How would we increase the rotation rate of very large structures in deep space?

[nonsense333]

Homework Statement

Determining applications of kinetic energy under specific guidelines.

Relevant Equations

KE = 0.5 x mv^2

Imagine a 400-meter-long pipe with a 1600-meter diameter, floating in inter-planetary space. It is spinning at 0.5 gravity along its major axis and there are no secondary-axes spins. We need to increase rotation to 0.85 g. Its density is a uniform 2.3 kg/m³ and it weighs 49,120,056 kg.
Thanks to Izzy Newton, we can calculate the kinetic energy needed for the acceleration. KE = 0.5 x mv2. Applying the formula, 0.5 x 49,120,056 kg x 6724 (82 m/s2), resolves to 165,000,000,000 J.
But what does that mean?
Japan’s nuclear power plant at Kashiwazaki-Kariwa has 7 reactors with a total net capacity of 8 billion J (watts per second). Assuming reactors equivalent to each of those in Japan, 145 would power up our pipe to the stated rotation… but over what duration? Over one second? The pipe’s density means it is far from fragile, but neither could it take substantial torque, so imagine that we’ve solved the reinforcement problems using today’s technology and will set its acceleration in the midrange of solutions.
I have no training, education, nor necessary experience in a related field, so I’m hesitant to assert that I’ve made no mistakes so far. However, given the above, how would we power up a reinforced structure of the stated dimensions from 0.5 g to 0.85 g without tearing it apart? I want to tether an appropriate number of nuclear-powered rockets to the pipe to handle the transition, but my understanding of physics is pale. Would 10 rockets do the job over 6 months? 50 rockets over a year?

 

[RPinPA]

OK, I'm going to rewrite this slightly, as I hate all those zeros and commas and need to see these numbers in scientific notation with an appropriate number of significant figures to get a feel for them.

Its density is a uniform 2.3 kg/m³ and it weighs 49,120,056 kg.

I just want to check if this is reasonable. The mass is 4.9 x 10$^7$ kg. So the volume is 2.1 x 10$^7$ m$^3$. The outer area of a cylindrical tube of length 400 m and radius 800 m is $2 \pi r L$ so that's 2.0 x 10$^6$ m$^2$, meaning your pipe has a thickness of (2.1 x 10$^7$)/(2.0 x 10$^6$) or about 10 and a half meters.

OK, plausible.

Thanks to Izzy Newton, we can calculate the kinetic energy needed for the acceleration. KE = 0.5 x mv2.

Uh, I'm not sure what you're doing here. Now, we would normally calculate this as rotational kinetic energy, but since your pipe is thin (10 m thick compared to a radius of 800 m), that approach is OK. But what value are you using for v? You have two different v's to consider, the one at 0.5 g and the one at 0.85 g, and then you want the difference in energy between those. Also, you didn't show your calculation for going from the gravity value to the speed. So let's do that.

$a = v^2/r$ so $v^2 = a r$ and KE = $(1/2)mv^2 = (1/2)m a r$. So at 0.5 g you've got a KE of $(1/2)(4.9 \times 10^7)(0.5 * 9.8)*800$ or $9.6 \times 10^10$ J initially, and at your final speed you've got a KE of $(1/2)(4.9 \times 10^7)(0.85 * 9.8)*800$ or $16.3 \times 10^10$ J. So the increase is $6.7 \times 10^10 J$.

Same order of magnitude as your answer but only about 1/3 as much. Not that that really changes the question.

8 billion J (watts per second)

That's slightly confused. Watts are the number of Joules per second. If the capacity is 8 GW, that means it produces $8 \times 10^9$ Joules every second. To produce $67 \times 10^9$ Joules would therefore take 8.4 seconds.

The issue here really is force and momentum, more than energy. In order to give the additional momentum to this large structure, something is going to have to have equal and opposite momentum. Force is equal to the rate of change of momentum, so you could keep the forces small by transferring momentum slowly. I wouldn't think there's any reason to try to accomplish this change in a couple of seconds. Why not take a week?

But how? That's your design problem. You are going from 63 to 82 m/s tangential velocity, adding $4.9 \times 10^7$ kg * 19 m/s momentum. If you have rockets which eject fuel material at, say, 100 m/s to provide thrust then you'll need about 10 million kg of fuel, whether you do it quickly or slowly.

The only way I can think of around this would be to provide counter-rotation to something. Maybe you're spinning up a second pipe in the opposite direction at the same time. Twice the energy but you save a lot in momentum. You wouldn't need to burn up any mass.

 

[nonsense333]

your pipe has a thickness of (2.1 x 1077^7)/(2.0 x 1066^6) or about 10 and a half meters.

OUCH!
Actually, I calculated my thickness as 110 meters, an essential dimension, but I'm sure your calculations are correct, so I'll need to modify something. Quoting myself, this is another "essential:"

I have no training, education, nor necessary experience in a related field, so I’m hesitant

