because its almost 1 AM here and I'm tired) I should be able to make the trip in about 110 hours with a continuous acceleration to halfway and then deceleration the rest of the way. Meaning I would make the trip in roughly 4.5 days, which isn't ideal. I would appreciate if someone could check my math, although it might be correct since I did find an article stating the same thing on Earth might take around 7.5 days.

Is it reasonable to assume this is the fastest method of space elevator travel?

I'm open to suggestions on design changes, but would like to clarify that there is no artificial gravity tech.

Further background for clarification: This is suppose to take place roughly 400 years from now, however technology has probably only advanced 100-150 years give or take due to the destruction of Earth around 2100 AD, leaving Mars as the new center of human life. Computer science slowed with the end of Moore's law, but jumped up again with early quantum computing. Medical science has advanced to counteract most of the negative affects of living in low G through strict exercise regiments and medication to promote bone and muscle development. Materials science has made a few leaps to allow for the issues with maximum speed or just building a space elevator in the first place.

Edit: Previously had written Mars geostationary orbit as 20,400 km instead of 17,000 km. The reason for confusion is that the previous number was the distance from Mars' center of mass, not the Martian surface. That said, my calculations were definitely wrong since they used the first number.

Edit: My calculations were way off because I forgot to convert 0.5 G back to 4.9 m/s^2, so I just used 0.5 and got way wrong numbers.

Answer 1

TL:DR: 62 minutes, shorter than that, or "it depends, but longer" are are possible answers.

An Hour

Okay, it looks like from this question here, that 4.9m/s2 of additional acceleration caused no major problems for seven days. So we should be fine at a half-gee.

Great. We can do this. Now, how long will it take? Assuming we've got a maglev with a theoretically unlimited speed, how long will it take to get up there? $$ 8\,500\,000 \text{ m} = \frac{4.9 \text{ m/s}^2 * t^2}{2} $$

Well, given an acceleration of 4.9m/s2, it will take 1863 seconds (31 minutes) to get up to the halfway point, at which point it will be moving at almost 10 kilometers per second. Slowing down will take an equal amount of time and distance, so a bit over an hour.

... is the slow time

But wait, that's the simplified explanation. See, the strength of Mars's gravity is decreasing, while the centripetal effect on the train is increasing. So you can start accelerating even faster.

While accelerating up (start of upward trip, end of downward trip,) we accelerate at rate a: $$ a = 8.6 \text{ m/s}^2 - \frac{GM}{r^2} + \omega^2r $$ And while accelerating down (end of upward trip, start of downward trip,) we accelerate at rate a: $$ a = 8.6 \text{ m/s}^2 + \frac{GM}{r^2} - \omega^2r $$

Where G is the universal gravitational constant, M is the mass of Mars, r is the distance to Mars's center, and ω is Mars's rotational speed, in radians per second. The first term is the "gravity" the passengers will feel. The second term is Mars's gravity. The third term is the centripetal effect on the rail.

Now figuring out how to use those equations to get actual answers (total travel time, time until flip (it isn't halfway,) top speed) is going to require knowing how to handle differential equations, so... you can just stick with the 62 minute answer if you want.

Hang on a second

This is, of course, assuming that your maglev has an unlimited top speed. If it doesn't, then that becomes the limiting factor. The "multiple days to get to GEO is based off a climber going at ~80 meters per second, based off the speed of the fastest trains. at that speed, it takes about ~60 hours to travel 17,000 km.

A Maglev is actually a good idea for this, since it won't run into the problem of wheels exploding when you try to spin them up to 10km/s. Still, there will be technical problems with trying to get a maglev to work at those speeds. Not unsolvable, but still something to keep in mind.

Even if you could get a maglev to go that fast, people might not consider it safe, and limit it to slower speeds.

Answer 2

A discussion of climbers or the vehicles travelling up and down space elevators proposed the following limits factors for the speed of a climber.

Climber speed would be limited by the Coriolis force, available power, and by the need to ensure the climber's accelerating force does not break the cable. Climbers would also need to maintain a minimum average speed in order to move material up and down economically and expeditiously.[citation needed] At the speed of a very fast car or train of 300 km/h (190 mph) it will take about 5 days to climb to geosynchronous orbit.

Source: Space elevator

The five days cited above applies to a space elevator for planet Earth.

However, this paper by L M Weinstein, NASA, has an estimate for an ascent on a Martian space elevator (where the space elevator is connected to the moon Phobos as part of the system).

As the payload is moved higher above Mars, it tries to move laterally from the elevator, since it is moving to an altitude with a different rotational velocity. However, since the payload is attached to the elevator, it gains orbital velocity by the constraining side force imposed by the tether itself. Lifting a payload from Mars to Phobos at an average speed of 133 km/hr (arbitrarily selected reasonable speed) would take about two days, so the side force can be considered a very weak acceleration continuously applied over that long time. The side force on the elevator would only be about 5 N for a 500-kg load, and would be easily constrained.

This paper above gives an answer of two days to travel a Martian space elevator.

REFERENCE:

2003: Space Colonization Using Space-Elevators from Phobos, by Leonard M. Weinstein, NASA Langley Research Center. 9 pages.

Author: Leonard M. Weinstein Advanced Measurement and Diagnostics Branch, NASA Langley Research Center, Hampton, VA 23681, USA

E-mail: [email protected]

Answer 3

Your idea to compare to Maglevs is smart, but you have a problem with timescales. Those trains only take a minute or so to get up to top speed, at which point they stop accelerating. Your elevator will take days. There have been a few experiments with astronauts in a centrifuge at 1.5G for approximately a week. (https://space.stackexchange.com/questions/6154/maximum-survivable-long-term-g-forces) They didn't have any ill effects but they're also the some of the fittest people on the planet.

