Writing: Input Wanted Lunar Orbital Rings: Calculating Height and Rotation for Earth-like Gravity

[Chatterton]

I'm working on a story set on the moon post-industrialization. The moon has an orbital ring with a spinning exterior to simulate Earth gravity. People work on the surface in lunar grav, then go up to live on the ring under conditions more favorable for human bodies.

Two questions I need to know:

1) How high up should the ring be?
2) How long would it take to rotate around the ring to simulate Earth gravity? Would it be so fast that it created problems? Dizziness, nausea, etc.

Thanks!

 

[Buzz Bloom]

Two questions I need to know:

Hi Chatterton:

I would like to help you, but I have no clear concept of what you mean by an "orbital ring" and by a "spinning exterior". Please describe these concepts with much more detail.

Regards,
Buzz

 

[Chatterton]

Thanks, Buzz Bloom.

Sorry for the confusion. I borrowed the terminology from the SFIA YouTube channel:


The orbital ring is a continuous megastructure around the equator in whatever the lunar equivalent of geosynchronous orbit is (lunasynchronous?) likely tethered by a series of space elevators.

The ring is built in two parts. The inner part is fixed, the outer part rotates to generate spin gravity.

Clear as mud?

 

[RPinPA]

Found a pretty good discussion of "lunar stationary orbits" here:
https://astronomy.stackexchange.com...-stable-lunarstationary-orbit-around-the-moon

You'll find a calculation that the orbital radius is 88417 km, and also an explanation why that's too far to be a stable orbit around the moon. So I'm not sure your engineers should count on being able to achieve that.

Let's look at your second ring, the one with artificial gravity. The formula for centripetal acceleration in terms of rotation speed is $a = \omega^2 r$ so $\omega = \sqrt{a/r}$ in radians per second, or $\omega = (180/\pi) \sqrt{a/r}$ deg/s.

You want $a = 9.8 \text{ m/s}^2$. Plugging in various values of $r$ that gives the following:

$r = 10000 \text { km} \implies \omega = 0.0567$ deg/s or one rotation every 6349 second (1h45m49s)
$r = 20000 \text { km} \implies \omega = 0.0401$ deg/s or one rotation every 8978 second (2h29m38s)
$r = 30000 \text { km} \implies \omega = 0.0327$ deg/s or one rotation every 11009 second (3h3m29s)

Note that $r$ is the distance from the center and the distance from the surface will be 1737 km less.

Edit to add: the comment below mentioning "centrifugal force" made me realize I'd neglected the moon's gravity. The analysis of the artificial gravity on this ring would be equivalent to the analysis of the top of the motion when you are swinging something in a vertical circle, i.e., the apparent gravity is $g$ minus the lunar gravity at that altitude.

So to achieve $g$ the equation you need to solve has to include a lunar gravity term and the speeds will be a little higher than what I worked out. Since 10000 km is 6 lunar radii and the lunar gravity is 1/36 of the lunar surface (and even less at the higher altitudes) it's not a major correction.

 

[hmmm27]

Clear as mud?

Okay, let me get up to speed...

- the geostationary ring is attached to the ground. Since the Moon's rotational period is a month long, free-falling geostationary positioning isn't going to work (the Earth captures it) - so, the ring is going to experience pretty much the usual falling.

- which is where the artifical gravity ring comes in, its centrifugal force holds it and the geo ring's mass up (or down, if you happen to be standing on it)...

- which is the cool bit, because you're getting 1g net by pulling (or pushing) on the geo ring, not the tethers - which are mostly there for station-keeping and to provide a convenient rope ladder to climb up and down on.

So, it's really neat, but there's some basic gotchas...
- the geo ring's speed is in feet per second, the grav ring in miles per second
- depending on how close to the surface, an equatorial orbit you have to deal with weighing more or less at different times of the day (because of Earth's influence)
- anything other than an equatorial orbit and there's precessional forces to deal with

Of course, the distance from the Moon's surface relies entirely on mass ratio between grav ring and geo ring(+tethers), and the required orbital time. You could build it right at the surface.

 

[Chatterton]

You could build it right at the surface.