However, my underlying intention is to show how we can take our next steps into space at a fraction of the cost, a fraction of the risk, and a fraction of the time otherwise used settling Mars. While my end product is an alternative-history novel spanning 1969-2010 that assumes that JFK's influence survived the societal inertia, I want my science to be credible. In line with that, I'm writing either a very short book or a very long article of factual information that justifies the story line.
Although I attempted to enlist the help of NASA sources, sources at two local universities and other organizations, my misunderstandings were likely enough to alarm most knowledgeable people into not responding--which is an obtuse way of saying "thank you" to you. Sans mentor, I relied on online calculators and sources to get as far as I have, not without blunders, but I think most of my calculations and assertions beyond the extent of issues presented here are correct.
I don't fully understand your paragraph with the bulk of your calculations, but one of my earlier, discarded projections on the increase in KE matched your 6.7 × 1010 J so I'm comfortable. Plus, my math error might explain the difference between your 10.5 meters and my 110-meter thickness.
While I've largely understood most of your response, I don't understand most of the "force and momentum" concepts or conclusion. Can you briefly elaborate?
My space object is a reinforced structure, but it has the density of sandstone, so it needs to be spun to speed carefully and slowly. Two years isn't unreasonable, but is six months remotely possible?
My assumptions included that nuclear reactors built and used in space will be remarkably cheap, given that they can't easily hurt anybody or anything--there are no neighbors--so most of the expensive safety controls become unnecessary. But when you wrote that we'd need 10 million kg of fuel, did you mean fissile materials--e.g., uranium? The problem is that the percentage of radioactive materials in asteroids remains undefined and debated, although its abundance on the Moon has been roughly estimated since 2007. (Part of the 10 million kg of fuel can come from Earth in the form of spent fuel rods, etc., but not without a fight.)
The solution to my "design problem," was tethered reactor-powered rockets using proven reactors or magical ion drives or radioisotope thermoelectric generators. Spinning up two pipes is only slightly less doable--and intriguing. Thus, as concerns your "Twice the energy but you save a lot in momentum. You wouldn't need to burn up any mass," can you elaborate a bit on how twice the energy is more costly overall than is increased momentum?
Thank you.

 

[RPinPA]

I don't fully understand your paragraph with the bulk of your calculations, but one of my earlier, discarded projections on the increase in KE matched your 6.7 × 1010 J so I'm comfortable. Plus, my math error might explain the difference between your 10.5 meters and my 110-meter thickness.

I could have made an error. As for the KE, you were calculating the total KE at 0.85g, but you said you were starting at 0.5g. So I calculated the difference, not the total.

Probably worth checking the thickness calculation, at least a sanity check. Let's see: I got a volume of $2 \times 10^7 m^3$ from your mass of $4.9 \times 10^7 kg$. Since you said the density is 2.3 kg/m^3, that's correct.

The volume of a thin shell is A * d where A = surface area and d = thickness. The surface area is $2\pi r L$ = about 6 times $8 \times 10^2$ times $4 \times 10^2$. 6 * 8 is about 50, so 6 * 8 * 4 is about 200, so that's 200 times $10^4$ or $2 \times 10^6 m^2$. So yes, the volume is about 10 times that number and the thickness is about 10 m. It checks.

While I've largely understood most of your response, I don't understand most of the "force and momentum" concepts or conclusion. Can you briefly elaborate?

How about if we talk in terms of acceleration? Suppose you are in space at 0 m/s and you want to get up to 2000 m/s. You could do it by accelerating at 200 m/s^2 for 10 seconds. That would be 20 g's. It would probably kill you. Or you could do it by accelerating at 0.02 m/s^2, 1/500 g, for 10000 seconds (a little under 3 hours). You probably wouldn't even feel that.

The endpoint is the same, you have the same final momentum and kinetic energy. But the forces on your body are very different. So all I'm saying is if you're worried about the stresses on your structure, you're talking about force, so you want to keep the forces small. Get there slowly.

The other part of my calculation was just how much reaction mass you'd need to make that change. Whether you do it quickly or slowly, if you're doing it with thrusters, then every bit of momentum you add to the ship comes from giving an equal and opposite momentum to the gas ejected from your thrusters. If you want to give a mass of 1 kg a boost of 10 m/s in velocity with thrusters, a change in momentum of 10 kg-m/s, then your ejected gas has to have m * v = 10 kg-m/s. For instance 0.1 kg moving at 100 m/s, or 10 kg moving at 1 m/s.

But when you wrote that we'd need 10 million kg of fuel, did you mean fissile materials--e.g., uranium?

No, I was purely talking about m * v on the assumption that you're boosting the rotation with thrusters, i.e. shooting out hot gas. That's how much mass m of gas is being ejected on some assumption of what v is, which I made up. I was just calculating an $mv$ which is equal to the $\Delta mv$ of the ship.

Heat the gas up with nuclear reactors or steam engines, it doesn't matter. If you're using thrusters to boost the ship, then $mv$ of the gas is $mv$ of the ship.

 

[RPinPA]

That would be 20 g's. It would probably kill you. Or you could do it by accelerating at 0.02 m/s^2, 1/500 g, for 10000 seconds (a little under 3 hours).

Slipped a decimal point. $(2 \times 10^3 m/s) / (2 \times 10^{-2} m/s^2) = 10^5 s$, about 28 hours.

The endpoint is the same, you have the same final momentum and kinetic energy. But the forces on your body are very different. So all I'm saying is if you're worried about the stresses on your structure, you're talking about force, so you want to keep the forces small. Get there slowly.

Also you want the forces as evenly distributed as possible, so having small thrusters distributed all around the perimeter and length is probably a good idea.

 

[nonsense333]

I'm struggling with your insightful (and gracious) analysis, but while trying to make my numbers coordinate with yours, I realized I gave you misleading data. The "pipe" object is indeed 110m thick, but I neglected to mention that 80% of its original mass (at 2.3kg/m^3 density) has been hollowed out. Essential data, huh.
I will continue reworking the problem, but thanks to your clarifications I was able to discover another of my mistakes. I think I'm in good standing now, so thank you again.

 

[RPinPA]

I was going to mention that 2.3 kg/m^3 is kind of low. More like a gas than a solid. Air is 1 kg/m^3. Water is 1000 kg/m^3.

 

[nonsense333]

Good catch. Somewhere along the line I moved the decimal from sandstone's 2,300 kg to the aberrant 2.3 kg. This drastically damages my perception, but better now than later.