I'm also assuming the people riding this elevator have grown up on Earth. Nobody really knows exactly what limits a person who grew up in reduced gravity can have. If these people grew up on Mars, the most specific answer anyone will be able to give is "Less than someone who grew up on Earth".

The surface gravity of Mars is 0.376G, which means you get 0.624G for free. A person riding the elevator up would just feel the same force as that of gravity on Earth. They will notice the acceleration, but it won't strain them at all. Assuming the tests I mentioned above are valid for everyone, not just astronauts in peak fitness, this means your elevator can accelerate safely at 2.124G at least, and to the rider this will only feel like 1.5G. You may also get a lot higher if you change the rider's posture. The seats in fighter jets are pretty reclined, they're almost laying supine in them. This is so they feel the G's pressing down towards their back, not towards their feet, reducing how hard it is on their cardiovascular system. Fighter pilots only pull these G's for a few seconds at a time, so I can't say any specifics as to how much, if at all, this will help your riders, but it's something to consider.

The return trip can be even quicker, since Mars' gravity is working with you. You should be able to safely get 2.5G assuming the previous figures.

When calculating how long the trip will take, remember to keep Jerk low. Also remember your cars will probably have speed limits near the stations. These combine to mean that any trip will likely be a little longer than your calculations suggest.

In conclusion, I can't say it will be a fun ride, but the passengers won't die or be maimed.

Answer 4

TL;DR: 47 minutes and 20 seconds at constant one-gee felt by the passengers.

An important thing to consider in this sort of problem is that there are, in the rotating reference frame, five separate forces we need to consider: gravity, centrifugal force, Coriolis force, drive, and the constraining force of the cable.

Given as constant the mass of Mars and its sidereal period, and assuming a spherical Mars, this gives us the forces varying as follows:

Gravity varies inversely with the square of the distance from the center of Mars ("radius") and acts downward.

Centrifugal force varies directly with the radius and acts upward.

Coriolis force varies directly with the speed of the car and acts laterally.

Drive is controllable and acts upward or downward as needed.

Constraining force is exactly counter the Coriolis force and is the reaction of the cable against the Coriolis force produced by the car's velocity.

Now let us assume that the passengers will prefer to feel one-gee pressing them to the floor the entire trip. This does not mean that they are accelerating at that rate. Just consider vertical forces momentarily. Gravity acts against that one-gee, weakening with altitude, and centrifugal force acts with the drive force to pull the car upward. Once the car begins to decelerate, gravity helps despite its weakening strength, and centrifugal force pulls against the floor of the car. Now by definition, areostationary orbit (the point we are trying to reach) is at the altitude where gravity and centrifugal force exactly counter each other, so we will see the total surface-ward force on the car before applying the drive vary from Martian surface gravity down to zero at our resting point.

Taking into account only those vertical forces brings us very close to the 47-minute figure already mentioned. However, by numerically integrating (one second intervals on a trapezoidal integration) and examining at various points I found that the velocity of the car produces a Coriolis force at maximum speed of around 12 km/s comparable to Martian surface gravity. That's a significant force pulling the car away from the cable, and the constraining force is felt by the passengers, with the vector sum of that with the drive force at higher than one-gee, angled away from the floor slightly. So instead of a constant one-gee drive, why not have a car that can angle itself (it would have to anyways to flip and decelerate), and have the car's drive vary as needed so that the total force on the passengers is one-gee?

By integrating all these forces over time, and using the simplification that rotating the car takes only one second, we find that at 26:00 the car reaches maximum speed of approximately 12.5 km/s, with a lateral force component of 1.78 m/s^2 (0.18 g), flips and begins decelerating, then comes to a stop at the areostationary orbital point of the elevator at 47:20.

As to the issue of a tram being able to travel that fast, and it being considered safe - this is in essence a space ship constrained to travel in two directions, and 12.5 km/s is comparable to other velocities of space travel.

Answer 5

What is the maximum G-forces someone can stand COMFORTABLY?

I posit that in order for someone to experience comfort that they must remain conscious, so the method by which people are transported having been rendered unconscious is out.

I would suggest a G-Suit.

(More accuratley an Anti-G-Suit)

It is designed to prevent a black-out and g-LOC (g-induced loss of consciousness) caused by the blood pooling in the lower part of the body when under acceleration, thus depriving the brain of blood.

A g-suit does not so much increase the g-threshold, but makes it possible to sustain high g longer without excessive physical fatigue. The resting g-tolerance of a typical person is anywhere from 3-5 g depending on the person. A g-suit will typically add 1 g of tolerance to that limit...

So that's 4g acceleration for the average (non-infirm) adult, but...

Comfort.

High g is not comfortable, even with a g-suit.

So, give everybody happy pills. This, from the marketing perspective would have the benefit of generating lots of repeat customers by it's effects:

(it gives) the user feelings of euphoria, intense relaxation, and decreased perception of pain.

Well that (from a marketing department's perspective) sounds comfortable.

Maximum velocity and time.

Just plug 4g into the equations. I'd say somewhere in the order of 17 minutes either way, and roughly 36Km/s at turnaround.

Issues.

We can assume that the problems of maglev at that sort of speed, such as magnetic hysteresis (the time taken to magnetise and de-magnetise a material) - a suitable technique or material to compensate would have been found.

Compensating for the force of wind at low altitudes and the tendency of atmospheric buffeting to de-stabilise the vehicle will all have been resolved by then. Possibly by enclosing the base of the tether and the first 7 Km or so:

The upper stratosphere model is used for altitudes above 7,000 meters. In the upper atmosphere the temperature decreases linearly and the pressure decreases exponentially.

On your suggestion, the whole thing could be enclosed by the tether as an elevator - that would help.

Quick turnaround, happy customers and another plus - you get to market g-suits with your branding on.