That's also a possibility. But I had thought that the ring would be multi-functional, though, including a shipyard. Mine stuff on the moon, run it up to the ring on the space elevator, refine it, send the product to the shipyard. Fewer rockets, minimize fuel costs.

 

[hmmm27]

Fewer rockets, minimize fuel costs

On the ground, there's the advantage of not having to deal with tether-weight and just use columns rated for supporting the grav ring and its encasement at +1/6 to -1 Earth g's so a bit of downtime could be had, occasionally. It's not as spectacular though, of course.

Either scenario it'd be a pain to dock a spaceship with, unless there was an orbital-velocity (0g) ring, though I'll grant that not plowing into a mountain at a few km/sec if something goes wrong could be construed as an advantage for a high-altitude ring

I've my own pet Lunar development, of course : equatorial, encircling train tracks : one carrying Earth sidereal trains for tourists, government and farms/ranches, running @ 200'ish mph to match Earth's diurnal period, and a Sol sidereal one @ 10mph for a pair of trains : one always under the sun, the other always facing the stars. For this scenario, the permanent residents can spend their off-shift time "at home" in rotating-wheel stations either on the ground, or in Lunar orbit, where the shipyards are.

Oh yes, and a tangential takeoff/landing spaceport on top of an equatorial mountain range on the far side, for shuttlecraft.

Not that I've thought about it much, of course

 

[hmmm27]

But, I digress. Getting back to the original questions ; in reverse order...

2) How long would it take to rotate around the ring to simulate Earth gravity? Would it be so fast that it created problems? Dizziness, nausea, etc.

RPinPA covered the approximate times in post #4. You wouldn't feel any inner-ear related disorientation.

1) How high up should the ring be?

Good question. Remember how the orbital ring works : the push outwards from the grav-ring matches the push inwards from the geo-ring, so there's no stresses being put either on the tethers or on the ring structure itself, except the 1g where the two rings meet, and that mechanism could be as simple as that to "levitate" a maglev train on Earth, and no more powerful (per unit of distance) either.

This means that the mass of the geo-ring, under the influence (mostly) of the relatively weak gravity of the Moon, has to mass more than the mass of the 1g centrifuge of the grav-ring. If it were near the ground, it would be 6x as much. The further off the surface, the less the influence of the Moon's gravity, so the higher you go the greater the ratio of the mass of the geo-ring to the mass of the grav-ring (it doesn't have to be structural - just pile rocks on top). So you get to choose how high you want to go just by changing that parameter.

Except, if you go too high you get in range of the Earth Moon L1 & L2 LaGrangian points, where all the gravities and centripetal forces do some cancelling out and an object can just "hover" directly above the Moon without moving (relative to the Moon). There, the grav-ring still has pull, but the geo-ring doesn't. So, not that high (23,000 km for L1). As is, no matter what height you choose, you'll want to weighten or lighten some sections of the geo-ring, like balance weights on a car wheel, to compensate for things like mascons and the aforementioned combined gravity and centripetal vectors.

Maybe, pick an altitude such that, if you drop an item (or a load of metal) through the "floor" of the grav-ring, it will sail up all on its own to the orbit of where your spaceship factory is.

 

[hmmm27]

Wikipedia has an entry on Orbital Rings.

The proposals seem rather crude, compared with what you envision.

 

[Nik_2213]

Tiny problem: IIRC, the Moon's many mascons will so wobble the ring's orbit, require constant adjustment...

 

[Tiran]

Instead of a space port, why not spin the ring close to the moon's surface and use its close passage to points on the ground to magnetically accelerate things on the surface to either transfer them to the ring or to fling them out of moon orbit, since the ring's velocity is 9.8m/s2 above the moon's escape velocity. How much of that velocity you impart to the surface object determines where it ends up - and that could be used to transfer to space stations in standard lunar or geo orbits, the Lagrange points or out of the Earth system.

Of course, with a ring even as small as the moon's diameter you are going to need a fictional material to keep the ring together. Even if you use some sort of magnetic support, those magnets are going to require tethers of impossibly strong material.

