Creating Artificial Gravity In A Smaller Craft Where Energy Was Not An Issue - Energy Required To Do So?

Question

If you had a craft say the size of the spaceship in Guardian Of The Galaxy, big enough for a few rooms, etc. Now say you wanted to create gravity(artificial using centrifugal force) and energy was not a limiting factor(Yes this is definitely for fiction). Assuming if you are traveling towards some point at say close to c(The Speed Limit Of Causality in the universe) say .99*c (Where * is the multiplication operator under vector field hopefully * is not odd for multiplication operators. Anyway...)

And we can assume the direction of overall motion has a unit direction vector D.
0.99*c*D

Then if you say wanted to produce a gravity roughly that of the moon, you could rotate perpendicular to D in a circle that had a radius of about say 6000m once a minute. Or say 3000m twice a minute etc.
I assume that a wider circle would be advantageous as far as the feeling for the people on board such a craft not feeling nauseous. I don't know which would require more energy to produce a large or smarler circle? I guess I could try to work that out, that is actually interesting as well to know the answer to that. But there is a different question I am confused about....

So now say if you have two ships exactly the same moving in the direct D, but one moving very close to the speed of light, and the other very slowly, not anywhere near speeds where relativity would start to show effects, again in terms of overall velocity over time... So ship A say has overall velocity...

.99*c*D

and B have a velocity of....

0.000001*c*D

What I am unsure about is to create artificial gravity in both these situations, meaning create a constant acceleration perpendicular to D and such that you are producing that circle of say 3000m radius. Is there a difference in the energy required for ship A to produce that rotating motion compared to ship B.

I know alternatively if say the ship was itself had a radius of 3000m and just rotated then I believe the answer is that the energy required to produce the spin is the same in either case. But I was not sure if that is true for a ship that is rather moving in in a circle perpendicular to the overall velocity and must constantly create a changing direction of acceleration to produce that gravity.

I guess I am asking if you are moving close to the speed of light relative to the stars/space around you, do you essentially act like you have more mass no matter which direction you intend to accelerate in, or do you only essentially see an increase in relative mass in the direction D if you were trying to change your velocity.

As you can tell I don't have a great deal of experience with complex relativistic motion, Lagrange transforms/mechanics are still a mystery to me. But even just a simple one-line explanation me in the right direction would very much appreciate As right now I don't even know where to find a good explanation of such a case?

Answer 1

In your spacecraft's helical trajectory, the components of the circular motion are perpendicular to the direction D of forward motion, and so the circular motion is independent of the forward motion. That is, the energy required for the circular motion is identical for ship A and ship B.

This independence of perpendicular vectors is a very important principle in physics (and mathematics). There's a nice introduction to this topic on The Physics Classroom.

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I don't know how you calculated that 6 km circle per minute, but that gives way too much acceleration. The formula for centripetal acceleration is

$$a_c = v^2/r$$

We don't know $v$, but we can substitute $v = 2\pi r/T$, where $T$ is the period of one revolution. So

$$a_c = 4\pi^2r/T^2$$

With $r$ = 894 m and $T$ = 1 minute, that gives $a_c$ = 9.8038 m/s², just slightly less than $g$.