The only way around that I can think of is to have 6 times the ring's mass stacked outside its circumference magnetically pushing down, due to the moon only providing 1/6 the attraction of the ring's rotating 'expansion'. And the amount of electrical power to operate that maglev will also be enormous.Big rings are somewhat impractical compared to several small rings in low lunar orbit.

 

[hmmm27]

What ? no.

The interface between stationary and g-rings are well within current materials science - there, a pretty bog standard maglev, done ; it doesn't even have to be any more powerful than the one moving kids around Disneyworld : 7/6g is not a problem. Ditto ac/decelerating cargo/passengers to transfer from one ring to the other (we haven't tried it yet, but existing engineering is well up to the task).

Since the stationary and g-ring conspire together to be weightless, tethers are necessary, not for support but for simple stationkeeping. And that's where we would run into a material-science problem, depending on how high off the surface the ring-assembly is.

 

[Tiran]

What ? no.Since the stationary and g-ring conspire together to be weightless, tethers are necessary, not for support but for simple stationkeeping. And that's where we would run into a material-science problem, depending on how high off the surface the ring-assembly is.

No? What?

I said the tether's would have to be fictionally strong, and you seem to be agreeing with that. What are you saying no to?

 

[hmmm27]

No? What?

I said the tether's would have to be fictionally strong, and you seem to be agreeing with that. What are you saying no to?

It may have been the order i read it, but...

Of course, with a ring even as small as the moon's diameter you are going to need a fictional material to keep the ring together. Even if you use some sort of magnetic support, those magnets are going to require tethers of impossibly strong material.

... jumped out at me since, for a ring built on or near the surface, the supporting structure need be only as strong as a suspended train track under a bridge on Earth (but upside down, of course).

 

[Tiran]

jumped out at me since, for a ring built on or near the surface, the supporting structure need be only as strong as a suspended train track under a bridge on Earth (but upside down, of course).

Except a train trestle is in compression not tension. But the real issue is whether your spinning ring has the average mass of a train or something much larger.

If it is just a train, put it upside down in a tunnel.

 

[stefan r]

...
If it is just a train, put it upside down in a tunnel.

I believe that is still an orbital ring. A ring that is mostly above the surface could also be a tunnel at mountains. A completely subsurface orbital ring is more believable on Phobos, Europa, or Charon.

If you really just want spin gravity for medical purposes you can spin around the wall of a crater.

 

[Tiran]

I believe that is still an orbital ring. A ring that is mostly above the surface could also be a tunnel at mountains. A completely subsurface orbital ring is more believable on Phobos, Europa, or Charon.

If you really just want spin gravity for medical purposes you can spin around the wall of a crater.

I didn't say it isn't an orbital ring.

As for your suggestion, have you ever been in a centrifuge where you have both angular momentum and gravity? It is not pleasant.In 1/6 gravity you could set up a vertical system that spins around an off center point closer to the bottom. You could effectively cancel the moon's gravity with this kind of eccentric "orbit".

 

[stefan r]

I didn't say it isn't an orbital ring.

As for your suggestion, have you ever been in a centrifuge where you have both angular momentum and gravity? It is not pleasant.In 1/6 gravity you could set up a vertical system that spins around an off center point closer to the bottom. You could effectively cancel the moon's gravity with this kind of eccentric "orbit".

Amusement park rides that loop overhead are uncomfortable. There is a ride where you stand against the wall and they remove the floor. I felt perfectly comfortable. A merry-go-round would be fine except that the rpm is too high. Over the short distance your fluids sense a difference. You would notice nothing if it was a 10 km train track with no windows.

 

[Tiran]

Amusement park rides that loop overhead are uncomfortable. There is a ride where you stand against the wall and they remove the floor. I felt perfectly comfortable. A merry-go-round would be fine except that the rpm is too high. Over the short distance your fluids sense a difference. You would notice nothing if it was a 10 km train track with no windows.

If you had moved your head out of plane with the floor drop ride you would have noticed a very disorienting feeling.

Overhead loops are uncomfortable because apparent gravity keeps shifting. But if your rotation speed is variable and actual gravity is lower than 1G, you can bias the angular momentum to be additive or subtractive with the moons gravity to produce an apparent 1G throughout the loop.